第 12 类 RD Sharma 解决方案 – 第 31 章概率 – 练习 31.1

问题 1：将编号为 1 到 10 的十张牌放入一个盒子中，彻底混合，然后随机抽取一张牌。如果已知抽出的牌上的数字大于 3，那么它是偶数的概率是多少？

解决方案：

Here,
Sample space (S) = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
Let, A = the number on the drawn card is even
So, A = {2, 4, 6, 8, 10}
n(A) = 5
and B = the number on the drawn card is more than 3
So, B = {4, 5, 6, 7, 8, 9, 10}
n(B) = 7
Now, A ∩ B = {4, 6, 8, 10}
n(A ∩ B) = 4
Thus, the required probability is –
P(A/B) = n(A ∩ B)/ n(B) = 4/7

问题 2：假设每个出生的孩子都是男孩或女孩的可能性相同。如果一个家庭有两个孩子，假设两个孩子都是女孩的条件概率是多少——

(i) 最小的是女孩。

(ii) 至少一个是女孩。

解决方案：

Let b and g represent the boy and the girl child respectively. If a family has two children, the sample space will be –
S = {(b, b), (b, g), (g, b), (g, g)}
n(S) = 4
Let A be the event that both children are girls.
So, A = {(g, g)}
n(A) = 1

(i) Let B be the event that the youngest child is a girl.
So, B = {(b, g), (g, g)}
n(B) = 2
Now, A ∩ B = {(g, g)}
n(A ∩ B) = 1
So, P(B) = n(B)/ n(S) = 2/4 = 1/2
and P(A ∩ B) = n(A ∩ B)/ n(S) = 1/4

Now, the conditional probability that both are girls, given that the youngest child is a girl, is –
P(A/B) = P(A ∩ B)/ P(B) = (1/4)/ (1/2) = 1/2
Thus, the required probability is 1/2.

(ii) Let C the event that at least one child is a girl.
So, C = {(b, g), (g, b), (g, g)}
n(C) = 3
Now, A ∩ C = {(g, g)}
n(A ∩ C) = 1
So, P(C) = n(C)/ n(S) = 3/4
and P(A ∩ C) = n(A ∩ C)/ n(S) = 1/4

Now, the conditional probability that both are girls, given that the youngest child is a girl, is –
P(A/C) = P(A ∩ C)/ P(C) = (1/4)/ (3/4) = 1/3
Thus, the required probability is 1/3.

问题 3：假设掷两个骰子时出现的两个数字不同。找出事件“骰子上的数字之和为 4”的概率。

解决方案：

Let A be the event of having two different numbers on the dice.
So, A = {(1, 2), (1, 3), (1, 4), (1, 5), (1, 6),(2, 1), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 4), (3, 5), (3, 6),
(4, 1), (4, 2), (4, 3), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5)}
n(A) = 30
and B be the event getting a sum of 4 on the dice.
So, B = {(1, 3), (2, 2), (3, 1)}
n(B) = 3
Now, A ∩ B = {(1, 3), (3, 1)}
n(A ∩ B) = 2
Thus, the required conditional probability is –
P(B/A) = n(A ∩ B)/ n(A) = 2/30 = 1/15

问题 4：抛硬币 3 次，如果前两次出现正面，求第三次出现正面的概率。

解决方案：

Let A be the event of a head appearing on the first two tosses.
So, A = {HHT, HHH}
n(A) = 2
and B be the event of getting a head on the third toss.
So, B = {HHH, HTH, THH, TTH}
n(B) = 4
Now, A ∩ B = {HHH}
n(A ∩ B) = 1
Thus, the required conditional probability is –
P(B/A) = n(A ∩ B)/ n(A) = 1/2

问题 5：掷骰子 3 次，如果给定 6 和 5 分别出现在前两次掷中，求 4 在第三次掷中出现的概率。

解决方案：

Let A be the event of 4 appearing on the third toss, if a die is thrown three times.
So, A = {(1, 1, 4), (1, 2, 4), (1, 3, 4), (1, 4, 4), (1, 5, 4), (1, 6, 4),
(2, 1, 4), (2, 2, 4), (2, 3, 4), (2, 4, 4), (2, 5, 4), (2, 6, 4)
(3, 1, 4), (3, 2, 4), (3, 3, 4), (3, 4, 4), (3, 5, 4), (3, 6, 4)
(4, 1, 4), (4, 2, 4), (4, 3, 4), (4, 4, 4), (4, 5, 4), (4, 6, 4)
(5, 1, 4), (5, 2, 4), (5, 3, 4), (5, 4, 4), (5, 5, 4), (5, 6, 4)
(6, 1, 4), (6, 2, 4), (6, 3, 4), (6, 4, 4), (6, 5, 4), (6, 6, 4)}
n(A) = 36
and B be the event of 6 and 5 appearing respectively on first two tosses, if the die is tossed three times.
So, B = {(6, 5, 1), (6, 5, 2), (6, 5, 3), (6, 5, 4), (6, 5, 5), (6, 5, 6)}
n(B) = 6
Now, A ∩ B = {(6, 5, 4)}
n(A ∩ B) = 1
Thus, the required probability is –
P(A/B) = n(A ∩ B)/ n(B) = 1/6

问题 6：计算 P(A/B)，如果 P(B) = 0.5 且 P(A ∩ B) = 0.32

解决方案：

Given, P(B) = 0.5 and P(A ∩ B) = 0.32
We know that, P(A/B) = P(A ∩ B)/ P(B) = 0.32/ 0.5 = 16/25
Thus, P(A/B) = 16/25

问题 7：如果 P(A) = 0.4，P(B) = 0.3 和 P(B/A) = 0.5，求 P(A ∩ B) 和 P(A/B)。

解决方案：

Given, P(A) = 0.4, P(B) = 0.3 and P(B/A) = 0.5
We know that, 
P(B/A) = P(A ∩ B)/ P(A)
0.5 = P(A ∩ B)/ 0.4
P(A ∩ B) = 0.5 × 0.4 
Thus, P(A ∩ B) = 0.2

Now, P(A/B) = P(A ∩ B)/ P(B)= 0.2/ 0.3
Thus, P(A/B) = 2/3

问题 8：如果 A 和 B 是两个事件，使得 P(A) = 1/3，P(B) = 1/5 和 P(A ∪ B) = 11/30，求 P(A/B) 和 P (B/A)。

解决方案：

Given, P(A) = 1/3, P(B) = 1/5 and P(A ∪ B) = 11/30
We know that,
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
11/30 = 1/3 + 1/5 – P(A ∩ B)
P(A ∩ B) = 1/3 + 1/5 – 11/30
P(A ∩ B) = (10 + 6 – 11)/ 30
P(A ∩ B) = 5/30 = 1/6
Now,
P(A/B) = P(A ∩ B)/ P(B)
P(A/B) = (1/6)/ (1/5)
P(A/B)= 5/6

and P(B/A) = P(A ∩ B)/ P(A)
P(B/A) = (1/6)/ (1/3)
P(B/A) = 3/6 = 1/2

Thus, P(A/B) = 5/6 and P(B/A) = 1/2.

问题 9：一对夫妇有两个孩子。找出两个孩子都是的概率——

(i) 男性，如果知道至少有一个孩子是男性。

(ii) 女性，如果知道大孩子是女性。

解决方案：

Let m and f represent the male and the female child respectively.

(i) Let A be the event that both are males. 
So, A = {(m, m)}
n(A) = 1
and B be the event that at least one is male. 
So, B = {(m, m), (m, f), (f, m)}
n(B) = 3
Now, A ∩ B = {(m, m)}
n(A ∩ B) = 1
Thus, the required probability is –
P(A/B) = n(A ∩ B) / n(B) = 1/3

(ii) Let C be the event that both are females. 
So, C = {(f, f)}
n(C) = 1
and D be the event that elder child is female. 
So, D = {(f, m), (f, f)}
n(D) = 2
Now, C ∩ D = {(f, f)}
n(C ∩ D) = 1
Thus, the required probability is –
P(C/D) = n(C ∩ D) / n(D) = 1/2