[{"id":"math-counting-001","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"counting","type":"single-choice","difficulty":1,"stem":"A student owns 3 different shirts and 4 different pairs of trousers. How many different outfits can be made by choosing one shirt and one pair of trousers?","stemZh":"某学生有 3 件不同的衬衫和 4 条不同的裤子，选一件衬衫和一条裤子搭配，共有多少种不同穿法？","options":[{"id":"A","text":"7"},{"id":"B","text":"24"},{"id":"C","text":"12"},{"id":"D","text":"10"}],"answer":"C","explanation":"This needs two steps $(shirt$, then $trousers)$, so use the multiplication $principle: 3 \\times 4 = 12$. Adding $(3 + 4 = 7)$ answers a different question — choosing a shirt OR a pair of trousers.","explanationZh":"$这件事分两步完成（先选衬衫，再选裤子），用分步乘法计数原理：3 \\times 4 = 12。若用加法 3 + 4 = 7，回答的是‘选一件衬衫或一条裤子’的问题。$","tags":["easy","multiplication-principle"],"status":"published"},{"id":"math-counting-002","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"counting","type":"single-choice","difficulty":2,"stem":"How many three-digit numbers can be formed from the digits 1, 2, 3, 4 if no digit may be repeated?","stemZh":"用数字 1、2、3、4 组成没有重复数字的三位数，共可组成多少个？","options":[{"id":"A","text":"24"},{"id":"B","text":"64"},{"id":"C","text":"12"},{"id":"D","text":"6"}],"answer":"A","explanation":"We arrange 3 of the 4 digits, and order $matters: A(4,3) = 4 \\times 3 \\times 2 = 24$. The value $4^{3} = 64$ allows repetition, and $C(4,3) = 12$ ignores the order of the digits.","explanationZh":"从 4 个数字中取 $3 个排成一列，与顺序有关：A(4,3) = 4 \\times 3 \\times 2 = 24。4^{3} = 64 是允许重复的情形，C(4,3) = 12$ 则忽略了数字的顺序。","tags":["easy","permutation"],"status":"published"},{"id":"math-counting-003","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"counting","type":"single-choice","difficulty":1,"stem":"In how many ways can 5 different books be arranged in a row on a shelf?","stemZh":"5 本不同的书排成一排放在书架上，共有多少种排法？","options":[{"id":"A","text":"25"},{"id":"B","text":"120"},{"id":"C","text":"60"},{"id":"D","text":"24"}],"answer":"B","explanation":"Arranging all 5 objects is a full $permutation: 5! = 5 \\times 4 \\times 3 \\times 2 \\times 1 = 120$. The slips $5^{2} = 25$ and $4! = 24$ are common but wrong.","explanationZh":"$5 个不同元素的全排列：5! = 5 \\times 4 \\times 3 \\times 2 \\times 1 = 120。常见错误是写成 5^{2} = 25 或 4! = 24。$","tags":["easy","factorial","permutation"],"status":"published"},{"id":"math-counting-004","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"counting","type":"single-choice","difficulty":1,"stem":"The value of $C(6,2)$ is","stemZh":"$C(6,2)$ 的值为","options":[{"id":"A","text":"30"},{"id":"B","text":"36"},{"id":"C","text":"12"},{"id":"D","text":"15"}],"answer":"D","explanation":"$C(6,2) = 6!/(2!·4!) = (6 \\times 5)/2 = 15$. The value 30 is $A(6,2)$, which counts each pair twice because it keeps the order.","explanationZh":"$C(6,2) = 6!/(2!·4!) = (6 \\times 5)/2 = 15。30 是 A(6,2)，它保留了顺序，把同一组重复计算了两次。$","tags":["easy","combination"],"status":"published"},{"id":"math-counting-005","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"counting","type":"single-choice","difficulty":2,"stem":"The value of $A(5,3)$ is","stemZh":"$A(5,3)$ 的值为","options":[{"id":"A","text":"15"},{"id":"B","text":"20"},{"id":"C","text":"60"},{"id":"D","text":"125"}],"answer":"C","explanation":"$A(5,3) = 5 \\times 4 \\times 3 = 60$. Using $C(5,3) = 15$ ignores the order, and $5^{3} = 125$ is the case where repetition is allowed.","explanationZh":"$A(5,3) = 5 \\times 4 \\times 3 = 60。C(5,3) = 15 忽略了顺序，5^{3} = 125$ 是允许重复选取的情形。","tags":["easy","permutation"],"status":"published"},{"id":"math-counting-006","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"counting","type":"single-choice","difficulty":2,"stem":"Using the identity $C(n,m) = C(n,n−m)$, the value of $C(8,6)$ is","stemZh":"利用公式 $C(n,m) = C(n,n−m)，C(8,6)$ 的值为","options":[{"id":"A","text":"28"},{"id":"B","text":"56"},{"id":"C","text":"48"},{"id":"D","text":"40320"}],"answer":"A","explanation":"$C(8,6) = C(8,2) = (8 \\times 7)/2 = 28$. The value $56 = A(8,2)$ counts $order, 48 = 8 \\times 6$ is plain multiplication, and $40320 = 8$! is a full permutation.","explanationZh":"$C(8,6) = C(8,2) = (8 \\times 7)/2 = 28。56 = A(8,2) 把顺序算进去了，48 = 8 \\times 6 只是简单相乘，40320 = 8$! 是全排列。","tags":["easy","combination","symmetry"],"status":"published"},{"id":"math-counting-007","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"counting","type":"single-choice","difficulty":1,"stem":"To travel from town A to town B one may take one of 3 buses or one of 2 trains. How many different ways are there to make the trip?","stemZh":"从 A 城到 B 城可以乘 3 路公交车中的任意一班，或乘 2 列火车中的任意一列，共有多少种走法？","options":[{"id":"A","text":"6"},{"id":"B","text":"5"},{"id":"C","text":"3"},{"id":"D","text":"2"}],"answer":"B","explanation":"The two types of transport are alternatives, so use the addition $principle: 3 + 2 = 5$. Multiplying $(3 \\times 2 = 6)$ would only be right if the trip required a bus AND a train.","explanationZh":"$公交与火车是彼此独立的两类方案，任选其一即可，用分类加法计数原理：3 + 2 = 5。只有‘先乘公交再乘火车’这类分步问题才用乘法 3 \\times 2 = 6。$","tags":["easy","addition-principle"],"status":"published"},{"id":"math-counting-008","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"counting","type":"single-choice","difficulty":1,"stem":"From 5 students, in how many ways can 2 of them be chosen to form a committee?","stemZh":"从 5 名学生中选出 2 人组成一个小组，共有多少种选法？","options":[{"id":"A","text":"20"},{"id":"B","text":"25"},{"id":"C","text":"7"},{"id":"D","text":"10"}],"answer":"D","explanation":"The two members have no distinct roles, so order does not $matter: C(5,2) = (5 \\times 4)/2 = 10. A(5,2) = 20$ counts every pair twice.","explanationZh":"$小组成员没有职务差别，与顺序无关：C(5,2) = (5 \\times 4)/2 = 10。A(5,2) = 20$ 会把同一组重复计算两次。","tags":["easy","combination"],"status":"published"},{"id":"math-counting-009","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"counting","type":"single-choice","difficulty":2,"stem":"How many two-digit numbers can be formed from the digits 1, 2, 3, 4, 5 if digits may be repeated?","stemZh":"用数字 1、2、3、4、5 组成两位数，数字允许重复使用，共可组成多少个？","options":[{"id":"A","text":"20"},{"id":"B","text":"10"},{"id":"C","text":"25"},{"id":"D","text":"32"}],"answer":"C","explanation":"Each of the two positions can be filled in 5 ways, so $5 \\times 5 = 25$. The value $A(5,2) = 20$ is the answer when repetition is NOT allowed.","explanationZh":"十位和个位各有 5 种填法，故 $5 \\times 5 = 25。A(5,2) = 20$ 是不允许重复数字时的答案。","tags":["easy","multiplication-principle","repetition"],"status":"published"},{"id":"math-counting-010","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"counting","type":"single-choice","difficulty":2,"stem":"The coefficient of $x^{2}$ in the expansion of $(1 + x)^{4}$ is","stemZh":"$(1 + x)^{4}$ 的展开式中 $x^{2}$ 的系数是","options":[{"id":"A","text":"6"},{"id":"B","text":"4"},{"id":"C","text":"12"},{"id":"D","text":"16"}],"answer":"A","explanation":"By the binomial theorem the term is $C(4,2)·1^{2}·x^{2} = 6x^{2}$, so the coefficient is 6. The exponent 4 and the value $4^{2} = 16$ are not coefficients.","explanationZh":"由二项式定理，该项为 $C(4,2)·1^{2}·x^{2} = 6x^{2}，故系数为 6。指数 4$ 以及 $4^{2} = 16$ 都不是系数。","tags":["easy","binomial-theorem"],"status":"published"},{"id":"math-counting-011","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"counting","type":"single-choice","difficulty":3,"stem":"Six people stand in a row. If Ann and Ben must stand next to each other, how many arrangements are possible?","stemZh":"6 人站成一排，若甲、乙两人必须相邻，共有多少种排法？","options":[{"id":"A","text":"720"},{"id":"B","text":"120"},{"id":"C","text":"48"},{"id":"D","text":"240"}],"answer":"D","explanation":"Bind Ann and Ben into one block: 5 objects give $5! = 120 orders$, and inside the block they can swap $(2! = 2)$. Total $= 2 \\times 120 = 240$.","explanationZh":"把甲、乙捆成一个整体：5 个对象全排列有 $5! = 120 种，整体内部两人可交换位置（2! = 2）。故共 2 \\times 120 = 240$ 种。","tags":["medium","permutation","adjacent"],"status":"published"},{"id":"math-counting-012","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"counting","type":"single-choice","difficulty":3,"stem":"Six people stand in a row and Ann must stand in the first position. How many arrangements are possible?","stemZh":"6 人站成一排，若甲必须站在第一个位置，共有多少种排法？","options":[{"id":"A","text":"720"},{"id":"B","text":"120"},{"id":"C","text":"600"},{"id":"D","text":"24"}],"answer":"B","explanation":"Ann occupies a fixed place, so only the other 5 people are $arranged: 5! = 120$. The value $6! = 720$ ignores the restriction.","explanationZh":"甲的位置固定，只需排列其余 $5 人：5! = 120。6! = 720$ 是没有考虑限制条件的排法数。","tags":["medium","permutation","restriction"],"status":"published"},{"id":"math-counting-013","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"counting","type":"single-choice","difficulty":3,"stem":"In how many ways can 5 people be seated around a round table, if arrangements that differ only by rotation count as the same?","stemZh":"5 人围坐在一张圆桌旁，若仅由旋转得到的坐法视为相同，共有多少种坐法？","options":[{"id":"A","text":"120"},{"id":"B","text":"20"},{"id":"C","text":"24"},{"id":"D","text":"5"}],"answer":"C","explanation":"Fix one person to remove the rotational symmetry, then arrange the remaining $4: (5 − 1)! = 4! = 24$. The value $5! = 120$ counts each seating 5 times.","explanationZh":"先固定一人以消除旋转带来的重复，再排列其余 $4 人：(5 − 1)! = 4! = 24。5! = 120$ 会把每种坐法重复计算 5 次。","tags":["medium","circular-permutation"],"status":"published"},{"id":"math-counting-014","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"counting","type":"single-choice","difficulty":3,"stem":"How many different arrangements can be made using all the letters of the word BANANA?","stemZh":"用单词 BANANA 的全部字母进行排列，共有多少种不同的排法？","options":[{"id":"A","text":"60"},{"id":"B","text":"720"},{"id":"C","text":"120"},{"id":"D","text":"30"}],"answer":"A","explanation":"BANANA has 6 letters, with A appearing 3 times and N twice. Divide by the $repeats: 6!/(3!·2!) = 720/12 = 60$.","explanationZh":"BANANA 共 6 个字母，其中 A 出现 3 次、N 出现 $2 次，需除以重复元素的全排列：6!/(3!·2!) = 720/12 = 60。$","tags":["medium","repeated-elements","permutation"],"status":"published"},{"id":"math-counting-015","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"counting","type":"single-choice","difficulty":3,"stem":"A group of 3 is chosen from 4 boys and 3 girls and must contain exactly 1 girl. How many such groups are there?","stemZh":"从 4 名男生和 3 名女生中选出 3 人，要求恰好有 1 名女生，共有多少种选法？","options":[{"id":"A","text":"35"},{"id":"B","text":"30"},{"id":"C","text":"9"},{"id":"D","text":"18"}],"answer":"D","explanation":"Choose the girl $(C(3,1) = 3)$ and the two boys $(C(4,2) = 6)$, then $multiply: 3 \\times 6 = 18$. The value $35 = C(7,3)$ is the total without the restriction.","explanationZh":"先选女生 $C(3,1) = 3$ 种，再选两名男生 $C(4,2) = 6$ 种，相乘得 $3 \\times 6 = 18。35 = C(7,3)$ 是没有限制条件时的总数。","tags":["medium","combination","classification"],"status":"published"},{"id":"math-counting-016","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"counting","type":"single-choice","difficulty":3,"stem":"How many diagonals does a convex hexagon have?","stemZh":"一个凸六边形共有多少条对角线？","options":[{"id":"A","text":"15"},{"id":"B","text":"9"},{"id":"C","text":"12"},{"id":"D","text":"6"}],"answer":"B","explanation":"$C(6,2) = 15$ counts all segments joining two vertices; 6 of these are sides of the hexagon, leaving $15 − 6 = 9$ diagonals.","explanationZh":"$C(6,2) = 15$ 表示任取两个顶点连成的线段总数，其中 6 条是六边形的边，故对角线为 $15 − 6 = 9$ 条。","tags":["medium","combination","geometry"],"status":"published"},{"id":"math-counting-017","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"counting","type":"single-choice","difficulty":3,"stem":"How many three-digit numbers with no repeated digits can be formed from the digits 0, 1, 2, 3, 4?","stemZh":"用数字 0、1、2、3、4 组成没有重复数字的三位数，共可组成多少个？","options":[{"id":"A","text":"60"},{"id":"B","text":"36"},{"id":"C","text":"48"},{"id":"D","text":"24"}],"answer":"C","explanation":"The hundreds digit cannot be 0, so it has 4 choices. Then 4 digits remain for the tens place and 3 for the $units: 4 \\times 4 \\times 3 = 48$. Using $A(5,3) = 60$ wrongly allows a leading zero.","explanationZh":"百位不能为 0，有 4 种选法；十位从剩下 4 个数字中选，个位从剩下 $3 个中选：4 \\times 4 \\times 3 = 48。A(5,3) = 60$ 错误地允许了首位为 0。","tags":["medium","permutation","restriction"],"status":"published"},{"id":"math-counting-018","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"counting","type":"single-choice","difficulty":3,"stem":"From 10 candidates, in how many ways can a president and a vice-president be selected?","stemZh":"从 10 名候选人中选出一名主席和一名副主席，共有多少种选法？","options":[{"id":"A","text":"45"},{"id":"B","text":"100"},{"id":"C","text":"20"},{"id":"D","text":"90"}],"answer":"D","explanation":"The two posts are different, so order $matters: A(10,2) = 10 \\times 9 = 90$. Using $C(10,2) = 45$ would treat the two posts as identical.","explanationZh":"$主席与副主席是两个不同的职位，与顺序有关：A(10,2) = 10 \\times 9 = 90。若用 C(10,2) = 45，相当于把两个职位看成相同的。$","tags":["medium","permutation"],"status":"published"},{"id":"math-counting-019","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"counting","type":"single-choice","difficulty":3,"stem":"In how many ways can 5 different gifts be given to 3 children, if each gift goes to exactly one child?","stemZh":"5 件不同的礼物分给 3 个小孩，每件礼物恰好分给一个小孩，共有多少种分法？","options":[{"id":"A","text":"243"},{"id":"B","text":"125"},{"id":"C","text":"60"},{"id":"D","text":"15"}],"answer":"A","explanation":"Each of the 5 gifts independently has 3 possible recipients, so $3^{5} = 243$. The values $5^{3} = 125$ and $3 \\times 5 = 15$ swap the roles of gifts and children.","explanationZh":"每件礼物都有 $3 种分配对象，5 件礼物相互独立：3^{5} = 243。5^{3} = 125 与 3 \\times 5 = 15$ 都弄混了礼物与小孩的角色。","tags":["medium","multiplication-principle"],"status":"published"},{"id":"math-counting-020","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"counting","type":"single-choice","difficulty":3,"stem":"The coefficient of $x^{3}$ in the expansion of $(1 + x)^{7}$ is","stemZh":"$(1 + x)^{7}$ 的展开式中 $x^{3}$ 的系数是","options":[{"id":"A","text":"21"},{"id":"B","text":"35"},{"id":"C","text":"210"},{"id":"D","text":"343"}],"answer":"B","explanation":"The general term is $C(7,k)·x^k$, so the coefficient of $x^{3}$ is $C(7,3) = 35$. Using $A(7,3) = 210$ wrongly treats the coefficient as an arrangement count.","explanationZh":"通项为 $C(7,k)·x^k，故 x^{3}$ 的系数为 $C(7,3) = 35。A(7,3) = 210$ 是错把系数当成了排列数。","tags":["medium","binomial-theorem"],"status":"published"},{"id":"math-counting-021","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"counting","type":"single-choice","difficulty":4,"stem":"Seven people stand in a row and Ann, Ben and Carl must all stand together. How many arrangements are possible?","stemZh":"7 人站成一排，若甲、乙、丙三人必须站在一起，共有多少种排法？","options":[{"id":"A","text":"5040"},{"id":"B","text":"120"},{"id":"C","text":"720"},{"id":"D","text":"1440"}],"answer":"C","explanation":"Bind the three into a single block: 5 objects give $5! = 120 orders$, and inside the block the three can be ordered in $3! = 6$ ways. Total $= 6 \\times 120 = 720$.","explanationZh":"把三人捆成一个整体：5 个对象全排列有 $5! = 120$ 种，整体内部三人有 $3! = 6$ 种排法。故共 $6 \\times 120 = 720$ 种。","tags":["hard","permutation","adjacent"],"status":"published"},{"id":"math-counting-022","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"counting","type":"single-choice","difficulty":4,"stem":"Six people stand in a row and Ann and Ben must NOT stand next to each other. How many arrangements are possible?","stemZh":"6 人站成一排，若甲、乙两人不能相邻，共有多少种排法？","options":[{"id":"A","text":"240"},{"id":"B","text":"360"},{"id":"C","text":"600"},{"id":"D","text":"480"}],"answer":"D","explanation":"Use the complement: all $6! = 720$ orders minus the $2!·5! = 240$ orders in which Ann and Ben are adjacent, giving $720 − 240 = 480$.","explanationZh":"用补集法：全体排法 $6! = 720，减去甲、乙相邻的 2!·5! = 240$ 种，得 $720 − 240 = 480$ 种。","tags":["hard","permutation","complement"],"status":"published"},{"id":"math-counting-023","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"counting","type":"single-choice","difficulty":4,"stem":"A team of 4 is chosen from 5 boys and 4 girls and must contain at least one girl. How many such teams are there?","stemZh":"从 5 名男生和 4 名女生中选出 4 人，要求至少有一名女生，共有多少种选法？","options":[{"id":"A","text":"126"},{"id":"B","text":"121"},{"id":"C","text":"120"},{"id":"D","text":"84"}],"answer":"B","explanation":"Take all $C(9,4) = 126$ possible teams and remove the all-boy teams $C(5,4) = 5: 126 − 5 = 121$. Directly counting 1, 2, 3 or 4 girls gives the same result but takes longer.","explanationZh":"先取全体 $C(9,4) = 126$ 种，再减去全是男生的 $C(5,4) = 5 种：126 − 5 = 121。分类数 1、2、3、4$ 名女生也可得到同样结果，但计算更繁琐。","tags":["hard","combination","at-least-one"],"status":"published"},{"id":"math-counting-024","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"counting","type":"single-choice","difficulty":5,"stem":"How many four-digit even numbers with no repeated digits can be formed from the digits 0, 1, 2, 3, 4, 5?","stemZh":"用数字 0、1、2、3、4、5 组成没有重复数字的四位偶数，共可组成多少个？","options":[{"id":"A","text":"156"},{"id":"B","text":"300"},{"id":"C","text":"180"},{"id":"D","text":"108"}],"answer":"A","explanation":"Split on the units digit. If it is 0, the other three places are filled from the remaining $5 digits: A(5,3) = 60$. If it is 2 or $4 (2 choices)$, the thousands digit has 4 choices $(not 0$ and not the units digit) and the middle two places give $A(4,2) = 12: 2 \\times 4 \\times 12 = 96$. Total $= 60 + 96 = 156$.","explanationZh":"按个位分类。个位为 0 时，其余三位从剩下 $5 个数字中选：A(5,3) = 60；个位为 2 或 4 时（2$ 种），千位有 4 种选法（不能为 $0，也不能与个位重复），中间两位有 A(4,2) = 12$ 种，共 $2 \\times 4 \\times 12 = 96。合计 60 + 96 = 156。$","tags":["hard","permutation","restriction"],"status":"published"},{"id":"math-counting-025","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"counting","type":"single-choice","difficulty":5,"stem":"Six people stand in a row. In how many arrangements is Ann to the left of Ben (not necessarily adjacent)?","stemZh":"6 人站成一排，甲站在乙的左边（不要求相邻）的排法共有多少种？","options":[{"id":"A","text":"720"},{"id":"B","text":"120"},{"id":"C","text":"360"},{"id":"D","text":"240"}],"answer":"C","explanation":"Of the $6! = 720 orders$, exactly half have Ann left of Ben and half have her right of him (swap their names to pair them $up)$, so $720/2 = 360$.","explanationZh":"$在 6! = 720$ 种排法中，甲在乙左与甲在乙右的情况一一对应、各占一半，故 $720/2 = 360$ 种。","tags":["hard","permutation","symmetry"],"status":"published"},{"id":"math-counting-026","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"counting","type":"single-choice","difficulty":4,"stem":"How many different arrangements can be made using all the letters of the word SUCCESS?","stemZh":"用单词 SUCCESS 的全部字母进行排列，共有多少种不同的排法？","options":[{"id":"A","text":"5040"},{"id":"B","text":"840"},{"id":"C","text":"210"},{"id":"D","text":"420"}],"answer":"D","explanation":"SUCCESS has 7 letters, with S appearing 3 times and $C twice: 7!/(3!·2!) = 5040/12 = 420$. Dividing by only one of the repeats gives 840 or 1680.","explanationZh":"SUCCESS 共 7 个字母，其中 S 出现 3 次、C 出现 $2 次：7!/(3!·2!) = 5040/12 = 420。只除以其中一个重复因子会得到 840$ 等错误结果。","tags":["hard","repeated-elements","permutation"],"status":"published"},{"id":"math-counting-027","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"counting","type":"single-choice","difficulty":5,"stem":"In how many ways can 4 different books be distributed to 3 students so that every student receives at least one book?","stemZh":"4 本不同的书分给 3 名学生，要求每名学生至少分到一本书，共有多少种分法？","options":[{"id":"A","text":"36"},{"id":"B","text":"81"},{"id":"C","text":"72"},{"id":"D","text":"24"}],"answer":"A","explanation":"Without the condition there are $3^{4} = 81$ ways. Subtract the ways with a chosen student $empty, C(3,1)·2^{4} = 48$, then add back those with two chosen students $empty, C(3,2)·1^{4} = 3: 81 − 48 + 3 = 36. (Equivalently$: the split must be 2+1+1 — choose who gets 2 books $(3 ways)$, choose the pair $(C(4,2) = 6)$ and hand out the rest $(2!): 3 \\times 6 \\times 2 = 36.)$","explanationZh":"不加限制有 $3^{4} = 81$ 种；减去指定一名学生为空的 $C(3,1)·2^{4} = 48$ 种，再加回指定两名学生为空的 $C(3,2)·1^{4} = 3 种：81 − 48 + 3 = 36。（也可按 2+1+1$ 分配：选拿到 2 本的人 3 种，选这 2 本书 $C(4,2) = 6$ 种，其余分法 $2! 种，3 \\times 6 \\times 2 = 36。）$","tags":["hard","inclusion-exclusion","distribution"],"status":"published"},{"id":"math-counting-028","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"counting","type":"single-choice","difficulty":4,"stem":"The coefficient of $x^{2}$ in the expansion of $(2 + x)^{5}$ is","stemZh":"$(2 + x)^{5}$ 的展开式中 $x^{2}$ 的系数是","options":[{"id":"A","text":"40"},{"id":"B","text":"80"},{"id":"C","text":"10"},{"id":"D","text":"32"}],"answer":"B","explanation":"The term is $C(5,2)·2^{3}·x^{2} = 10 \\times 8 \\times x^{2} = 80x^{2}$, so the coefficient is 80. Taking only $C(5,2) = 10$ forgets the power of the constant 2.","explanationZh":"该项为 $C(5,2)·2^{3}·x^{2} = 10 \\times 8 \\times x^{2} = 80x^{2}，故系数为 80。只取 C(5,2) = 10$ 是漏掉了常数 2 的幂次。","tags":["hard","binomial-theorem"],"status":"published"},{"id":"math-counting-029","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"counting","type":"single-choice","difficulty":5,"stem":"In how many ways can 6 people be split into 3 unordered pairs?","stemZh":"把 6 个人分成 3 组，每组 2 人且组与组之间不分顺序，共有多少种分法？","options":[{"id":"A","text":"120"},{"id":"B","text":"45"},{"id":"C","text":"15"},{"id":"D","text":"30"}],"answer":"C","explanation":"Line the 6 people up $(6!)$, then divide by the order inside each pair $(2!·2!·2!)$ and by the order of the 3 pairs $(3!): 720/48 = 15$.","explanationZh":"先把 $6 人排成一列（6!），再除以每组内部的顺序（2!·2!·2!）和 3 组之间的顺序（3!）：720/48 = 15。$","tags":["hard","grouping","combination"],"status":"published"},{"id":"math-counting-030","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"counting","type":"single-choice","difficulty":4,"stem":"How many non-empty subsets does a set with 5 elements have?","stemZh":"含有 5 个元素的集合共有多少个非空子集？","options":[{"id":"A","text":"32"},{"id":"B","text":"25"},{"id":"C","text":"10"},{"id":"D","text":"31"}],"answer":"D","explanation":"Each of the 5 elements is independently either in or out, giving $2^{5} = 32 subsets$; deleting the empty set leaves 31.","explanationZh":"5 个元素各自都有‘取’与‘不取’两种选择，共有 $2^{5} = 32$ 个子集；去掉空集后为 31 个。","tags":["hard","subsets","counting"],"status":"published"},{"id":"math-classical-probability-001","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"classical-probability","type":"single-choice","difficulty":1,"stem":"A fair die is rolled once. What is the probability of getting an even score?","stemZh":"掷一枚均匀的骰子一次，出现偶数点的概率是多少？","options":[{"id":"A","text":"1/2"},{"id":"B","text":"1/6"},{"id":"C","text":"1/3"},{"id":"D","text":"2/3"}],"answer":"A","explanation":"Three of the six equally likely outcomes $(2, 4, 6)$ are even, so $P = 3/6 = 1/2$.","explanationZh":"6 个等可能结果中，2、4、6 共 3 个是偶数，故 $P = 3/6 = 1/2。$","tags":["easy","classical-probability","dice"],"status":"published"},{"id":"math-classical-probability-002","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"classical-probability","type":"single-choice","difficulty":2,"stem":"A fair coin is tossed twice. What is the probability of getting at least one head?","stemZh":"一枚均匀硬币连掷两次，至少出现一次正面的概率是多少？","options":[{"id":"A","text":"1/4"},{"id":"B","text":"1/2"},{"id":"C","text":"3/4"},{"id":"D","text":"1"}],"answer":"C","explanation":"The four equally likely outcomes are HH, HT, TH, TT; only TT has no head, so $P = 3/4 (equivalently 1 − 1/4 = 3/4)$.","explanationZh":"4 个等可能结果为 HH、HT、TH、TT，只有 TT 没有正面，故 $P = 3/4（即 1 − 1/4 = 3/4）。$","tags":["easy","complement","coins"],"status":"published"},{"id":"math-classical-probability-003","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"classical-probability","type":"single-choice","difficulty":1,"stem":"A bag contains 3 red balls and 2 blue balls. One ball is drawn at random. What is the probability that it is red?","stemZh":"袋中有 3 个红球和 2 个蓝球，随机取出一个，取到红球的概率是多少？","options":[{"id":"A","text":"2/5"},{"id":"B","text":"3/5"},{"id":"C","text":"1/5"},{"id":"D","text":"3/2"}],"answer":"B","explanation":"All 5 balls are equally likely to be drawn and 3 of them are red, so $P = 3/5$.","explanationZh":"5 个球被取到的机会相同，其中红球 3 个，故 $P = 3/5。$","tags":["easy","classical-probability","balls"],"status":"published"},{"id":"math-classical-probability-004","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"classical-probability","type":"single-choice","difficulty":1,"stem":"One card is drawn at random from a standard 52-card deck. What is the probability that it is a heart?","stemZh":"从一副 52 张的扑克牌中随机抽取一张，抽到红桃的概率是多少？","options":[{"id":"A","text":"1/52"},{"id":"B","text":"1/13"},{"id":"C","text":"4/13"},{"id":"D","text":"1/4"}],"answer":"D","explanation":"13 of the 52 cards are hearts, so $P = 13/52 = 1/4$. The value 1/13 is the probability of one particular rank being drawn.","explanationZh":"52 张牌中有 13 张红桃，故 $P = 13/52 = 1/4。1/13$ 是抽到某一特定点数的概率。","tags":["easy","classical-probability","cards"],"status":"published"},{"id":"math-classical-probability-005","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"classical-probability","type":"single-choice","difficulty":2,"stem":"A fair coin is tossed three times. What is the probability that all three tosses are heads?","stemZh":"一枚均匀硬币连掷三次，三次都是正面的概率是多少？","options":[{"id":"A","text":"1/8"},{"id":"B","text":"1/6"},{"id":"C","text":"1/4"},{"id":"D","text":"3/8"}],"answer":"A","explanation":"There are $2^{3} = 8$ equally likely outcomes and only HHH qualifies, so $P = 1/8$; by independence this is also $(1/2)^{3} = 1/8$.","explanationZh":"共有 $2^{3} = 8$ 个等可能结果，只有 HHH 符合，故 $P = 1/8；由独立性也可得 (1/2)^{3} = 1/8。$","tags":["easy","coins","independent-events"],"status":"published"},{"id":"math-classical-probability-006","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"classical-probability","type":"single-choice","difficulty":1,"stem":"A number is chosen at random from 1, 2, …, 10. What is the probability that it is a multiple of 3?","stemZh":"从 1, 2, …, 10 中随机取一个数，它是 3 的倍数的概率是多少？","options":[{"id":"A","text":"1/3"},{"id":"B","text":"1/10"},{"id":"C","text":"3/10"},{"id":"D","text":"3/9"}],"answer":"C","explanation":"The multiples of 3 in the list are 3, 6 and 9, so 3 of the 10 numbers are $favourable: P = 3/10$.","explanationZh":"1 到 10 中 3 的倍数有 3、6、9，共 3 个数，故 $P = 3/10。$","tags":["easy","classical-probability"],"status":"published"},{"id":"math-classical-probability-007","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"classical-probability","type":"single-choice","difficulty":1,"stem":"A box contains 4 white balls and 6 black balls. One ball is drawn at random. What is the probability that it is white?","stemZh":"盒中有 4 个白球和 6 个黑球，随机取出一个，取到白球的概率是多少？","options":[{"id":"A","text":"4/6"},{"id":"B","text":"2/5"},{"id":"C","text":"1/5"},{"id":"D","text":"3/5"}],"answer":"B","explanation":"There are 10 balls in total, of which 4 are white, so $P = 4/10 = 2/5$. The ratio 4/6 compares white with black instead of with the total.","explanationZh":"共 10 个球，其中白球 4 个，故 $P = 4/10 = 2/5。4/6$ 是白球与黑球之比，分母不是总数。","tags":["easy","classical-probability","balls"],"status":"published"},{"id":"math-classical-probability-008","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"classical-probability","type":"single-choice","difficulty":1,"stem":"A fair die is rolled once. What is the probability that the score is greater than 4?","stemZh":"掷一枚均匀的骰子一次，点数大于 4 的概率是多少？","options":[{"id":"A","text":"1/6"},{"id":"B","text":"1/2"},{"id":"C","text":"2/3"},{"id":"D","text":"1/3"}],"answer":"D","explanation":"Only the scores 5 and 6 qualify, so $P = 2/6 = 1/3$. Using a strict inequality matters: 4 itself does not count.","explanationZh":"只有 5 点和 6 点符合，故 $P = 2/6 = 1/3。注意是严格大于 4，点数 4$ 本身不算。","tags":["easy","dice","classical-probability"],"status":"published"},{"id":"math-classical-probability-009","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"classical-probability","type":"single-choice","difficulty":2,"stem":"A bag contains 2 red balls and 3 blue balls. Two balls are drawn at random without replacement. What is the probability that both are red?","stemZh":"袋中有 2 个红球和 3 个蓝球，不放回地随机取出 2 个，两个都是红球的概率是多少？","options":[{"id":"A","text":"1/10"},{"id":"B","text":"4/25"},{"id":"C","text":"1/5"},{"id":"D","text":"2/5"}],"answer":"A","explanation":"$P = C(2,2)/C(5,2) = 1/10$, or step by step $(2/5) \\times (1/4) = 1/10$. The value $4/25 = (2/5)^{2}$ is the answer when the first ball is replaced.","explanationZh":"$P = C(2,2)/C(5,2) = 1/10，或分步计算 (2/5) \\times (1/4) = 1/10。4/25 = (2/5)^{2}$ 是有放回抽取时的答案。","tags":["easy","without-replacement","balls"],"status":"published"},{"id":"math-classical-probability-010","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"classical-probability","type":"single-choice","difficulty":1,"stem":"A family has two children. Assuming a boy and a girl are equally likely and the births are independent, what is the probability that both children are boys?","stemZh":"某家庭有两个孩子，假设生男生女等可能且相互独立，两个孩子都是男孩的概率是多少？","options":[{"id":"A","text":"1/2"},{"id":"B","text":"1/3"},{"id":"C","text":"1/4"},{"id":"D","text":"3/4"}],"answer":"C","explanation":"The four outcomes BB, BG, GB, GG are equally likely and only BB qualifies, so $P = 1/4$. Treating BG and GB as one case would wrongly give 1/3.","explanationZh":"BB、BG、GB、GG 四种结果等可能，只有 BB 符合，故 $P = 1/4。若把$ BG 与 GB 合并成一种情况，会错得 1/3。","tags":["easy","classical-probability","equiprobable"],"status":"published"},{"id":"math-classical-probability-011","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"classical-probability","type":"single-choice","difficulty":3,"stem":"Two fair dice are rolled. What is the probability that the sum of the scores is 7?","stemZh":"同时掷两枚均匀的骰子，点数之和为 7 的概率是多少？","options":[{"id":"A","text":"1/12"},{"id":"B","text":"1/6"},{"id":"C","text":"5/36"},{"id":"D","text":"1/9"}],"answer":"B","explanation":"There are $6 \\times 6 = 36$ equally likely ordered pairs, and 6 of them give the sum $7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$. So $P = 6/36 = 1/6$.","explanationZh":"共有 $6 \\times 6 = 36$ 个等可能的有序数对，其中和为 7 的有 $(1,6)、(2,5)、(3,4)、(4,3)、(5,2)、(6,1) 共 6$ 个，故 $P = 6/36 = 1/6。$","tags":["medium","dice","classical-probability"],"status":"published"},{"id":"math-classical-probability-012","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"classical-probability","type":"single-choice","difficulty":3,"stem":"A bag contains 5 red balls and 3 white balls. Two balls are drawn at random without replacement. What is the probability that both are white?","stemZh":"袋中有 5 个红球和 3 个白球，不放回地随机取出 2 个，两个都是白球的概率是多少？","options":[{"id":"A","text":"9/64"},{"id":"B","text":"1/4"},{"id":"C","text":"3/8"},{"id":"D","text":"3/28"}],"answer":"D","explanation":"$P = C(3,2)/C(8,2) = 3/28$, or $(3/8) \\times (2/7) = 6/56 = 3/28$. The value $9/64 = (3/8)^{2}$ assumes the first ball is put back.","explanationZh":"$P = C(3,2)/C(8,2) = 3/28，或分步计算 (3/8) \\times (2/7) = 6/56 = 3/28。9/64 = (3/8)^{2}$ 是把第一个球放回后的结果。","tags":["medium","without-replacement","balls"],"status":"published"},{"id":"math-classical-probability-013","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"classical-probability","type":"single-choice","difficulty":3,"stem":"Two cards are drawn at random from a standard 52-card deck without replacement. What is the probability that both are aces?","stemZh":"从一副 52 张的扑克牌中不放回地随机抽取 2 张，两张都是 A 的概率是多少？","options":[{"id":"A","text":"1/221"},{"id":"B","text":"1/169"},{"id":"C","text":"1/13"},{"id":"D","text":"1/52"}],"answer":"A","explanation":"$P = C(4,2)/C(52,2) = 6/1326 = 1/221$, or $(4/52) \\times (3/51) = 1/221$. The value $1/169 = (4/52)^{2}$ treats the two draws as independent (with replacement).","explanationZh":"$P = C(4,2)/C(52,2) = 6/1326 = 1/221，或分步计算 (4/52) \\times (3/51) = 1/221。1/169 = (4/52)^{2}$ 是把两次抽取当作相互独立（有放回）的结果。","tags":["medium","without-replacement","cards"],"status":"published"},{"id":"math-classical-probability-014","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"classical-probability","type":"single-choice","difficulty":3,"stem":"A fair coin is tossed four times. What is the probability of getting exactly two heads?","stemZh":"一枚均匀硬币连掷四次，恰好出现两次正面的概率是多少？","options":[{"id":"A","text":"1/2"},{"id":"B","text":"1/4"},{"id":"C","text":"3/8"},{"id":"D","text":"5/16"}],"answer":"C","explanation":"There are $2^{4} = 16$ equally likely outcomes, of which $C(4,2) = 6$ have exactly two heads, so $P = 6/16 = 3/8$.","explanationZh":"共有 $2^{4} = 16$ 个等可能结果，其中恰好两次正面的是从 4 次中选 $2 次：C(4,2) = 6，故 P = 6/16 = 3/8。$","tags":["medium","coins","combination"],"status":"published"},{"id":"math-classical-probability-015","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"classical-probability","type":"single-choice","difficulty":3,"stem":"A bag contains 4 red, 3 blue and 2 green balls. One ball is drawn at random. What is the probability that it is red or blue?","stemZh":"袋中有 4 个红球、3 个蓝球和 2 个绿球，随机取出一个，取到红球或蓝球的概率是多少？","options":[{"id":"A","text":"4/9"},{"id":"B","text":"7/9"},{"id":"C","text":"2/9"},{"id":"D","text":"5/9"}],"answer":"B","explanation":"Red and blue cannot occur together, so the events are mutually exclusive and the probabilities $add: 4/9 + 3/9 = 7/9 (i.e. 7\\text{ of the }9 balls)$.","explanationZh":"$红球与蓝球不可能同时取到，两事件互斥，概率相加：4/9 + 3/9 = 7/9（即 9$ 个球中有 7 个符合）。","tags":["medium","mutually-exclusive","addition-rule"],"status":"published"},{"id":"math-classical-probability-016","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"classical-probability","type":"single-choice","difficulty":3,"stem":"Events A and B are mutually exclusive, with $P(A) = 0.3$ and $P(B) = 0.4$. What is $P(A ∪ B)$?","stemZh":"事件 A 与 B 互斥，且 $P(A) = 0.3，P(B) = 0.4，则 P(A ∪ B)$ 等于多少？","options":[{"id":"A","text":"0.12"},{"id":"B","text":"0.1"},{"id":"C","text":"1"},{"id":"D","text":"0.7"}],"answer":"D","explanation":"For mutually exclusive events the probabilities simply $add: P(A ∪ B) = 0.3 + 0.4 = 0.7$. Multiplying $(0.12)$ applies to independent events, not to a union.","explanationZh":"$互斥事件的概率可以直接相加：P(A ∪ B) = 0.3 + 0.4 = 0.7。相乘得 0.12$ 是独立事件同时发生的算法，不适用于求并事件。","tags":["medium","mutually-exclusive","addition-rule"],"status":"published"},{"id":"math-classical-probability-017","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"classical-probability","type":"single-choice","difficulty":3,"stem":"A shooter hits the target with probability 0.8 on each shot, and shots are independent. What is the probability that he misses at least once in two shots?","stemZh":"某射手每发命中目标的概率为 0.8，各发相互独立，则他两发中至少有一发脱靶的概率是多少？","options":[{"id":"A","text":"0.36"},{"id":"B","text":"0.04"},{"id":"C","text":"0.64"},{"id":"D","text":"0.2"}],"answer":"A","explanation":"Use the complement: P(at least one $miss) = 1 − P(both hit) = 1 − 0.8^{2} = 1 − 0.64 = 0.36$. The value $0.04 = 0.2^{2}$ is the probability of missing both shots.","explanationZh":"$用对立事件：P(至少一发脱靶) = 1 − P(两发都命中) = 1 − 0.8^{2} = 1 − 0.64 = 0.36。0.04 = 0.2^{2}$ 是两发都脱靶的概率。","tags":["medium","complement","independent-events"],"status":"published"},{"id":"math-classical-probability-018","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"classical-probability","type":"single-choice","difficulty":3,"stem":"A and B are independent events with $P(A) = 1/2$ and $P(B) = 1/3$. What is $P(A ∩ B)$?","stemZh":"事件 A 与 B 相互独立，且 $P(A) = 1/2，P(B) = 1/3，则 P(A ∩ B)$ 等于多少？","options":[{"id":"A","text":"5/6"},{"id":"B","text":"1/3"},{"id":"C","text":"1/6"},{"id":"D","text":"2/3"}],"answer":"C","explanation":"For independent events the probabilities $multiply: P(A ∩ B) = (1/2) \\times (1/3) = 1/6$. The value 5/6 is $P(A ∪ B)$, which is a different question.","explanationZh":"$独立事件同时发生的概率相乘：P(A ∩ B) = (1/2) \\times (1/3) = 1/6。5/6 是 P(A ∪ B)，问的是另一个问题。$","tags":["medium","independent-events","multiplication-rule"],"status":"published"},{"id":"math-classical-probability-019","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"classical-probability","type":"single-choice","difficulty":3,"stem":"A three-digit number is formed using only the digits 1 to 9, with repetition allowed. What is the probability that all three digits are odd?","stemZh":"只用数字 1 到 9 组成一个三位数（数字可重复），三个数字全为奇数的概率是多少？","options":[{"id":"A","text":"5/9"},{"id":"B","text":"25/81"},{"id":"C","text":"1/8"},{"id":"D","text":"125/729"}],"answer":"D","explanation":"There are $9^{3} = 729$ possible numbers and $5^{3} = 125$ in which every digit is odd (chosen from $1, 3, 5, 7, 9)$, so $P = 125/729 = (5/9)^{3}$. The value $25/81 = (5/9)^{2}$ uses only two positions.","explanationZh":"共有 $9^{3} = 729$ 个三位数，每位都从 1、3、5、7、9 中选的三位数有 $5^{3} = 125$ 个，故 $P = 125/729 = (5/9)^{3}。25/81 = (5/9)^{2}$ 只算了两位的情形。","tags":["medium","classical-probability","multiplication-principle"],"status":"published"},{"id":"math-classical-probability-020","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"classical-probability","type":"single-choice","difficulty":3,"stem":"Two students are chosen at random from a group of 5 boys and 4 girls. What is the probability that both are girls?","stemZh":"从 5 名男生和 4 名女生中随机选取 2 人，两人都是女生的概率是多少？","options":[{"id":"A","text":"16/81"},{"id":"B","text":"1/6"},{"id":"C","text":"2/9"},{"id":"D","text":"4/9"}],"answer":"B","explanation":"$P = C(4,2)/C(9,2) = 6/36 = 1/6$. The value $16/81 = (4/9)^{2}$ would be correct only if the first student were replaced before the second was chosen.","explanationZh":"$P = C(4,2)/C(9,2) = 6/36 = 1/6。16/81 = (4/9)^{2}$ 只有在‘选完第一人后再放回’时才成立。","tags":["medium","without-replacement","combination"],"status":"published"},{"id":"math-classical-probability-021","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"classical-probability","type":"single-choice","difficulty":4,"stem":"A bag contains 6 red balls and 4 white balls. Three balls are drawn at random without replacement. What is the probability that at least one is white?","stemZh":"袋中有 6 个红球和 4 个白球，不放回地随机取出 3 个，至少有一个白球的概率是多少？","options":[{"id":"A","text":"1/6"},{"id":"B","text":"2/5"},{"id":"C","text":"5/6"},{"id":"D","text":"1/2"}],"answer":"C","explanation":"Use the $complement: P(no white) = C(6,3)/C(10,3) = 20/120 = 1/6$, so P(at least one $white) = 1 − 1/6 = 5/6$. Counting 1, 2 or 3 white balls directly gives the same value.","explanationZh":"$用对立事件：P(全是红球) = C(6,3)/C(10,3) = 20/120 = 1/6，故 P(至少一个白球) = 1 − 1/6 = 5/6。直接分 1、2、3$ 个白球计算也得到同样结果。","tags":["hard","complement","at-least-one"],"status":"published"},{"id":"math-classical-probability-022","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"classical-probability","type":"single-choice","difficulty":4,"stem":"A and B are independent events with $P(A) = 0.6$ and $P(B) = 0.5$. What is $P(A ∪ B)$?","stemZh":"事件 A 与 B 相互独立，且 $P(A) = 0.6，P(B) = 0.5，则 P(A ∪ B)$ 等于多少？","options":[{"id":"A","text":"0.8"},{"id":"B","text":"1.1"},{"id":"C","text":"0.3"},{"id":"D","text":"0.9"}],"answer":"A","explanation":"$P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 0.6 + 0.5 − (0.6 \\times 0.5) = 1.1 − 0.3 = 0.8$. Simply adding gives 1.1, which forgets to remove the overlap and is impossible for a probability.","explanationZh":"$P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 0.6 + 0.5 − 0.6 \\times 0.5 = 1.1 − 0.3 = 0.8。直接相加得 1.1，既漏掉了重叠部分，也超过了概率的取值范围。$","tags":["hard","independent-events","addition-rule"],"status":"published"},{"id":"math-classical-probability-023","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"classical-probability","type":"single-choice","difficulty":4,"stem":"Two fair dice are rolled. What is the probability that the product of the two scores is 12?","stemZh":"同时掷两枚均匀的骰子，点数之积为 12 的概率是多少？","options":[{"id":"A","text":"1/6"},{"id":"B","text":"1/12"},{"id":"C","text":"1/18"},{"id":"D","text":"1/9"}],"answer":"D","explanation":"The ordered pairs with product 12 are $(2,6), (6,2), (3,4)$ and $(4,3) — 4$ out of 36 — so $P = 4/36 = 1/9$. Counting only unordered pairs would give $2/36 = 1/18$.","explanationZh":"积为 12 的有序数对为 $(2,6)、(6,2)、(3,4)、(4,3)，共 4$ 个，故 $P = 4/36 = 1/9。若只按无序数对计算会得到 2/36 = 1/18。$","tags":["hard","dice","classical-probability"],"status":"published"},{"id":"math-classical-probability-024","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"classical-probability","type":"single-choice","difficulty":5,"stem":"A box contains 3 red, 2 white and 1 black ball. Two balls are drawn at random without replacement. What is the probability that they have different colours?","stemZh":"盒中有 3 个红球、2 个白球和 1 个黑球，不放回地随机取出 2 个，两个球颜色不同的概率是多少？","options":[{"id":"A","text":"4/15"},{"id":"B","text":"11/15"},{"id":"C","text":"1/3"},{"id":"D","text":"2/5"}],"answer":"B","explanation":"$P(same colour) = [C(3,2) + C(2,2)]/C(6,2) = (3 + 1)/15 = 4/15 (black\\text{ cannot pair with }itself)$, so $P(different colours) = 1 − 4/15 = 11/15$.","explanationZh":"$P(同色) = [C(3,2) + C(2,2)]/C(6,2) = (3 + 1)/15 = 4/15（黑球只有 1$ 个，无法同色配对），故 $P(异色) = 1 − 4/15 = 11/15。$","tags":["hard","complement","balls"],"status":"published"},{"id":"math-classical-probability-025","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"classical-probability","type":"single-choice","difficulty":4,"stem":"In a class of 40 students, 22 study French, 15 study German and 7 study both. What is the probability that a randomly chosen student studies at least one of the two languages?","stemZh":"某班 40 名学生中，22 人学法语，15 人学德语，7 人两门都学。随机抽取一名学生，他至少学一门外语的概率是多少？","options":[{"id":"A","text":"37/40"},{"id":"B","text":"1/4"},{"id":"C","text":"3/4"},{"id":"D","text":"7/40"}],"answer":"C","explanation":"By $inclusion–exclusion, 22 + 15 − 7 = 30$ students study at least one language, so $P = 30/40 = 3/4$. Adding $22 + 15 = 37$ double-counts the 7 students who study both.","explanationZh":"由容斥原理，至少学一门的人数为 $22 + 15 − 7 = 30，故 P = 30/40 = 3/4。直接算 22 + 15 = 37$ 会把两门都学的 7 人重复计算一次。","tags":["hard","inclusion-exclusion","addition-rule"],"status":"published"},{"id":"math-classical-probability-026","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"classical-probability","type":"single-choice","difficulty":4,"stem":"A fair die is rolled once. Given that the score is even, what is the probability that it is greater than 4?","stemZh":"掷一枚均匀的骰子一次，已知点数为偶数，则点数大于 4 的概率是多少？","options":[{"id":"A","text":"1/3"},{"id":"B","text":"1/6"},{"id":"C","text":"1/2"},{"id":"D","text":"2/3"}],"answer":"A","explanation":"Given that the score is even, the reduced sample space is {2, 4, 6} and only 6 exceeds 4, so $P = 1/3$. By the $formula, P(>4 | even) = P(>4\\text{ and }even)/P(even) = (1/6)/(1/2) = 1/3$.","explanationZh":"已知点数为偶数，样本空间缩减为 ${2, 4, 6}，其中只有 6$ 大于 $4，故 P = 1/3。用条件概率公式：P(>4 | 偶数) = P(大于 4 且为偶数)/P(偶数) = (1/6)/(1/2) = 1/3。$","tags":["hard","conditional-probability","dice"],"status":"published"},{"id":"math-classical-probability-027","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"classical-probability","type":"single-choice","difficulty":5,"stem":"Two cards are drawn at random from a standard 52-card deck without replacement. What is the probability that both are red?","stemZh":"从一副 52 张的扑克牌中不放回地随机抽取 2 张，两张都是红色牌的概率是多少？","options":[{"id":"A","text":"1/4"},{"id":"B","text":"25/51"},{"id":"C","text":"13/51"},{"id":"D","text":"25/102"}],"answer":"D","explanation":"$P = (26/52) \\times (25/51) = (1/2) \\times (25/51) = 25/102$, or $C(26,2)/C(52,2) = 325/1326 = 25/102$. The value $1/4 = (26/52)^{2}$ is the answer with replacement, and 25/51 forgets the first factor.","explanationZh":"$P = (26/52) \\times (25/51) = (1/2) \\times (25/51) = 25/102，或 C(26,2)/C(52,2) = 325/1326 = 25/102。1/4 = (26/52)^{2} 是有放回抽取的结果，25/51$ 漏掉了第一个因子。","tags":["hard","without-replacement","cards"],"status":"published"},{"id":"math-classical-probability-028","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"classical-probability","type":"single-choice","difficulty":4,"stem":"A box contains 4 good items and 2 defective items. Two items are drawn at random without replacement. What is the probability that exactly one is defective?","stemZh":"盒中有 4 件正品和 2 件次品，不放回地随机取出 2 件，恰好有 1 件次品的概率是多少？","options":[{"id":"A","text":"4/15"},{"id":"B","text":"8/15"},{"id":"C","text":"1/3"},{"id":"D","text":"1/5"}],"answer":"B","explanation":"$P = C(2,1)·C(4,1)/C(6,2) = 8/15$. Computing only $(2/6) \\times (4/5) = 4/15$ forgets that the defective item can be drawn first or second; both orders give $8/30 + 8/30 = 8/15$.","explanationZh":"$P = C(2,1)·C(4,1)/C(6,2) = 8/15。只算 (2/6) \\times (4/5) = 4/15，会漏掉次品在第二次被取到的情形；两种顺序相加为 8/30 + 8/30 = 8/15。$","tags":["hard","without-replacement","balls"],"status":"published"},{"id":"math-classical-probability-029","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"classical-probability","type":"single-choice","difficulty":4,"stem":"A fair coin is tossed five times. What is the probability of getting exactly three heads?","stemZh":"一枚均匀硬币连掷五次，恰好出现三次正面的概率是多少？","options":[{"id":"A","text":"3/8"},{"id":"B","text":"1/2"},{"id":"C","text":"5/16"},{"id":"D","text":"5/32"}],"answer":"C","explanation":"There are $2^{5} = 32$ equally likely outcomes, of which $C(5,3) = 10$ contain exactly three heads, so $P = 10/32 = 5/16$.","explanationZh":"共有 $2^{5} = 32$ 个等可能结果，其中恰好三次正面的为从 5 次中选 $3 次：C(5,3) = 10，故 P = 10/32 = 5/16。$","tags":["hard","coins","combination"],"status":"published"},{"id":"math-classical-probability-030","subject":"math","moduleId":"probability-statistics","unitId":"probability","topicId":"classical-probability","type":"single-choice","difficulty":5,"stem":"Bag 1 contains 3 red and 2 blue balls; bag 2 contains 1 red and 4 blue balls. A bag is chosen at random and then one ball is drawn from it. What is the probability that the ball is red?","stemZh":"1 号袋中有 3 个红球和 2 个蓝球，2 号袋中有 1 个红球和 4 个蓝球。先随机选一个袋子，再从该袋中取出一个球，取到红球的概率是多少？","options":[{"id":"A","text":"2/5"},{"id":"B","text":"3/5"},{"id":"C","text":"1/5"},{"id":"D","text":"3/10"}],"answer":"A","explanation":"By the law of total $probability, P = (1/2)(3/5) + (1/2)(1/5) = 3/10 + 1/10 = 4/10 = 2/5$. Taking 3/5 or 1/5 uses only one bag, and 3/10 counts just the first branch.","explanationZh":"由全概率公式：P $= (1/2)(3/5) + (1/2)(1/5) = 3/10 + 1/10 = 4/10 = 2/5。只取 3/5 或 1/5 是只考虑了一个袋子，3/10$ 只算了第一条分支。","tags":["hard","total-probability","conditional-probability"],"status":"published"},{"id":"math-descriptive-001","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"descriptive","type":"single-choice","difficulty":1,"stem":"Find the mean of the data set 2, 4, 6, 8, 10.","stemZh":"求数据 2, 4, 6, 8, 10 的平均数。","options":[{"id":"A","text":"4"},{"id":"B","text":"6"},{"id":"C","text":"7"},{"id":"D","text":"8"}],"answer":"B","explanation":"Sum $= 2 + 4 + 6 + 8 + 10 = 30$ and $n = 5$, so $\\bar{x} = 30/5 = 6$.","explanationZh":"总和 $2+4+6+8+10 = 30，n = 5，所以 \\bar{x} = 30/5 = 6。$","tags":["easy","mean"],"status":"published"},{"id":"math-descriptive-002","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"descriptive","type":"single-choice","difficulty":1,"stem":"Find the median of the data set 3, 1, 7, 5, 9.","stemZh":"求数据 3, 1, 7, 5, 9 的中位数。","options":[{"id":"A","text":"5"},{"id":"B","text":"7"},{"id":"C","text":"1"},{"id":"D","text":"3"}],"answer":"A","explanation":"Sort the data first: 1, 3, 5, 7, 9. With $n = 5$ the median is the 3rd value, namely 5. Taking 7 means reading off the middle term without sorting.","explanationZh":"$先排序：1, 3, 5, 7, 9。n = 5，中位数是第 3$ 个数，即 5。选 7 是没有排序直接取中间项。","tags":["easy","median","sorting"],"status":"published"},{"id":"math-descriptive-003","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"descriptive","type":"single-choice","difficulty":1,"stem":"Find the mode of the data set 2, 3, 3, 5, 7, 3, 8.","stemZh":"求数据 2, 3, 3, 5, 7, 3, 8 的众数。","options":[{"id":"A","text":"2"},{"id":"B","text":"5"},{"id":"C","text":"3"},{"id":"D","text":"7"}],"answer":"C","explanation":"The value 3 occurs three times, more often than any other value, so the mode is 3.","explanationZh":"3 出现 3 次，次数最多，所以众数是 3。","tags":["easy","mode"],"status":"published"},{"id":"math-descriptive-004","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"descriptive","type":"single-choice","difficulty":1,"stem":"Find the range of the data set 4, 9, 2, 11, 6.","stemZh":"求数据 4, 9, 2, 11, 6 的极差。","options":[{"id":"A","text":"11"},{"id":"B","text":"7"},{"id":"C","text":"2"},{"id":"D","text":"9"}],"answer":"D","explanation":"Range $=$ maximum − minimum $= 11 − 2 = 9$. Option A only quotes the maximum; option B subtracts the first value $(11 − 4)$ instead of the minimum.","explanationZh":"极差 $=$ 最大值 − 最小值 $= 11 − 2 = 9。A$ 只取了最大值；B 用第一个数 4 代替了最小值。","tags":["easy","range"],"status":"published"},{"id":"math-descriptive-005","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"descriptive","type":"single-choice","difficulty":2,"stem":"Find the variance $s^{2}$ of the data set 1, 2, 3, 4, 5.","stemZh":"求数据 1, 2, 3, 4, 5 的方差 $s^{2}。$","options":[{"id":"A","text":"2"},{"id":"B","text":"$\\sqrt{2}$"},{"id":"C","text":"10"},{"id":"D","text":"2.5"}],"answer":"A","explanation":"$\\bar{x} = 15/5 = 3. Deviations: −2, −1, 0, 1, 2$, so $Σ(x_{i} − \\bar{x})^{2} = 4 + 1 + 0 + 1 + 4 = 10$ and $s^{2} = 10/5 = 2. \\sqrt{2}$ is the standard deviation, and 2.5 comes from dividing by $n − 1 = 4$ instead of n.","explanationZh":"$\\bar{x} = 15/5 = 3，离差为 −2, −1, 0, 1, 2，Σ(x_{i} − \\bar{x})^{2} = 10，方差 s^{2} = 10/5 = 2。\\sqrt{2} 是标准差；2.5$ 是误除以 $n − 1 = 4。$","tags":["easy","variance"],"status":"published"},{"id":"math-descriptive-006","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"descriptive","type":"single-choice","difficulty":2,"stem":"Find the standard deviation s of the data set 1, 2, 3, 4, 5.","stemZh":"求数据 1, 2, 3, 4, 5 的标准差 s。","options":[{"id":"A","text":"2"},{"id":"B","text":"$\\sqrt{2}$"},{"id":"C","text":"1"},{"id":"D","text":"$\\sqrt{10}$"}],"answer":"B","explanation":"$\\bar{x} = 3$ and $Σ(x_{i} − \\bar{x})^{2} = 10$, so $s^{2} = 10/5 = 2$ and $s = \\sqrt{2}$. The value 2 is the variance, not the standard deviation.","explanationZh":"$\\bar{x} = 3，Σ(x_{i} − \\bar{x})^{2} = 10，所以 s^{2} = 10/5 = 2，s = \\sqrt{2}。2$ 是方差而不是标准差。","tags":["easy","standard-deviation"],"status":"published"},{"id":"math-descriptive-007","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"descriptive","type":"single-choice","difficulty":2,"stem":"A data set has mean $\\bar{x} = 5$. Every value x is replaced by $y = 2x + 3$. Find the mean of the new data set.","stemZh":"一组数据的平均数 $\\bar{x} = 5，把每个数据 x$ 换成 $y = 2x + 3，求新数据的平均数。$","options":[{"id":"A","text":"8"},{"id":"B","text":"10"},{"id":"C","text":"13"},{"id":"D","text":"16"}],"answer":"C","explanation":"Under $y =$ ax + b the mean becomes $ȳ = a·\\bar{x} + b = 2·5 + 3 = 13$. Adding 3 only $(8)$, doubling only $(10)$ or doubling after adding $(16)$ are all wrong.","explanationZh":"$y =$ ax + b 时平均数变为 $ȳ = a·\\bar{x} + b = 2\\times5 + 3 = 13。只加 3、只乘 2$ 或先加后乘都不对。","tags":["easy","transformation","mean"],"status":"published"},{"id":"math-descriptive-008","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"descriptive","type":"single-choice","difficulty":2,"stem":"Find the median of the data set 8, 2, 6, 4.","stemZh":"求数据 8, 2, 6, 4 的中位数。","options":[{"id":"A","text":"4"},{"id":"B","text":"6"},{"id":"C","text":"7"},{"id":"D","text":"5"}],"answer":"D","explanation":"Sort first: 2, 4, 6, 8. With $n = 4$ the median is the average of the two middle $values: (4 + 6)/2 = 5$. Taking 7 averages the middle two of the unsorted list $(8$ and $6)$.","explanationZh":"$先排序：2, 4, 6, 8。n = 4，中位数是中间两个数的平均数：(4 + 6)/2 = 5。取 7$ 是没有排序就把中间两项 8 和 6 求平均。","tags":["easy","median","sorting"],"status":"published"},{"id":"math-descriptive-009","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"descriptive","type":"single-choice","difficulty":2,"stem":"A sample of size 10 is summarised as: value $1 (frequency 2)$, value $2 (frequency 3)$, value $3 (frequency 5)$. Find the mean of the sample.","stemZh":"容量为 10 的样本：数值 1（频数 2）、数值 2（频数 3）、数值 3（频数 5），求样本平均数。","options":[{"id":"A","text":"2"},{"id":"B","text":"2.3"},{"id":"C","text":"7.7"},{"id":"D","text":"23"}],"answer":"B","explanation":"$\\bar{x} = (1·2 + 2·3 + 3·5)/10 = (2 + 6 + 15)/10 = 23/10 = 2.3$. Option A averages only the three distinct $values, (1 + 2 + 3)/3 = 2$, ignoring the frequencies.","explanationZh":"$\\bar{x} = (1\\times2 + 2\\times3 + 3\\times5)/10 = 23/10 = 2.3。A$ 是只对三个不同数值求平均 $(1+2+3)/3 = 2，忽略了频数。$","tags":["easy","frequency-table","mean"],"status":"published"},{"id":"math-descriptive-010","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"descriptive","type":"single-choice","difficulty":1,"stem":"Which of the following quantities measures how spread out a data set is?","stemZh":"下列哪个量用来刻画一组数据的离散（波动）程度？","options":[{"id":"A","text":"Standard deviation"},{"id":"B","text":"Mean"},{"id":"C","text":"Median"},{"id":"D","text":"Mode"}],"answer":"A","explanation":"The standard deviation (and the variance) measure spread. The mean, median and mode measure the centre or the most typical value of the data.","explanationZh":"标准差（以及方差）刻画离散程度；平均数、中位数、众数刻画数据的集中趋势。","tags":["easy","standard-deviation","concept"],"status":"published"},{"id":"math-descriptive-011","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"descriptive","type":"single-choice","difficulty":3,"stem":"Find the variance $s^{2}$ of the data set 2, 4, 6, 8.","stemZh":"求数据 2, 4, 6, 8 的方差 $s^{2}。$","options":[{"id":"A","text":"$\\sqrt{5}$"},{"id":"B","text":"20"},{"id":"C","text":"5"},{"id":"D","text":"20/3"}],"answer":"C","explanation":"$\\bar{x} = 20/4 = 5. Deviations: −3, −1, 1, 3$, so $Σ(x_{i} − \\bar{x})^{2} = 9 + 1 + 1 + 9 = 20$ and $s^{2} = 20/4 = 5. \\sqrt{5}$ is the standard deviation, 20 forgets the division, and 20/3 divides by n − 1.","explanationZh":"$\\bar{x} = 20/4 = 5，Σ(x_{i} − \\bar{x})^{2} = 9+1+1+9 = 20，s^{2} = 20/4 = 5。\\sqrt{5} 是标准差，20$ 忘了除以 n，20/3 是除以 n − 1。","tags":["medium","variance"],"status":"published"},{"id":"math-descriptive-012","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"descriptive","type":"single-choice","difficulty":3,"stem":"A data set has mean 4 and variance 9. Every value x is replaced by $y = 3x − 2$. Find the variance of the new data set.","stemZh":"一组数据平均数 4、方差 9，把每个 x 换成 $y = 3x − 2，求新数据的方差。$","options":[{"id":"A","text":"9"},{"id":"B","text":"25"},{"id":"C","text":"27"},{"id":"D","text":"81"}],"answer":"D","explanation":"Under $y =$ ax + b the variance is multiplied by $a^{2}$ and the shift b has no $effect: s^{2}_y = 3^{2}·9 = 81$. Applying the linear rule to the variance $(3·9 − 2 = 25)$ or multiplying by a instead of $a^{2} (27)$ are common errors.","explanationZh":"$y =$ ax + b 时方差乘以 $a^{2}，平移 b 不影响方差：s^{2}_y = 3^{2}\\times9 = 81。把线性关系直接套到方差上（3\\times9 − 2 = 25）或只乘 a$ 不乘 $a^{2}（27）都是常见错误。$","tags":["medium","transformation","variance"],"status":"published"},{"id":"math-descriptive-013","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"descriptive","type":"single-choice","difficulty":3,"stem":"A data set has standard deviation 5. Every value x is replaced by $y = −2x + 1$. Find the standard deviation of the new data set.","stemZh":"一组数据标准差为 5，把每个 x 换成 $y = −2x + 1，求新数据的标准差。$","options":[{"id":"A","text":"10"},{"id":"B","text":"−10"},{"id":"C","text":"5"},{"id":"D","text":"20"}],"answer":"A","explanation":"Under $y =$ ax + b the standard deviation is scaled by $|a|: s_y = |−2|·5 = 10. A$ standard deviation is never negative, and adding 1 does not change the spread.","explanationZh":"$y =$ ax + b 时标准差变为 $s_y = |a|·s = |−2|\\times5 = 10。标准差不可能为负，加 1$ 不改变离散程度。","tags":["medium","transformation","standard-deviation"],"status":"published"},{"id":"math-descriptive-014","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"descriptive","type":"single-choice","difficulty":3,"stem":"A frequency table has classes $[0, 10)$ with frequency $3, [10, 20)$ with frequency 7 and $[20, 30)$ with frequency 10. Using class midpoints, estimate the mean.","stemZh":"$频数分布表：[0,10)$ 频数 $3，[10,20)$ 频数 $7，[20,30)$ 频数 10。用组中值估计平均数。","options":[{"id":"A","text":"15"},{"id":"B","text":"18.5"},{"id":"C","text":"20"},{"id":"D","text":"25"}],"answer":"B","explanation":"Midpoints are 5, 15, 25 and $n = 3 + 7 + 10 = 20$, so $\\bar{x} ≈ (5·3 + 15·7 + 25·10)/20 = (15 + 105 + 250)/20 = 370/20 = 18.5$. Option A averages the three midpoints and ignores the frequencies.","explanationZh":"组中值为 $5, 15, 25，n = 3+7+10 = 20，\\bar{x} ≈ (5\\times3 + 15\\times7 + 25\\times10)/20 = 370/20 = 18.5。A$ 只对三个组中值求平均，忽略了频数。","tags":["medium","grouped-data","mean"],"status":"published"},{"id":"math-descriptive-015","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"descriptive","type":"single-choice","difficulty":3,"stem":"Student A scores 78, 82, 80, 80, 80 and student B scores 70, 90, 75, 85, 80 on five tests. Whose results are more stable?","stemZh":"甲五次成绩为 78, 82, 80, 80, 80，乙五次成绩为 70, 90, 75, 85, 80，谁的成绩更稳定？","options":[{"id":"A","text":"Student B"},{"id":"B","text":"It cannot be determined"},{"id":"C","text":"Student A"},{"id":"D","text":"They are equally stable"}],"answer":"C","explanation":"Both means are 80, so compare variances. $A: Σ(x_{i} − 80)^{2} = 4 + 4 + 0 + 0 + 0 = 8, s^{2} = 8/5 = 1.6. B: Σ(x_{i} − 80)^{2} = 100 + 100 + 25 + 25 + 0 = 250, s^{2} = 250/5 = 50$. The smaller variance belongs to A, so A is more stable.","explanationZh":"两人平均数都是 $80，比较方差即可。甲：Σ(x_{i} − 80)^{2} = 8，s^{2} = 1.6；乙：Σ(x_{i} − 80)^{2} = 250，s^{2} = 50。甲方差更小，成绩更稳定。$","tags":["medium","stability","variance"],"status":"published"},{"id":"math-descriptive-016","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"descriptive","type":"single-choice","difficulty":3,"stem":"The scores of 10 students are summarised $as: 60 (2 students), 70 (3 students), 80 (4 students), 90 (1 student)$. Find the median score.","stemZh":"10 名学生的成绩：60 分 2 人，70 分 3 人，80 分 4 人，90 分 1 人，求中位数。","options":[{"id":"A","text":"70"},{"id":"B","text":"80"},{"id":"C","text":"74"},{"id":"D","text":"75"}],"answer":"D","explanation":"In order the scores are 60, 60, 70, 70, 70, 80, 80, 80, 80, 90. With $n = 10$ the median is the average of the 5th and $6th values: (70 + 80)/2 = 75$. The value 74 is the $mean, (60·2 + 70·3 + 80·4 + 90·1)/10 = 740/10$.","explanationZh":"排序后为 $60, 60, 70, 70, 70, 80, 80, 80, 80, 90。n = 10，中位数是第 5、6 个数的平均数：(70 + 80)/2 = 75。74$ 是平均数。","tags":["medium","median","frequency-table"],"status":"published"},{"id":"math-descriptive-017","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"descriptive","type":"single-choice","difficulty":3,"stem":"A data set has 5 values with mean 8. Four of the values are 6, 8, 10 and 7. Find the missing value.","stemZh":"一组 5 个数据的平均数为 8，其中四个是 6, 8, 10, 7，求第五个数。","options":[{"id":"A","text":"9"},{"id":"B","text":"8"},{"id":"C","text":"10"},{"id":"D","text":"7"}],"answer":"A","explanation":"The total is $5·8 = 40$, and the four known values sum to 31, so the missing value is $40 − 31 = 9$.","explanationZh":"总和为 $5\\times8 = 40，已知四个数之和为 31，故第五个数为 40 − 31 = 9。$","tags":["medium","mean","missing-value"],"status":"published"},{"id":"math-descriptive-018","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"descriptive","type":"single-choice","difficulty":3,"stem":"A histogram has three classes of equal width whose bar heights are in the ratio 3 : 7 : 5. If the total frequency is 60, what is the frequency of the first class?","stemZh":"某频率分布直方图三个组距相同，条形高之比为 3 : 7 : 5，总频数为 60，求第一组的频数。","options":[{"id":"A","text":"20"},{"id":"B","text":"12"},{"id":"C","text":"15"},{"id":"D","text":"28"}],"answer":"B","explanation":"With equal class widths the frequency is proportional to the bar height, so the frequencies are in the ratio 3 : 7 : 5 and the total is $3 + 7 + 5 = 15$ parts. The first class gets $60·(3/15) = 12$.","explanationZh":"组距相同时频数与条形高成正比，频数之比为 3 : 7 : 5，共 15 份，第一组频数 $= 60 \\times 3/15 = 12。$","tags":["medium","histogram","frequency"],"status":"published"},{"id":"math-descriptive-019","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"descriptive","type":"single-choice","difficulty":3,"stem":"Find the variance $s^{2}$ of the data set 2, 5, 5, 8.","stemZh":"求数据 2, 5, 5, 8 的方差 $s^{2}。$","options":[{"id":"A","text":"$\\sqrt{4.5}$"},{"id":"B","text":"18"},{"id":"C","text":"4.5"},{"id":"D","text":"6"}],"answer":"C","explanation":"$\\bar{x} = 20/4 = 5. Deviations: −3, 0, 0, 3$, so $Σ(x_{i} − \\bar{x})^{2} = 9 + 0 + 0 + 9 = 18$ and $s^{2} = 18/4 = 4.5. \\sqrt{4.5}$ is the standard deviation, and 6 comes from dividing by $n − 1 = 3$.","explanationZh":"$\\bar{x} = 20/4 = 5，Σ(x_{i} − \\bar{x})^{2} = 9+0+0+9 = 18，s^{2} = 18/4 = 4.5。\\sqrt{4.5} 是标准差，6$ 是除以 $n − 1 = 3$ 的结果。","tags":["medium","variance"],"status":"published"},{"id":"math-descriptive-020","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"descriptive","type":"single-choice","difficulty":3,"stem":"Class A has 30 students with mean score 70 and class B has 20 students with mean score 85. Find the mean score of all 50 students.","stemZh":"A 班 30 人平均 70 分，B 班 20 人平均 85 分，求全体 50 人的平均分。","options":[{"id":"A","text":"77.5"},{"id":"B","text":"70"},{"id":"C","text":"85"},{"id":"D","text":"76"}],"answer":"D","explanation":"The combined mean is $(30·70 + 20·85)/50 = (2100 + 1700)/50 = 3800/50 = 76$. It is not $(70 + 85)/2 = 77.5$, because the two classes have different sizes.","explanationZh":"总平均 $= (30\\times70 + 20\\times85)/50 = 3800/50 = 76。不能直接用 (70 + 85)/2 = 77.5，因为两班人数不同。$","tags":["medium","weighted-mean","mean"],"status":"published"},{"id":"math-descriptive-021","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"descriptive","type":"single-choice","difficulty":4,"stem":"A data set has mean 10 and variance 4. Every value x is replaced by $y = −3x + 5$. Find the standard deviation of the new data set.","stemZh":"一组数据平均数 10、方差 4，把每个 x 换成 $y = −3x + 5，求新数据的标准差。$","options":[{"id":"A","text":"6"},{"id":"B","text":"36"},{"id":"C","text":"−6"},{"id":"D","text":"12"}],"answer":"A","explanation":"The original standard deviation is $s = \\sqrt{4} = 2$. Under $y =$ ax $+ b, s_y = |a|·s = |−3|·2 = 6$. The value 36 is the new variance $(a^{2}·s^{2} = 9·4)$, and a standard deviation can never be negative.","explanationZh":"原标准差 $s = \\sqrt{4} = 2，y =$ ax $+ b 时 s_y = |a|·s = |−3|\\times2 = 6。36 是新方差（a^{2}·s^{2} = 9\\times4），标准差不可能为负。$","tags":["hard","transformation","standard-deviation"],"status":"published"},{"id":"math-descriptive-022","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"descriptive","type":"single-choice","difficulty":4,"stem":"For the data set 1, 3, 5, 7, find the value of $(1/n)·Σ(x_{i} − \\bar{x})^{2}$, where n is the number of data values.","stemZh":"对数据 $1, 3, 5, 7，求 (1/n)·Σ(x_{i} − \\bar{x})^{2}$ 的值（n 为数据个数）。","options":[{"id":"A","text":"5"},{"id":"B","text":"20/3"},{"id":"C","text":"$\\sqrt{5}$"},{"id":"D","text":"20"}],"answer":"A","explanation":"$\\bar{x} = 16/4 = 4$, so $Σ(x_{i} − \\bar{x})^{2} = 9 + 1 + 1 + 9 = 20$ and $(1/n)·20 = 20/4 = 5$. The value 20/3 would come from dividing by n − 1, which is not the convention used here: the divisor is n.","explanationZh":"$\\bar{x} = 16/4 = 4，Σ(x_{i} − \\bar{x})^{2} = 9+1+1+9 = 20，(1/n)\\times20 = 20/4 = 5。20/3$ 是除以 n − 1 的结果，本题除以 n。","tags":["hard","variance","convention"],"status":"published"},{"id":"math-descriptive-023","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"descriptive","type":"single-choice","difficulty":5,"stem":"For a data set of $5 values, Σx = 40$ and $Σx^{2} = 340$. Find the variance $s^{2}$.","stemZh":"一组 5 个数据满足 $Σx = 40，Σx^{2} = 340，求方差 s^{2}。$","options":[{"id":"A","text":"4"},{"id":"B","text":"68"},{"id":"C","text":"2"},{"id":"D","text":"64"}],"answer":"A","explanation":"Use $s^{2} = (Σx^{2})/n − \\bar{x}^{2}$. Here $\\bar{x} = 40/5 = 8$, so $s^{2} = 340/5 − 8^{2} = 68 − 64 = 4$. Option B stops at $Σx^{2}/n$ and option D is just $\\bar{x}^{2}$, both forgetting to subtract.","explanationZh":"用公式 $s^{2} = (Σx^{2})/n − \\bar{x}^{2}。\\bar{x} = 40/5 = 8，所以 s^{2} = 340/5 − 8^{2} = 68 − 64 = 4。B$ 只算到 $Σx^{2}/n，D$ 只是 $\\bar{x}^{2}，都没有相减。$","tags":["hard","variance","computational-formula"],"status":"published"},{"id":"math-descriptive-024","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"descriptive","type":"single-choice","difficulty":4,"stem":"Find the median of the data set 12, 5, 9, 5, 20, 9.","stemZh":"求数据 12, 5, 9, 5, 20, 9 的中位数。","options":[{"id":"A","text":"7"},{"id":"B","text":"9"},{"id":"C","text":"10"},{"id":"D","text":"5"}],"answer":"B","explanation":"Sorted: 5, 5, 9, 9, 12, 20. With $n = 6$ the median is $(9 + 9)/2 = 9$. The value 7 averages the middle two of the unsorted list $(9$ and $5), 10$ is the mean, and 5 is the mode.","explanationZh":"$排序：5, 5, 9, 9, 12, 20。n = 6，中位数为 (9 + 9)/2 = 9。7$ 是未排序时中间两项 9 与 5 的平均；10 是平均数；5 是众数。","tags":["hard","median","sorting"],"status":"published"},{"id":"math-descriptive-025","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"descriptive","type":"single-choice","difficulty":5,"stem":"Class A has 20 students with mean score 80, and class B has mean score 90. If the mean of the two classes combined is 85, how many students are in class B?","stemZh":"A 班 20 人平均 80 分，B 班平均 90 分，两班合并平均 85 分，求 B 班人数。","options":[{"id":"A","text":"10"},{"id":"B","text":"20"},{"id":"C","text":"25"},{"id":"D","text":"30"}],"answer":"B","explanation":"Let class B have n students. Then $(20·80 + 90n)/(20 + n) = 85$, i.e. $1600 + 90n = 1700 + 85n$, so $5n = 100$ and $n = 20$.","explanationZh":"设 B 班 n 人，则 $(20\\times80 + 90n)/(20 + n) = 85，即 1600 + 90n = 1700 + 85n，得 5n = 100，n = 20。$","tags":["hard","weighted-mean","equation"],"status":"published"},{"id":"math-descriptive-026","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"descriptive","type":"single-choice","difficulty":4,"stem":"In the data set 4, 5, 6, 7, 8 the largest value 8 is replaced by 48. By how much does the median change?","stemZh":"数据 4, 5, 6, 7, 8 中最大值 8 被改成 48，中位数变化了多少？","options":[{"id":"A","text":"Increases by 8"},{"id":"B","text":"It does not change"},{"id":"C","text":"Increases to 14"},{"id":"D","text":"Increases by 6"}],"answer":"B","explanation":"The new data are 4, 5, 6, 7, 48, whose median is still the 3rd value, 6 — unchanged. The mean does move, from 6 to $70/5 = 14$, an increase of 8, but the median is insensitive to this extreme value.","explanationZh":"新数据为 4, 5, 6, 7, 48，中位数仍是第 3 个数 6，没有变化。平均数由 6 变为 $70/5 = 14（增加 8），但中位数不受极端值影响。$","tags":["hard","median","outlier"],"status":"published"},{"id":"math-descriptive-027","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"descriptive","type":"single-choice","difficulty":4,"stem":"A data set has standard deviation 5. Every value x is replaced by $y = 4x − 1$. Find the standard deviation of the new data set.","stemZh":"一组数据标准差为 5，把每个 x 换成 $y = 4x − 1，求新数据的标准差。$","options":[{"id":"A","text":"19"},{"id":"B","text":"25"},{"id":"C","text":"20"},{"id":"D","text":"5"}],"answer":"C","explanation":"Only the multiplier changes the $spread: s_y = |4|·5 = 20$. The shift −1 has no effect on the standard deviation, so neither $4·5 − 1 = 19$ nor the unchanged value 5 is right.","explanationZh":"$只有倍数改变离散程度：s_y = |4|\\times5 = 20。平移 −1$ 不影响标准差，所以 $4\\times5 − 1 = 19$ 和不变的 5 都不对。","tags":["hard","transformation","standard-deviation"],"status":"published"},{"id":"math-descriptive-028","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"descriptive","type":"single-choice","difficulty":4,"stem":"The mean of the four numbers 1, 5, x, 7 is 4. Find the variance of the completed data set.","stemZh":"四个数 1, 5, x, 7 的平均数为 4，求这组完整数据的方差。","options":[{"id":"A","text":"$\\sqrt{5}$"},{"id":"B","text":"4"},{"id":"C","text":"5"},{"id":"D","text":"20"}],"answer":"C","explanation":"The total is $4·4 = 16$, so $x = 16 − (1 + 5 + 7) = 3$. The data are 1, 3, 5, 7 with $\\bar{x} = 4$, so $Σ(x_{i} − \\bar{x})^{2} = 9 + 1 + 1 + 9 = 20$ and $s^{2} = 20/4 = 5$.","explanationZh":"总和为 $4\\times4 = 16，故 x = 16 − (1 + 5 + 7) = 3。数据为 1, 3, 5, 7，\\bar{x} = 4，Σ(x_{i} − \\bar{x})^{2} = 20，方差 s^{2} = 20/4 = 5。$","tags":["hard","missing-value","variance"],"status":"published"},{"id":"math-descriptive-029","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"descriptive","type":"single-choice","difficulty":5,"stem":"A data set has mean $\\bar{x}$ and variance $s^{2}$. Every value x is replaced by $y = 2x − 5$. Which statement about the new data set is correct?","stemZh":"一组数据平均数 $\\bar{x}、方差 s^{2}，把每个 x$ 换成 $y = 2x − 5，关于新数据的说法正确的是？$","options":[{"id":"A","text":"Mean $2\\bar{x} − 5$, variance $2s^{2} − 5$"},{"id":"B","text":"Mean $\\bar{x}$, variance $4s^{2}$"},{"id":"C","text":"Mean $2\\bar{x}$, variance $2s^{2}$"},{"id":"D","text":"Mean $2\\bar{x} − 5$, variance $4s^{2}$"}],"answer":"D","explanation":"Both parameters feel the same linear rule for the $centre: ȳ = 2\\bar{x} − 5$. But spread is scaled by the square of the multiplier and is unaffected by the $shift: s^{2}_y = 2^{2}·s^{2} = 4s^{2}$. The variance does not obey $y = 2x − 5$ itself.","explanationZh":"平均数按同样的线性关系变化：ȳ $= 2\\bar{x} − 5；离散程度只与倍数的平方有关，平移不影响：s^{2}_y = 2^{2}·s^{2} = 4s^{2}。方差本身不能代入 y = 2x − 5。$","tags":["hard","transformation","variance"],"status":"published"},{"id":"math-descriptive-030","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"descriptive","type":"single-choice","difficulty":5,"stem":"A data set of 10 values has mean 20 and variance 9. Two new values, both equal to 20, are added to the data set. What are the new mean and variance?","stemZh":"一组 10 个数据的平均数为 20、方差为 9，再加入两个都等于 20 的数据，新的平均数和方差是多少？","options":[{"id":"A","text":"Mean 20, variance 9"},{"id":"B","text":"Mean 21.8, variance 7.5"},{"id":"C","text":"Mean 20, variance 10.8"},{"id":"D","text":"Mean 20, variance 7.5"}],"answer":"D","explanation":"The new total is $200 + 40 = 240$, so the mean is $240/12 = 20 (unchanged)$. The two added values have deviation 0, so $Σ(x_{i} − 20)^{2}$ is still $10·9 = 90$ and the new variance is $90/12 = 7.5$.","explanationZh":"新总和 $200 + 40 = 240，平均数 240/12 = 20（不变）。新增两个数离差为 0，Σ(x_{i} − 20)^{2}$ 仍为 $10\\times9 = 90，新方差 = 90/12 = 7.5。$","tags":["hard","mean","variance"],"status":"published"},{"id":"math-normal-001","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"normal","type":"single-choice","difficulty":1,"stem":"For $X ~ N(3, 4)$, what is the mean $μ$?","stemZh":"已知 $X ~ N(3, 4)，均值 μ$ 是多少？","options":[{"id":"A","text":"3"},{"id":"B","text":"4"},{"id":"C","text":"2"},{"id":"D","text":"5"}],"answer":"A","explanation":"In $X ~ N(μ, σ^{2})$ the first parameter is the mean, so $μ = 3$. The second parameter is the $variance: σ^{2} = 4$, hence $σ = 2$, and $μ + σ = 5$.","explanationZh":"$X ~ N(μ, σ^{2})$ 中第一个参数是均值，故 $μ = 3；第二个参数是方差 σ^{2} = 4，所以 σ = 2，μ + σ = 5。$","tags":["easy","parameters","mean"],"status":"published"},{"id":"math-normal-002","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"normal","type":"single-choice","difficulty":1,"stem":"For $X ~ N(5, 9)$, what is the standard deviation $σ$?","stemZh":"已知 $X ~ N(5, 9)，标准差 σ$ 是多少？","options":[{"id":"A","text":"9"},{"id":"B","text":"3"},{"id":"C","text":"5"},{"id":"D","text":"81"}],"answer":"B","explanation":"The second parameter is the variance, so $σ^{2} = 9$ and $σ = 3$. The value 9 is $σ^{2}$ and 5 is the mean $μ$.","explanationZh":"第二个参数是方差，故 $σ^{2} = 9，σ = 3。9 是方差，5$ 是均值 $μ。$","tags":["easy","parameters","standard-deviation"],"status":"published"},{"id":"math-normal-003","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"normal","type":"single-choice","difficulty":1,"stem":"The density curve of a normal distribution $X ~ N(μ, σ^{2})$ is symmetric about which vertical line?","stemZh":"正态分布 $X ~ N(μ, σ^{2})$ 的密度曲线关于哪条竖直直线对称？","options":[{"id":"A","text":"$x = 0$"},{"id":"B","text":"$x = σ$"},{"id":"C","text":"$x = μ$"},{"id":"D","text":"$x = μ + σ$"}],"answer":"C","explanation":"The normal curve is bell-shaped and symmetric about the vertical line $x = μ$, the mean of the distribution.","explanationZh":"正态曲线呈钟形，关于直线 $x = μ（均值所在处）对称。$","tags":["easy","symmetry","curve"],"status":"published"},{"id":"math-normal-004","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"normal","type":"single-choice","difficulty":1,"stem":"What is the total area under a normal density curve?","stemZh":"正态密度曲线下方的总面积是多少？","options":[{"id":"A","text":"0.5"},{"id":"B","text":"0.6826"},{"id":"C","text":"2"},{"id":"D","text":"1"}],"answer":"D","explanation":"The area under any probability density curve equals 1; this is what makes the curve a probability distribution.","explanationZh":"任何概率密度曲线下方面积都等于 1，这正是概率分布的要求。","tags":["easy","area","concept"],"status":"published"},{"id":"math-normal-005","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"normal","type":"single-choice","difficulty":2,"stem":"For $X ~ N(μ, σ^{2})$, approximately what proportion of the values lie within one standard deviation of the mean, i.e. $P(μ − σ < X ≤ μ + σ)$?","stemZh":"$对 X ~ N(μ, σ^{2})，落在均值一个标准差范围内的比例 P(μ − σ < X ≤ μ + σ)$ 约为多少？","options":[{"id":"A","text":"0.9544"},{"id":"B","text":"0.6826"},{"id":"C","text":"0.9974"},{"id":"D","text":"0.3413"}],"answer":"B","explanation":"The empirical rule gives $P(μ − σ < X ≤ μ + σ) ≈ 0.6826$. The values 0.9544 and 0.9974 correspond to two and three standard deviations, and 0.3413 is only the right-hand half of the interval.","explanationZh":"由经验法则 $P(μ − σ < X ≤ μ + σ) ≈ 0.6826。0.9544 与 0.9974 分别对应两个、三个标准差，0.3413$ 只是区间右半部分。","tags":["easy","empirical-rule"],"status":"published"},{"id":"math-normal-006","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"normal","type":"single-choice","difficulty":2,"stem":"For $X ~ N(μ, σ^{2})$, evaluate $P(μ − 2σ < X ≤ μ + 2σ)$.","stemZh":"$对 X ~ N(μ, σ^{2})，求 P(μ − 2σ < X ≤ μ + 2σ)。$","options":[{"id":"A","text":"0.9544"},{"id":"B","text":"0.6826"},{"id":"C","text":"0.9974"},{"id":"D","text":"0.4772"}],"answer":"A","explanation":"The empirical rule gives $P(μ − 2σ < X ≤ μ + 2σ) ≈ 0.9544; 0.4772$ is only the half of that interval on one side of $μ$.","explanationZh":"经验法则给出 $P(μ − 2σ < X ≤ μ + 2σ) ≈ 0.9544；0.4772$ 只是 $μ$ 一侧的一半。","tags":["easy","empirical-rule"],"status":"published"},{"id":"math-normal-007","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"normal","type":"single-choice","difficulty":2,"stem":"For $X ~ N(10, 4)$, what is $P(X > 10)$?","stemZh":"已知 $X ~ N(10, 4)，求 P(X > 10)。$","options":[{"id":"A","text":"0.6826"},{"id":"B","text":"0.25"},{"id":"C","text":"0.5"},{"id":"D","text":"0.3174"}],"answer":"C","explanation":"The curve is symmetric about $x = μ = 10$, so half of the area lies on each $side: P(X > 10) = 0.5$.","explanationZh":"曲线关于 $x = μ = 10$ 对称，两侧面积各占一半，所以 $P(X > 10) = 0.5。$","tags":["easy","symmetry"],"status":"published"},{"id":"math-normal-008","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"normal","type":"single-choice","difficulty":2,"stem":"For $X ~ N(μ, σ^{2})$, by symmetry $P(X < μ − σ)$ equals $P(X > ?)$","stemZh":"$对 X ~ N(μ, σ^{2})，由对称性 P(X < μ − σ)$ 等于 $P(X > ?)$","options":[{"id":"A","text":"$μ$"},{"id":"B","text":"$μ − σ$"},{"id":"C","text":"$μ + 2σ$"},{"id":"D","text":"$μ + σ$"}],"answer":"D","explanation":"The points $μ − σ$ and $μ + σ$ are mirror images about $x = μ$, so the two tails $P(X < μ − σ)$ and $P(X > μ + σ)$ are equal.","explanationZh":"$μ − σ 与 μ + σ$ 关于 $x = μ$ 对称，所以两侧尾部概率 $P(X < μ − σ) 与 P(X > μ + σ)$ 相等。","tags":["easy","symmetry"],"status":"published"},{"id":"math-normal-009","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"normal","type":"single-choice","difficulty":2,"stem":"For $X ~ N(20, 25)$, what is the standardised value $z = (x − μ)/σ$ of $x = 20$?","stemZh":"已知 $X ~ N(20, 25)，x = 20$ 标准化后 $z = (x − μ)/σ$ 等于多少？","options":[{"id":"A","text":"0"},{"id":"B","text":"1"},{"id":"C","text":"−1"},{"id":"D","text":"5"}],"answer":"A","explanation":"Here $μ = 20$ and $σ = \\sqrt{25} = 5$, so $z = (20 − 20)/5 = 0$. The standardised value of the mean is always 0.","explanationZh":"$μ = 20，σ = \\sqrt{25} = 5，z = (20 − 20)/5 = 0。均值标准化后总是 0。$","tags":["easy","standardisation"],"status":"published"},{"id":"math-normal-010","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"normal","type":"single-choice","difficulty":1,"stem":"For a normal distribution, what happens to the curve if $σ$ is increased while $μ$ is kept fixed?","stemZh":"正态分布中保持 $μ$ 不变而增大 $σ，曲线会怎样变化？$","options":[{"id":"A","text":"It becomes taller and narrower"},{"id":"B","text":"It becomes flatter and wider"},{"id":"C","text":"It shifts to the right"},{"id":"D","text":"Its shape does not change"}],"answer":"B","explanation":"$σ$ controls the spread: a larger $σ$ spreads the same total area 1 over a wider interval, so the bell becomes flatter and wider. The centre stays at $μ$.","explanationZh":"$σ 决定离散程度：σ$ 越大，总面积 1 被摊到更宽的区间上，钟形曲线变得更矮更胖；对称轴仍在 $x = μ。$","tags":["easy","curve","standard-deviation"],"status":"published"},{"id":"math-normal-011","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"normal","type":"single-choice","difficulty":3,"stem":"For $X ~ N(μ, σ^{2})$, evaluate $P(μ < X ≤ μ + σ)$.","stemZh":"$对 X ~ N(μ, σ^{2})，求 P(μ < X ≤ μ + σ)。$","options":[{"id":"A","text":"0.6826"},{"id":"B","text":"0.4772"},{"id":"C","text":"0.1587"},{"id":"D","text":"0.3413"}],"answer":"D","explanation":"The interval $(μ − σ, μ + σ]$ carries probability 0.6826 and is symmetric about $μ$, so each half carries $0.6826/2 = 0.3413$.","explanationZh":"区间 $(μ − σ, μ + σ]$ 的概率为 $0.6826，且关于 μ$ 对称，所以每半边为 $0.6826/2 = 0.3413。$","tags":["medium","empirical-rule","symmetry"],"status":"published"},{"id":"math-normal-012","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"normal","type":"single-choice","difficulty":3,"stem":"For $X ~ N(μ, σ^{2})$, evaluate $P(X > μ + σ)$.","stemZh":"$对 X ~ N(μ, σ^{2})，求 P(X > μ + σ)。$","options":[{"id":"A","text":"0.3413"},{"id":"B","text":"0.1587"},{"id":"C","text":"0.3174"},{"id":"D","text":"0.8413"}],"answer":"B","explanation":"Outside $(μ − σ, μ + σ]$ the remaining probability is $1 − 0.6826 = 0.3174$, split equally between the two tails, so each tail is $0.3174/2 = 0.1587$.","explanationZh":"区间 $(μ − σ, μ + σ]$ 之外的概率为 $1 − 0.6826 = 0.3174，两侧尾部各占一半，故 P(X > μ + σ) = 0.3174/2 = 0.1587。$","tags":["medium","empirical-rule","tail"],"status":"published"},{"id":"math-normal-013","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"normal","type":"single-choice","difficulty":3,"stem":"For $X ~ N(μ, σ^{2})$, evaluate $P(μ − σ < X ≤ μ + 2σ)$.","stemZh":"$对 X ~ N(μ, σ^{2})，求 P(μ − σ < X ≤ μ + 2σ)。$","options":[{"id":"A","text":"0.9544"},{"id":"B","text":"0.4772"},{"id":"C","text":"0.8185"},{"id":"D","text":"0.3413"}],"answer":"C","explanation":"Split the interval at $μ: P(μ − σ < X ≤ μ) = 0.6826/2 = 0.3413$ and $P(μ < X ≤ μ + 2σ) = 0.9544/2 = 0.4772$. Their sum is 0.8185.","explanationZh":"$以 μ 为界分段：P(μ − σ < X ≤ μ) = 0.6826/2 = 0.3413，P(μ < X ≤ μ + 2σ) = 0.9544/2 = 0.4772，相加得 0.8185。$","tags":["medium","empirical-rule","symmetry"],"status":"published"},{"id":"math-normal-014","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"normal","type":"single-choice","difficulty":3,"stem":"For $X ~ N(50, 100)$, find the standardised value z of $x = 70$.","stemZh":"已知 $X ~ N(50, 100)，求 x = 70$ 的标准化值 z。","options":[{"id":"A","text":"2"},{"id":"B","text":"0.2"},{"id":"C","text":"20"},{"id":"D","text":"−2"}],"answer":"A","explanation":"$μ = 50$ and $σ = \\sqrt{100} = 10$, so $z = (70 − 50)/10 = 2$. Option B uses $σ^{2} = 100$ as the divisor.","explanationZh":"$μ = 50，σ = \\sqrt{100} = 10，z = (70 − 50)/10 = 2。B$ 是误用 $σ^{2} = 100$ 去除。","tags":["medium","standardisation"],"status":"published"},{"id":"math-normal-015","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"normal","type":"single-choice","difficulty":3,"stem":"For $X ~ N(8, 9)$, evaluate $P(5 < X ≤ 11)$.","stemZh":"已知 $X ~ N(8, 9)，求 P(5 < X ≤ 11)。$","options":[{"id":"A","text":"0.3413"},{"id":"B","text":"0.6826"},{"id":"C","text":"0.9544"},{"id":"D","text":"0.9974"}],"answer":"B","explanation":"$μ = 8$ and $σ = \\sqrt{9} = 3$, so $5 = μ − σ$ and $11 = μ + σ$. Hence $P(5 < X ≤ 11) = 0.6826$.","explanationZh":"$μ = 8，σ = \\sqrt{9} = 3，5 = μ − σ，11 = μ + σ，所以概率为 0.6826。$","tags":["medium","empirical-rule"],"status":"published"},{"id":"math-normal-016","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"normal","type":"single-choice","difficulty":3,"stem":"For $X ~ N(8, 9)$, evaluate $P(X > 11)$.","stemZh":"已知 $X ~ N(8, 9)，求 P(X > 11)。$","options":[{"id":"A","text":"0.8413"},{"id":"B","text":"0.3174"},{"id":"C","text":"0.3413"},{"id":"D","text":"0.1587"}],"answer":"D","explanation":"Here $11 = μ + σ$, so $P(X > 11) = (1 − 0.6826)/2 = 0.1587$. The value 0.3174 is the probability of lying outside $(μ − σ, μ + σ]$ altogether.","explanationZh":"$11 = μ + σ，所以 P(X > 11) = (1 − 0.6826)/2 = 0.1587。0.3174$ 是落在 $(μ − σ, μ + σ]$ 之外的总概率。","tags":["medium","empirical-rule","tail"],"status":"published"},{"id":"math-normal-017","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"normal","type":"single-choice","difficulty":3,"stem":"For $X ~ N(100, 64)$, evaluate $P(84 < X ≤ 116)$.","stemZh":"已知 $X ~ N(100, 64)，求 P(84 < X ≤ 116)。$","options":[{"id":"A","text":"0.9544"},{"id":"B","text":"0.6826"},{"id":"C","text":"0.4772"},{"id":"D","text":"0.9974"}],"answer":"A","explanation":"$μ = 100$ and $σ = \\sqrt{64} = 8$, so $84 = μ − 2σ$ and $116 = μ + 2σ$. Hence the probability is 0.9544.","explanationZh":"$μ = 100，σ = \\sqrt{64} = 8，84 = μ − 2σ，116 = μ + 2σ，故概率为 0.9544。$","tags":["medium","empirical-rule"],"status":"published"},{"id":"math-normal-018","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"normal","type":"single-choice","difficulty":3,"stem":"For $X ~ N(100, 64)$, evaluate $P(X ≤ 92)$.","stemZh":"已知 $X ~ N(100, 64)，求 P(X ≤ 92)。$","options":[{"id":"A","text":"0.8413"},{"id":"B","text":"0.1587"},{"id":"C","text":"0.3413"},{"id":"D","text":"0.3174"}],"answer":"B","explanation":"$92 = μ − σ$, and by symmetry the left tail equals the right $tail: P(X ≤ 92) = (1 − 0.6826)/2 = 0.1587$.","explanationZh":"$92 = μ − σ，由对称性左尾等于右尾：P(X ≤ 92) = (1 − 0.6826)/2 = 0.1587。$","tags":["medium","empirical-rule","symmetry"],"status":"published"},{"id":"math-normal-019","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"normal","type":"single-choice","difficulty":3,"stem":"Compare the density curves of $X ~ N(0, 1)$ and $Y ~ N(0, 4)$. Which statement is true?","stemZh":"比较 $X ~ N(0, 1) 与 Y ~ N(0, 4)$ 的密度曲线，哪个说法正确？","options":[{"id":"A","text":"The curve of X is flatter and wider than that of Y"},{"id":"B","text":"The two curves are identical"},{"id":"C","text":"The curve of Y is flatter and wider than that of X"},{"id":"D","text":"The curve of Y is shifted to the right of that of X"}],"answer":"C","explanation":"Both have mean 0, so both are centred at $x = 0$. But $σ_Y = \\sqrt{4} = 2$ while $σ_X = 1$, so Y spreads more and its bell is flatter and wider.","explanationZh":"两者均值都为 $0，对称轴都是 x = 0；但 σ_Y = \\sqrt{4} = 2，σ_X = 1，Y$ 离散程度更大，曲线更矮更胖。","tags":["medium","curve","standard-deviation"],"status":"published"},{"id":"math-normal-020","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"normal","type":"single-choice","difficulty":3,"stem":"For $X ~ N(70, 25)$, evaluate $P(X > 80)$.","stemZh":"已知 $X ~ N(70, 25)，求 P(X > 80)。$","options":[{"id":"A","text":"0.0456"},{"id":"B","text":"0.9772"},{"id":"C","text":"0.1587"},{"id":"D","text":"0.0228"}],"answer":"D","explanation":"$μ = 70$ and $σ = \\sqrt{25} = 5$, so $80 = μ + 2σ$ and $P(X > 80) = (1 − 0.9544)/2 = 0.0228$. The value 0.0456 is the total probability in both tails beyond $2σ$.","explanationZh":"$μ = 70，σ = \\sqrt{25} = 5，80 = μ + 2σ，P(X > 80) = (1 − 0.9544)/2 = 0.0228。0.0456$ 是两侧 $2σ$ 之外的总概率。","tags":["medium","empirical-rule","tail"],"status":"published"},{"id":"math-normal-021","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"normal","type":"single-choice","difficulty":4,"stem":"For $X ~ N(μ, σ^{2})$, evaluate $P(μ − 2σ < X ≤ μ − σ)$.","stemZh":"$对 X ~ N(μ, σ^{2})，求 P(μ − 2σ < X ≤ μ − σ)。$","options":[{"id":"A","text":"0.2718"},{"id":"B","text":"0.4772"},{"id":"C","text":"0.1359"},{"id":"D","text":"0.3413"}],"answer":"C","explanation":"Take the difference of the two empirical rules and halve $it: (0.9544 − 0.6826)/2 = 0.2718/2 = 0.1359$. The value 0.2718 counts the same band on both sides of $μ$.","explanationZh":"$用两条经验法则相减再取一半：(0.9544 − 0.6826)/2 = 0.2718/2 = 0.1359。0.2718 是 μ$ 两侧同一段的总和。","tags":["hard","empirical-rule","symmetry"],"status":"published"},{"id":"math-normal-022","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"normal","type":"single-choice","difficulty":4,"stem":"For $X ~ N(20, 16)$, evaluate $P(12 < X ≤ 24)$.","stemZh":"已知 $X ~ N(20, 16)，求 P(12 < X ≤ 24)。$","options":[{"id":"A","text":"0.4772"},{"id":"B","text":"0.3413"},{"id":"C","text":"0.9544"},{"id":"D","text":"0.8185"}],"answer":"D","explanation":"$μ = 20$ and $σ = \\sqrt{16} = 4$, so $12 = μ − 2σ$ and $24 = μ + σ$. Split at $μ: 0.9544/2 + 0.6826/2 = 0.4772 + 0.3413 = 0.8185$.","explanationZh":"$μ = 20，σ = \\sqrt{16} = 4，12 = μ − 2σ，24 = μ + σ。以 μ 分段：0.9544/2 + 0.6826/2 = 0.4772 + 0.3413 = 0.8185。$","tags":["hard","empirical-rule","symmetry"],"status":"published"},{"id":"math-normal-023","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"normal","type":"single-choice","difficulty":4,"stem":"For $X ~ N(20, 16)$, evaluate $P(X ≤ 12) + P(X > 28)$.","stemZh":"已知 $X ~ N(20, 16)，求 P(X ≤ 12) + P(X > 28)。$","options":[{"id":"A","text":"0.0456"},{"id":"B","text":"0.0228"},{"id":"C","text":"0.9544"},{"id":"D","text":"0.1359"}],"answer":"A","explanation":"Here $12 = μ − 2σ$ and $28 = μ + 2σ$, so the two tails together are the complement of $0.9544: 1 − 0.9544 = 0.0456. A$ single tail would be 0.0228.","explanationZh":"$12 = μ − 2σ，28 = μ + 2σ，两侧尾部之和为 0.9544 的补：1 − 0.9544 = 0.0456。单侧尾部才是 0.0228。$","tags":["hard","empirical-rule","tail"],"status":"published"},{"id":"math-normal-024","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"normal","type":"single-choice","difficulty":4,"stem":"For $X ~ N(μ, σ^{2})$, evaluate $P(X ≤ μ + σ)$.","stemZh":"$对 X ~ N(μ, σ^{2})，求 P(X ≤ μ + σ)。$","options":[{"id":"A","text":"0.6826"},{"id":"B","text":"0.8413"},{"id":"C","text":"0.3413"},{"id":"D","text":"0.1587"}],"answer":"B","explanation":"$P(X ≤ μ) = 0.5$ by symmetry, and $P(μ < X ≤ μ + σ) = 0.6826/2 = 0.3413$, so the total is $0.5 + 0.3413 = 0.8413$.","explanationZh":"由对称性 $P(X ≤ μ) = 0.5，且 P(μ < X ≤ μ + σ) = 0.6826/2 = 0.3413，合计 0.5 + 0.3413 = 0.8413。$","tags":["hard","empirical-rule","symmetry"],"status":"published"},{"id":"math-normal-025","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"normal","type":"single-choice","difficulty":5,"stem":"For $X ~ N(30, 4)$, evaluate $P(24 < X ≤ 30)$.","stemZh":"已知 $X ~ N(30, 4)，求 P(24 < X ≤ 30)。$","options":[{"id":"A","text":"0.9974"},{"id":"B","text":"0.4772"},{"id":"C","text":"0.4987"},{"id":"D","text":"0.3413"}],"answer":"C","explanation":"$μ = 30$ and $σ = \\sqrt{4} = 2$, so $24 = μ − 3σ$. By symmetry the left half of the $3σ$ band is $0.9974/2 = 0.4987 (not 0.5: a$ little probability lies beyond $3σ)$.","explanationZh":"$μ = 30，σ = \\sqrt{4} = 2，24 = μ − 3σ。由对称性，3σ$ 区间左半概率为 $0.9974/2 = 0.4987（不等于 0.5，因为 3σ$ 之外仍有少量概率）。","tags":["hard","empirical-rule","symmetry"],"status":"published"},{"id":"math-normal-026","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"normal","type":"single-choice","difficulty":5,"stem":"For $X ~ N(μ, σ^{2})$, which of the following statements is NOT correct?","stemZh":"$对 X ~ N(μ, σ^{2})，下列说法中不正确的是？$","options":[{"id":"A","text":"$P(μ − σ < X ≤ μ + σ) ≈ 0.6826$"},{"id":"B","text":"$P(X > μ) = 0.5$"},{"id":"C","text":"$P(μ − 3σ < X ≤ μ + 3σ) ≈ 0.9974$"},{"id":"D","text":"$P(μ − σ < X ≤ μ + 2σ) ≈ 0.9544$"}],"answer":"D","explanation":"The first three are the standard facts. For D, splitting at $μ$ gives $0.6826/2 + 0.9544/2 = 0.3413 + 0.4772 = 0.8185$, not 0.9544; 0.9544 belongs to the symmetric interval $(μ − 2σ, μ + 2σ]$.","explanationZh":"前三项都是标准结论。D 中以 $μ$ 分段得 $0.6826/2 + 0.9544/2 = 0.3413 + 0.4772 = 0.8185，而不是 0.9544；0.9544$ 对应对称区间 $(μ − 2σ, μ + 2σ]。$","tags":["hard","empirical-rule","error-distinction"],"status":"published"},{"id":"math-normal-027","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"normal","type":"single-choice","difficulty":4,"stem":"For $X ~ N(50, 25)$, evaluate $P(X ≤ 35)$.","stemZh":"已知 $X ~ N(50, 25)，求 P(X ≤ 35)。$","options":[{"id":"A","text":"0.0026"},{"id":"B","text":"0.0013"},{"id":"C","text":"0.9974"},{"id":"D","text":"0.0228"}],"answer":"B","explanation":"$μ = 50$ and $σ = \\sqrt{25} = 5$, so $35 = μ − 3σ$. The two tails beyond $3σ$ hold $1 − 0.9974 = 0.0026$ in total, so one tail is $0.0026/2 = 0.0013$.","explanationZh":"$μ = 50，σ = \\sqrt{25} = 5，35 = μ − 3σ。3σ$ 之外两侧共 $1 − 0.9974 = 0.0026，单侧为 0.0026/2 = 0.0013。$","tags":["hard","empirical-rule","tail"],"status":"published"},{"id":"math-normal-028","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"normal","type":"single-choice","difficulty":5,"stem":"For $X ~ N(12, 25)$, find the standardised value z of $x = 4.5$.","stemZh":"已知 $X ~ N(12, 25)，求 x = 4.5$ 的标准化值 z。","options":[{"id":"A","text":"−1.5"},{"id":"B","text":"1.5"},{"id":"C","text":"−0.3"},{"id":"D","text":"−7.5"}],"answer":"A","explanation":"$μ = 12$ and $σ = \\sqrt{25} = 5$, so $z = (4.5 − 12)/5 = −7.5/5 = −1.5$. Dividing by $σ^{2} = 25$ instead of $σ$ gives −0.3, and −7.5 is only the numerator.","explanationZh":"$μ = 12，σ = \\sqrt{25} = 5，z = (4.5 − 12)/5 = −7.5/5 = −1.5。若误除以 σ^{2} = 25 得 −0.3，−7.5$ 只是分子。","tags":["hard","standardisation"],"status":"published"},{"id":"math-normal-029","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"normal","type":"single-choice","difficulty":5,"stem":"Scores are distributed as $X ~ N(80, 36)$. Out of 1000 students, about how many score between 74 and 86?","stemZh":"成绩 $X ~ N(80, 36)。1000$ 名学生中，大约有多少人成绩在 74 到 86 之间？","options":[{"id":"A","text":"341"},{"id":"B","text":"954"},{"id":"C","text":"683"},{"id":"D","text":"318"}],"answer":"C","explanation":"$μ = 80$ and $σ = \\sqrt{36} = 6$, so $74 = μ − σ$ and $86 = μ + σ$. The probability is 0.6826, giving $1000·0.6826 = 682.6 ≈ 683$ students.","explanationZh":"$μ = 80，σ = \\sqrt{36} = 6，74 = μ − σ，86 = μ + σ，概率为 0.6826，人数约 1000 \\times 0.6826 = 682.6 ≈ 683$ 人。","tags":["hard","empirical-rule","expected-count"],"status":"published"},{"id":"math-normal-030","subject":"math","moduleId":"probability-statistics","unitId":"statistics","topicId":"normal","type":"single-choice","difficulty":5,"stem":"Scores are distributed as $X ~ N(80, 36)$. Out of 1000 students, about how many score below 68 or above 92?","stemZh":"成绩 $X ~ N(80, 36)。1000$ 名学生中，大约有多少人成绩低于 68 或高于 92？","options":[{"id":"A","text":"23"},{"id":"B","text":"456"},{"id":"C","text":"954"},{"id":"D","text":"46"}],"answer":"D","explanation":"$μ = 80$ and $σ = 6$, so $68 = μ − 2σ$ and $92 = μ + 2σ$. The two tails together carry $1 − 0.9544 = 0.0456$, giving $1000·0.0456 = 45.6 ≈ 46$ students. One tail alone would be about 23.","explanationZh":"$μ = 80，σ = 6，68 = μ − 2σ，92 = μ + 2σ。两侧尾部共 1 − 0.9544 = 0.0456，人数约 1000 \\times 0.0456 = 45.6 ≈ 46$ 人。只算单侧约 23 人。","tags":["hard","empirical-rule","expected-count"],"status":"published"}]

Probability & Statistics