问题15：扔一对骰子。如果已知第二个骰子始终显示质数，则求出获得7作为总和的概率。
解决方案：
A = Sum on two dice equals 7 = {(1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3)}
B = Second die always exhibits a prime number = {(1, 1), (2, 1), (3, 1), (4, 1), (5, 1), (6, 1),
(1, 3), (2, 3), (3, 3), (4, 3), (5, 3), (6, 3),
(1, 5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 5)}
(A ∩ B) = {(6, 1), (2, 5), (4, 3)}
P(A/B) = P(A ∩ B)/P(B)
= 3/18
= 1/6
问题16：掷骰子。如果结果是奇数，那么它是质数的概率是多少?
解决方案：
A die is rolled
A = A prime number on die = {2, 3, 5}
B = An odd number on die = {1, 3, 5}
(A ∩ B) = {3, 5}
P(A/B) = P(A ∩ B)/P(B)
= 2/3
问题17：掷出一对骰子。如果第一个骰子上出现4，则求出总和为8或更大的概率。
A pair of dice is thrown
A = Getting sum of 8 or more = {(2, 6), (6, 2), (3, 5), (5, 3), (4, 4), (3, 6), (6, 3)
(4, 5), (5, 4), (4, 6), (6, 4), (5, 5), (6, 5), (5, 6), (6, 6)}
B = 4 on first die = {(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)}
(A ∩ B) = {(4, 4), (4, 5), (4, 6)}
P(A/B) = P(A ∩ B)/P(B)
= 1/2
问题18.考虑到至少一个骰子不显示五个骰子，发现两个骰子上显示的数字总和为8的概率。
解决方案：
A pair of dice is thrown
A = Getting sum of 8 or more = {(2, 6), (6, 2), (3, 5), (5, 3), (4, 4)}
B = At least one die does not show 5 = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 6),
(2, 1), (2, 2), (2, 3), (2, 4), (2, 6),
(3, 1), (3, 2), (3, 3), (3, 4), (3, 6),
(4, 1), (4, 2), (4, 3), (4, 4), (4, 6),
(1, 1), (1, 2), (1, 3), (1, 4), (1, 6)}
(A ∩ B) = {(2, 6), (4, 6), (6, 2)}
P(A/B) = P(A ∩ B)/P(B)
= 3/25
问题19.从整数1到9中随机选择两个数字。如果总和为偶数，则找出两个数字均为奇数的可能性。
解决方案：
A = Both numbers are odd = {(3, 1), (5, 1), (7, 1), (9, 1), (3, 7), (5, 7), (9, 7)
(3, 9), (5, 9), (7, 9), (7, 3), (7, 5), (7, 9), (5, 3)
(9, 3), (1, 3), (1, 5), (1, 7)}
B = Sum of both numbers is even = {(1, 3), (1, 5), (2, 4), (1, 7), (2, 6), (3, 5), (1, 9),
(2, 8), (3, 7), (4, 6), (7, 5), (8, 4), (9, 3), (8, 6),
(9, 5), (9, 7)}
(A ∩ B) = {(1, 3), (1, 5), (1, 7), (3, 5), (1, 9), (3, 7), (7, 5), (9, 3), (9, 5), (9, 7)
P(A/B) = P(A ∩ B)/P(B)
= 10/16 = 5/8
问题20.掷骰子两次，观察到的数字总和为8。数字5至少出现一次的条件概率是多少?
A die is thrown twice
A = The number 5 has appeared at least once
A = {(1, 5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 6)}
B = Sum of numbers is 8 = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}
(A ∩ B) = {(3, 5), (5, 3)}
P(A/B) = P(A ∩ B)/P(B)
= 2/5
问题21：掷出两个骰子，并且已知第一个骰子显示为6。求出两个骰子上显示的数字之和为7的概率。
解决方案：
Two dice are thrown
A = Sum of the numbers is 7 = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}
B = First die shows a 6 = {(6, 1), ((6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
(A ∩ B) = {(6, 1)}
P(A/B) = P(A ∩ B)/P(B)
= 1/6
问题22：掷出一对骰子。令A为总和大于或等于10的事件，令F为“ 5在第一个骰子上出现”的事件。找出P(A / B)。如果B是事件“ 5个出现在至少一个骰子上”，则罚款P(A / B)。
解决方案：
A pair of dice is thrown
A = Sum is greater than or equal to 10 = {(4, 6), (6, 4), (5, 5), (5, 6), (6, 5), (6, 6)}
Case 1: B = 5 appears on first die = {(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}
(A ∩ B) = {(5, 5), (5, 6)}
P(A/B) = P(A ∩ B)/P(B)
= 2/6 = 1/3
Case 2: B = 5 appears on at least one die = {(1, 5), (2, 5), (3, 5), (4, 5), (6, 5),
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}
(A ∩ B) = {(5, 5), (5, 6), (6, 5)}
P(A/B) = P(A ∩ B)/P(B)
= 3/11
问题23：从班级中随机选择的学生通过数学的概率为4/5，而他/她在数学和计算机科学中通过的概率为1/2。如果已知他/她已通过数学，那么他/她将通过计算机科学的概率是多少?
解决方案：
Probability to pass mathematics: P(M) = 4/5
Probability to pass Mathematics(M) and Computer Science (C) = P(M ∩ C) = 1/2
We know that, P(C/M) = P(M ∩ C) /P(M)
= (1/2) ÷ (5/4) = 5/8
问题24.某人购买衬衫的概率为0.2，某人购买裤子的概率为0.3，在某人购买裤子的情况下，他购买衬衫的概率为0.4。找出他会同时购买衬衫和裤子的可能性。还要找到给定他购买衬衫的机会来购买裤子的可能性。
解决方案：
Probability that a person buys a shirt(S) = P(S) = 0.2
Probability that he buys a trouser(T) = P(T) = 0.3
P(S/T) = 0.4
We know that,
P(S/T) = P(S ∩ T)/P(T)
P(S ∩ T) = 0.4 × 0.3 = 0.12
P(T/S) = P(S ∩ T)/P(S)
= 0.12/0.2 = 0.6
问题25.一所学校有1000名学生，其中430名是女孩。众所周知，在430名学生中，有10％的女孩在XII类中学习。假设所选学生是一名女孩，则该学生选择在XII类中随机学习的概率是多少?
解决方案：
Total students = 1000
Number of girls = 430
Let A = Student chosen studies in class XII
B = Student chosen is a girl
Then P(B) = 430/1000
P(A ∩ B) = 43/1000
P(A/B) = P(A ∩ B)/P(B)
= 43/430 = 1/10
问题26.将十张编号为1到10的卡片放在一个盒子中，彻底混合，然后随机抽出一张卡片。如果知道抽奖卡上的数字大于3，那么该数字是偶数的概率是多少?
解决方案：
Total number of cards = 10
Let A = drawn number is more than 3
B = drawn number is even
P(B/A) = P(B ∩ A)/P(A)
P(A) = 7/10
P(A ∩ B) = 4/10
P(B/A) = 4/7
问题27.假设每个出生的孩子均可能是男孩或女孩。如果一个家庭有两个孩子，那么两个孩子都是女孩的条件概率是多少?鉴于
(i)最小的是一个女孩
(ii)至少有一个是女孩。
解决方案：
(i) Let ‘A’ be the event that both the children born are girls.
Let ‘B’ be the event that the youngest is a girl.
We have to find conditional probability P(A/B).
P(A/B) = P(A ∩ B)/P(B)
P(A ∩ B) = P(A) = P(GG)
= 1/2 × 1/2 = 1/4
P(B) = P(BG) + P(GG)
= 1/2 × 1/2 + 1/2 × 1/2 = 1/2
Hence, P(A/B) = (1/4) ÷ (1/2) = 1/2
(ii) Let ‘A’ be the event that both the children born are girls.
Let ‘B’ be the event that at least one is a girl.
We have to find the conditional probability P(A/B).
P(A/B) = P(A ∩ B)/P(B)
P(A ∩ B) = P(A) = P(GG)
= 1/2 × 1/2 = 1/4
P(B) = 1 – P(BB)
= 1 – 1/2 × 1/2 = 1 – 1/4 = 3/4
Hence, P(A/B) = (1/4) ÷ (3/4) = 1/3