第 12 类 RD Sharma 解决方案 - 第 31 章概率 - 练习 31.3 |设置 1
问题 1. 如果 P(A) = 7 /13,P(B) = 9/13 且 P(A ∩ B) = 4/13,求 P(A/B)。
解决方案:
Given: P(A) = 7/13, P(B) = 9/13, and P(A∩ B) = 4/13
We know that, P(A/B) = P(A ∩ B)/P(B)
= (4/13) ÷ (9/13)
= 4/9
问题 2. 如果 A 和 B 是 P(A) = 0.6、P(B) = 0.3 和 P(A ∩ B) = 0.2 的事件,求 P(A/B) 和 P(B/A)。
解决方案:
Given: P(A) = 0.6, P(B) = 0.3, P(A ∩ B) = 0.2
We know that P(A/B) = P(A ∩ B)/P(B)
= 0.2/0.3
= 2/3
and P(B/A) = P(A ∩ B)/P(A)
= 0.2/0.6
= 1/3
问题 3. 如果 A 和 B 是两个事件,使得 P(A ∩ B) = 0.32 和 P(B) = 0.5,求 P(A/B)。
解决方案:
Given: P(A ∩ B) = 0.32 and P(B) = 0.5
We know that P(A/B) = P(A ∩ B)/P(B)
= 0.32/0.5
= 0.64
问题 4. 如果 P(A) = 0.4,P(B) = 0.8,P(B/A) = 0.6,求 P(A/B) 和 P(A ∪ B)。
解决方案:
Given: P(A) = 0.4, P(B) = 0.8, P(B/A) = 0.6
We know that, P(B/A) = P(A ∩ B)/P(A)
P(A ∩ B) = P(B/A) × P(A)
P(A ∩ B) = 0.6 × 0.4
P(A ∩ B) = 0.24
Therefore, P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= 0.4 + 0.8 – 0.24
= 0.96
And, P(A/B) = P(A ∩ B)/P(B)
= 0.24/0.8
= 0.3
问题 5. 如果 A 和 B 是两个事件,使得
(i) P(A) = 1/3, P(B) = 1/4, P(A ∪ B) = 5/12,求P(A/B)和P(B/A)。
解决方案:
We know that, P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
P(A ∩ B) = P(A) + P(B) – P(A ∪ B)
P(A ∩ B) = 1/3 + 1/4 – 5/12 = 2/12
Therefore, P(A/B) = P(A ∩ B)/P(B) = (2/12) ÷ (1/4) = 2/3
and, P(B/A) = P(A ∩ B)/P(A) = (2/12) ÷ (1/3) = 1/2
(ii) P(A) = 6/11, P(B) = 5/11 和 P(A ∪ B) = 7/11, 求 P(A ∩ B), P(A/B) 和 P(B /一种)。
解决方案:
We know that, P(A∪ B) = P(A) + P(B) – P(A ∩ B)
P(A ∩ B)=P(A) + P(B) – P(A ∪ B)
P(A ∩ B) = 6/11 + 5/11 – 7/11
= 4/11
Therefore, P(A/B) = P(A ∩ B)/P(B)
= (4/11) ÷ (5/11) = 4/5
and, P(B/A) = P(A ∩ B)/P(A)
= (4/11) ÷ (6/11) = 2/3
(iii) P(A) = 7/13, P(B) = 9/13 和 P(A ∩ B) = 4/13,求 P(A'/B)。
解决方案:
We know that, P(A’ ∩ B) = P(B) – P(A ∩ B)
= 9/13 – 4/13
= 5/13
Therefore, P(A’/B) = P(A’ ∩ B)/P(B)
= (5/13) ÷ (9/13)
= 5/9
(iv) P(A) = 1/2, P(B) = 1/3 和 P(A ∩ B) = 1/4, 求 P(A/B), P(B/A), P(A '/B) 和 P(A'/B')。
解决方案:
P(A/B) = P(A ∩ B)/P(B)
= (1/4) ÷ (1/3)
= 3/4
P(B/A) = P(A ∩ B)/P(A)
= (1/4) ÷ (1/2)
= 1/2
P(A’/B) = (P(B) – P(A ∩ B))/P(B)
= (1/3 – 1/4) ÷ (1/3)
= 1/4
P(A’/B’) = P(A’ ∩ B’)/P(B’)
= (1 – P(A’ ∪ B’))/(P(A) – P(A ∩ B))
= (1 – P(A) – P(B) + P(A ∩ B))/(P(A) – P(A ∩ B))
= (1 – 1/2 – 1/3 + 1/4)/(1/2 – 1/4)
= 5/4
问题 6. 如果 A 和 B 是两个事件,使得 2 × P(A) = P(B) = 5/13 和 P(A/B) = 2/5,求 P(A ∪ B)。
解决方案:
Given, 2 × P(A) = P(B) = 5/13
P(A) = 5/26
P(A/B) = P(A ∩ B)/P(B)
2/5 = P(A ∩ B) ÷ (5/13)
P(A ∩ B) = (2/5) ÷ (5/13)
P(A ∩ B) = 2/13
We know that,
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= 5/26 + 5/13 – 2/13
= 11/26
问题 7. P(A) = 6/11, P(B) = 5/11 和 P(A ∪ B) = 7/11,求
(i) P(A∩B)
解决方案:
We know that, P(A∪ B) = P(A) + P(B) – P(A ∩ B)
P(A ∩ B) = P(A) + P(B) – P(A ∪ B)
= 6/11 + 5/11 – 7/11
= 4/11
(ii) P(A/B)
解决方案:
P(A/B) = P(A ∩ B)/P(B)
= (4/11) ÷ (5/11)
= 4/5
(iii) P(B/A)
解决方案:
P(B/A) = P(A ∩ B)/P(A)
= (4/11) ÷ (6/11)
= 2/3
问题 8. 一个硬币被抛了 3 次。在以下每一项中找到 P(A/B):
(i) A = 第三次抛正面,B = 前两次抛正面
(ii) A = 至少两个正面,B = 最多两个正面
(iii) A = 最多两条尾巴,B = 至少一条尾巴。
解决方案:
Sample space of three coins is given by
{HHH, HTH, THH, TTH, HHT, HTT, THT, TTT}
(i) A = Heads on third toss = {HHH, HTH, THH, TTH}
B = Heads on first two tosses = {HHH, HHT}
P(A ∩ B) = {HHH}
∴ P(A/B) = P(A ∩ B)/P(B) = 1/2
(ii) A = At least two heads = {HHH, HTH, THH, HHT}
B = At most two heads = {HHT, HTH, TTH, HHT, HTT, THT, TTT}
P(A ∩ B) = {HTH, THH, HHT}
∴ P(A/B) = P(A ∩ B)/P(B) = 3/7
(iii) A = At most two tails = {HHH, HTH, THH, TTH, HHT, HTT, THT}
B = At least one tail = {HTH, THH, TTH, HHT, HTT, THT, TTT}
P(A ∩ B) = {HTH, THH, TTH, HHT, HTT, THT}
∴ P(A/B) = P(A ∩ B)/P(B) = 6/7
问题 9. 两个硬币被抛一次。在以下每一项中找到 P(A/B):
(i) A = 尾出现在一枚硬币上,B = 一枚硬币出现正面
(ii) A = 不出现尾巴,B = 不出现头部
解决方案:
Sample space of two coins is given by
{HH, HT, TH, TT}
(i) A = Tail appears on one coin = {HT, TH}
B = One coin shows head = {HH, HT, TH}
P(A ∩ B) = {HT, TH}
∴ P(A/B) = P(A ∩ B)/P(B)
= 2/2
= 1
(ii) A = No tail appears = {HH}
B = No head appears= {TT}
P(A ∩ B) = { }
∴ P(A/B) = P(A ∩ B)/P(B) = 0
问题 10. 掷骰子 3 次。求 P(A/B) 和 P(B/A),如果
A = 4 出现在第三次投掷中,B = 6 和 5 分别出现在前两次投掷中。
解决方案:
A = 4 appears on the third toss = { (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)}
B = 6 and 5 appear respectively on first two tosses
B = {(6, 5, 1), (6, 5, 2), (6, 5, 3), (6, 5, 4), (6, 5, 5), (6, 5, 6)}
(A ∩ B) = {(6, 5, 4)}
∴ P(A/B) = P(A ∩ B)/P(B) = 1/6
∴ P(B/A) = P(A ∩ B)/P(A) = 1/36
问题 11. 母亲、父亲和儿子随机排队拍全家福。如果 A 和 B 是两个事件,由 A = Son 在一端,B = Father 在中间,找到 P(A/B) 和 P(B/A)。
解决方案:
Let Mother = M, Father = F, and son = S
Sample Space = {FMS, FSM, MFS, MSF, SFM, SMF}
A = Son on one end = {FMS, MFS, SFM, SMF}
B = Father in the middle = {MFS, SFM}
P(A ∩ B) = {MFS, SFM}
∴ P(A/B) = P(A ∩ B)/P(B) = 2/2 = 1
∴ P(B/A) = P(A ∩ B)/P(A) = 2/4 = 1/2
问题 12. 掷骰子两次,观察到出现的数字之和为 6。数字 4 至少出现一次的条件概率是多少?
解决方案:
Let the sample space of the experiment is {(1, 1), (1, 2), (1, 3), . . . .,(6, 6)} consisting of 36 outcomes
P(A) = P(Sum = 6) = 5/36
P(B) = P(4 has appeared at least once) = 11/36
P(B/A) = P(A ∩ B)/P(A) = (2/36) ÷ (5/36) = 2/5
问题 13. 掷了两个骰子。如果已知第二个骰子总是出现 4,则求出现数字总和为 8 的概率。
解决方案:
Two dice are thrown.
A = Sum on the dice is 8
A = {(2, 6), (6, 2), (3, 5), (5, 3), (4, 4)}
B = Second die always exhibits 4
= {(1, 4), (2, 4), (3, 4), (5, 4), (6, 4)}
P(A ∩ B) = {(4, 4)}
P(A/B) = P(A ∩ B)/P(B)
= 1/6
问题 14. 掷出一对骰子。如果已知第二个骰子总是显示奇数,则求总和为 7 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 an odd 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