问题13：在一个家庭中，丈夫(H)说谎的占30％，妻子(W)说谎的占35％。找出两者在同一事实上矛盾的可能性。
解决方案：
According to question:
It is given that,
In a family, Husband (H) tells a lie in 30% cases and Wife(W) tells a lie in 35% cases.
P(H) = 30/100, P(H‘) = 70/100
P(W) = 35/100, P(W‘) = 65/100
Now,
P(Both contradict on the same fact)
= [P(H ∩ W‘) ∪ P(H‘ ∩ W)]
= P(H ∩ W‘) + P(H‘ ∩ W)
= P(H) × P(W‘) + P(H‘) × P(W)
= 30/100 × 65/100 + 70/100 × 35/100
= 4400/10000 = 0.44
Hence, The required probability is 0.44 i.e., 44%
问题14.一对夫妻出现在同一职位的两个空缺面试中。丈夫选择的概率是1/7，妻子选择的概率是1/5。概率是多少
(a)他们都会被选中吗?
(b)将只选择其中之一?
(c)他们都不会被选中?
解决方案：
According to Question,
It is given that,
P(H) = 1/7 and, P(H‘) = 6/7
P(W) = 1/5 and, P(W‘) = 4/5
(i) P(Both of them will be selected)
= P(H∩W)
= P(H) × P(W)
= 1/7 × 1/5 = 1/35
Hence, The required probability = 1/35.
(ii) P(only one of them will be selected)
= P[(H ∩ W‘) ∪ (H‘ ∩ W)]
= P(H ∩ W‘) + P(H‘ ∩ W)
= P(H) × P(W‘) + P(H‘) × P(W)
= 1/7 × 4/5 + 6/7 × 1/5
= 4/35 + 6/35 = 10/35 = 2/7
Hence, The required probability = 2/7
(iii) P(none of them will be selected)
= P(H‘ ∩ W‘)
= P(H‘) × P(W‘)
= 6/7 × 4/5 = 24/35
Hence, The required probability = 24/35
问题15.一个袋子包含7个白色，5个黑色和4个红色的球。抽出四个球而无需更换。找出至少三个球是黑色的概率。
解决方案：
According to question,
It is given that,
A bag contains 7 white, 5 black, and 4 red balls.
And, four balls are drawn without replacement
Now,
P(at least three balls are black)
= P(exactly 3 black balls) + P(all 4 black balls)
= (11/16 × 5/15 × 4/14 × 3/13 × 4) + (5/16 × 4/15 × 3/14 × 2/13)
= 11/ (13 × 14) + 1/ (2 × 13× 14)
= 23/364
Hence, the required probability = 23/364
问题16. A，B和C是已知已发生事件的独立见证人。 A在四分之三的时间内讲真话，B在五分之四的讲真理，而C在六分之五的讲真理。三位证人中的大多数证人将真实地举报该事件的概率是多少?
解决方案：
According to question,
It is given that,
P(A speaks truth) = 3/4 and, P(A‘) = 1/4
P(B speaks truth) = 4/5 and, P(B‘) = 1/5
P(C speaks truth) = 5/6, and P(C‘) = 1/6
Now,
P(Reported truthfully by the majority of three witnesses)
= P(A) × P(B) × P(C‘) + P(A) × P(B‘) × P(C) + P(A‘) × P(B) × P(C)
= 3/4 × 4/5 × 1/6 + 3/4 × 1/5 × 5/6 + 1/4 × 4/5 × 5/6
= 107/120
Hence, the required probability = 107/120
问题17：一个袋子包含4个白球和2个黑球。另一个包含3个白球和5个黑球。如果从每个袋子中抽出一个球，求出
(i)都是白色的
(ii)都是黑色的
(iii)一种是白色，一种是黑色
解决方案：
According to question,
It is given that,
Bag 1 contains 4 white balls and 2 black balls.
Bag 2 contains 3 white balls and 5 black balls.
(i) P(Both are white)
= 4/6 × 3/8 = 12/48
= 1/4
Required probability = 1/4
(ii) P(Both are black)
= 2/6 × 5/8 = 10/48
= 5/24
Required probability = 5/24
(iii) P(One is white and one is black)
= 4/6 × 5/8 + 3/8 × 2/6
= 20/48 + 6/ 48
= 13/24
Required probability = 13/24
问题18.一个袋子包含4个白色，7个黑色和5个红色的球。抽出4个球进行替换。至少有两个是白人的概率是多少?
解决方案：
According to question,
It is given that,
A bag contains 4 white, 7 black, and 5 red balls.
Now,
P(at least two are white)
= 1 – P(Maximum 1 white ball)
= 1 – [P(no white) + P(exactly 1 white ball)]
= 1 – [12/16 × 12/16 × 12/16 × 12/16 + 4/16 × 12/16 × 12/16 × 12/16 × 4]
= 1 – [81/256 + 108/256]
= 1 – 189/256 = 67/256
Hence, the required probability = 67/256
问题19.从一张洗净的卡包中抽出三张卡，并进行替换。找出这些牌是国王，王后和杰克的概率。
解决方案：
According to question,
It is given that,
Three cards are drawn with replacement from a well-shuffled pack of cards.
Now,
P(King) = P(A) = 4/52
P(Queen) = P(B) = 4/52
P(Jack) = P(C) = 4/52
Now,
P(King, Queen and a Jack)
= 3! × P(A) × P(B) × P(C)
= 3 × 2 × 4/52 × 4/52 × 4/52
= 6/2197
Hence, the required probability = 6/2197.
问题20.一个袋子包含4个红色和5个黑色的球，第二个袋子包含3个红色和7个黑色的球。每个袋子随机抽出一个球；找出(i)球具有不同颜色(ii)球具有相同颜色的可能性。
解决方案：
According to question,
It is given that,
A bag contains 4 red and 5 black balls and,
a second bag contains 3 red and 7 black balls.
(i) P(balls are of different colors)
= P[(R1 ∩ B2) ∪ (B1 ∩ R2)]
= P(R1 ∩ B2) + P(B1 ∩ R2)
= P(R1) × P(B2) + P(B1) × P(R2)
= 4/9 × 7/10 + 5/9 × 3/10
= 28/90 + 15/90 = 43/90
Required probability = 43/90
(ii) P(balls are of the same color)
= P[(B1 ∩ B2) ∪ (R1 ∩ R2)]
= P(B1 ∩ B2) + P(R1 ∩ R2)
= P(B1) × P(B2) + P(R1) × P(R2)
= 5/9 × 7/10 + 4/9 × 3/10
= 47/90
Required probability = 47/90
问题21. A可以6击3次击中目标，B：6击2次，C：4击4次。他们修理齐射。命中至少2发的概率是多少?
解决方案：
According to question:
It is given that,
P(A hits a target) = 3/6 = 1/2, and, P(A‘) = 1/2
P(B hits a target) = 2/6 = 1/3 and, P(B‘) = 2/3
P(C hits a target) = 4/4 = 1
Now, we have to find that,
P(At least 2 shots hit) = P(Exactly two shot hit) + P(all three shot hit)
= 1/2 × 2/3 × 1 + 1/2 × 1/3 × 1 + 1/2 × 1/3 × (1 – 1) + 1/2 × 1/3 × 1
= 2/6 + 1/6 + 1/6 = 4/6 = 2/3
Hence, The required probability = 2/3
问题22：学生A通过考试的概率为2/9，学生B通过考试的概率为5/9。假设两个事件：“ A通过”，“ B通过独立”，找到(i)只有A通过考试(ii)只有其中一个通过考试的概率。
解决方案：
According to question,
It is given that,
P(A passing the examination) = 2/9, P(A) = 2/9 and, P(A‘) = 7/9
P(B passing the examination) = 5/9, P(B) = 5/9 and, P(B‘) = 4/9
Now,
(i) P(Only A passing the examination)
= P(A ∩ B‘)
= P(A) × P(B‘)
= 2/9 × 4/9 = 8/81
Required probability = 8/81
(ii) P(Only one of them passing the examination)
= P[(A ∩ B‘) ∪ (A‘ ∩ B)]
= P(A ∩ B’) + P(A‘ ∩ B)
= P(A)× P(B‘) + P(A‘) × P(B)
= 2/9 × 4/9 + 7/9 × 5/9
= 8/81 + 35/81 = 43/81
Required probability = 43/81
问题23.有三个骨灰盒A，B和C。骨灰盒A包含4个红球和3个黑球。 n B包含5个红球和4个黑球。 C包含4个红色和4个黑色的球。从这些骨灰盒中抽出一个球。 3个抽奖球由2个红球和1个黑球组成的概率是多少?
解决方案：
According to question,
It is given that,
Urn A contains 4 red balls and 3 black balls.
Urn B contains 5 red balls and 4 black balls.
Urn C contains 4 red and 4 black balls.
Now,
P(3 balls drawn consist of 2 red balls and a black ball)
= P(Black from urn A) + P(Black from urn B) + P(Black from Urn C)
= 3/7 × 5/9 × 4/8 + 4/7 × 4/9 × 4/8 + 4/7 × 5/9 × 4/8
= 204/504 = 17/42
Required probability = 17/42