问题1.一个袋子包含6个黑球和3个白球。另一个袋子包含5个黑球和4个白球。如果从每个袋子中抽出一个球，请找出这两个球颜色相同的可能性。
解决方案：
According to Question:
It is given that,
Bag 1 contains 6 black and 3 white balls.
Bag 2 contains 5 black and 4 white balls.
Now,
One ball is drawn from each bag
Then, P(one black ball from bag 1) = 6/9 and, P(one white ball from bag 1) = 3/9
P(one black ball from bag 2) = 5/9 and, P(one white ball from bag 2) = 4/9
Now,
P(Two balls are of same color) = P(Both are black) + P(Both are white)
= 6/9 × 5/9 + 3/9 × 4/9
= 30/81 + 12/81 = 42/81
= 14/27
Hence, The required probability = 14/27
问题2.一个袋子包含3个红色和5个黑色的球，另一个袋子包含6个红色和4个黑色的球。从每个袋子中抽出一个球。找出一个是红色而另一个是黑色的概率。
解决方案：
According to Question:
It is given that,
Bag 1 contains 3 red and 5 black balls.
Bag 2 contains 6 red and 4 black balls.
Now,
One ball is drawn from each bag
Then, P(one red ball from bag 1), P(R1) = 3/8 and, P(one black ball from bag 1), P(B1) = 5/8
P(one red ball from bag 2), P(R2) = 6/10 and, P(one black ball from bag 2), P(B2) = 4/10
Now,
We have to find that,
P(one is red and other is black)
= P(R1 ∩ B2) ∪ P(B1 ∩ R2)
= P(R1) × P(B2) + P(B1) × P(R2)
= 3/8 × 4/10 + 5/8 × 6/10 = 12/80 + 30/80 = 42/80
= 21/40
Hence, The required probability = 21/40
问题3.从装有10个黑色和8个红色球的盒子中随机抽取两个球。找出
(i)两个球均为红色(ii)第一个球为黑色，第二个球为红色。 (iii)其中一个是黑色，另一个是红色。
解决方案：
According to question:
It is given that,
A box contain 10 black and 8 red balls. And two balls are drawn at random with replacement.
Now,
(i) P(Both the balls are red)
= P(R1 ∩ R2)
= P(R1) × P(R2)
= 8/18 × 8/18 = 64/324
= 16/81
Required Probability = 16/81
(ii) P(The first ball is black and the second ball is red)
= P(B ∩ R)
= P(B) × P(R)
= 10/18 × 8/18 = 80/324
= 20/81
Required Probability = 20/81
(iii) P(One of them is black and the other is red)
= P((B ∩ R)∪ (R ∩ B))
= P(B ∩ R) + P(R ∩ B)
= P(B) × P(R) + P(R) × P(B)
= 10/18 × 8/18 + 8/18 × 10/18
= 20/81 + 20/81
= 40/81
Hence, The required probability = 40/81
问题4.连续抽出两张纸牌，而没有从洗净的纸牌堆中进行更换。找到恰好一张王牌的概率。
解决方案：
According to question,
It is given that,
Two cards are drawn successively without replacement. In a well – shuffled deck of cards there are total 4 ace.
Now,
P(Exactly one ace) = P(first card is ace) + P(Second card is ace)
= 4/52 × 48/51 + 48/52 × 4/51
= 96/663
= 32/221
Hence, The required probability = 32/221
问题5：在75％的案例中，A讲真话，而B在80％的情况下讲真话。在叙述同一事件的过程中，他们在多大比例的案例中可能会相互矛盾?
解决方案：
According to question:
It is given that,
A speaks truth in 75% cases.
B speaks truth in 80% cases.
Now, P(A) = 75/100 = 3/4 and, P(A‘) = 1 – 75/100 = 25/100 = 1/4
P(B) = 80/100 = 4/5 and, P(B‘) = 1 – 80/100 = 20/100 = 1/5
Now,
P(A and B contradict each other)
= P(A ∩ B‘) + P(A‘ ∩ B)
= P(A) × P(B‘) + P(A‘) × P(B)
= 3/4 × 1/5 + 1/4 × 4/5
= 3/20 + 4/20 = 7/20
= 0.35
= 35 %
Hence, The required probability = 35 %.
问题6。卡玛勒和莫妮卡出现了两个空缺的面试。卡马尔(Kamal)的选择概率为1/3，莫妮卡(Monika)的选择概率为1/5。找出
(i)他们都将被选中(ii)他们都不将被选中
(iii)将至少选择其中之一(iv)将仅选择其中之一。
解决方案：
According to question:
It is given that,
P(K) = 1/3 and, P(M) = 1/5
Now,
(i) P(Both of them are selected)
= P(K ∩ M) = P(K) × P(M)
= 1/3 × 1/5 = 1/15
The required probability = 1/15
(ii) P(none of them will be selected)
= P(K‘ ∩ M‘) = P(K‘) × P(M‘)
= 1 – 1/3 × 1 – 1/5 = 2/3 × 4/5
= 8/15
The required probability = 8/15
(iii) P(at least one of them will be selected)
= 1 – P(None of them is selected)
= 1 – 8/15 [From eq(ii)]
= 7/15
The required probability = 7/15
(iv) P(only one of them will be selected)
= P(K ∩ M‘) + P(K‘ ∩ M)
= P(K) × P(M‘) + P(K‘) × P(M)
= 1/3 × 4/5 + 2/3 × 1/5
= 4/15 + 2/15 = 6/15 = 2/5
Hence, The required probability = 2/5
问题7.一个袋子包含3个白色，4个红色和5个黑色的球。两个球一个接一个地抽出，无需更换。一个是白色而另一个是黑色的概率是多少?
解决方案：
According to question:
It is given that,
A bag contains 3 white, 4 red, and 5 black balls. And Two balls are
drawn one after the other, without replacement.
Now,
P(One is white and other is black)
= P((W ∩ B) ∪ (B ∩ W))
= P(W ∩ B) + P(B ∩ W)
= P(W) × P(B/W) + P(B) × P(W/B)
= 3/12 × 5/11 + 5/12 × 3/11
= 15/132 + 15/132
= 30/132 = 5/22
Hence, The required probability = 5/22.
问题8.一个袋子包含8个红色和6个绿色的球。三个球被一个接一个地抽出而无需更换。找出至少两个被抽成绿色的概率。
解决方案：
According to question:
It is given that,
A bag contains 8 red and 6 green balls. And Three balls are
drawn one after another without replacement.
Now,
P(at least two balls drawn are green)
= 1 – P(at most one ball is green)
= 1 – [P(first ball is green) + P(Second ball is green) + P(Third ball is green) + P(No green)]
= 1 – [6/14 × 8/13 × 7/12 + 8/14 × 6/13 × 7/12 + 8/14 × 7/13 × 6/12 + 8/14 × 7/13 × 6/12]
= 1 – [336/2184 + 336/2184 + 336/2184 + 336/2184]
= 1 – 1344/2184 = 840/2184
= 5/13
Hence, the required probability = 5/13
问题9。阿伦和塔伦出席了两个空缺的面试。阿伦被选中的概率为1/4，而塔伦被拒绝的概率为2/3。找出至少被选中的概率。
解决方案：
According to question:
It is given that,
P(Arun get selected), P(A) = 1/4 and, P(Arun get rejected), P(A‘) = 3/4
P(Tarun get rejected), P(T‘) = 2/3
P(Tarun get selected), P(T) = 1/3
Now,
P(at least one of them is selected)
= 1 – P(none of them is selected)
= 1 – P(A‘ ∩ T‘)
= 1 – P(A‘) × P(T‘)
= 1 – 3/4 × 2/3
= 1 – 6/12 = 1 – 1/2
= 1/2
Hence, The required probability is 1/2.
问题10：A和B交替投掷硬币，直到其中一个获得正面并赢得比赛为止；如果A开始游戏，则求出B赢得比赛的概率。
解决方案：
According to question:
Let E be the event occurring head.
P(E) = 1/2 and P(E‘) = 1/2
A wins the game in first, third and fifth throw,
P(A wins in first throw) = P(E) = 1/2
P(A wins in third throw) = P(E‘) × P(E‘) × P(E) = 1/2 × 1/2 × 1/2 = (1/2)3
Similarly, P(A wins in fifth throw) = (1/2)5
Now,
P(Wining of A)
= 1/2 + (1/2)3 + (1/2)5 + …
= 1/2 [1 + (1/2)2 + (1/2)4 + …]
= 1/2 [1/ (1 – (1/2)2] [Since, Sum of infinite term of g.p = a/1 – r]
= 1/2 [1/ 1 – 1/4]
= 1/2 × 4/3 = 2/3
Now, P(B wins) = 1 – P(A wins)
= 1 – 2/3 = 1/3
Hence, The required probability = 1/3
问题11.两张纸牌是从一组52张纸牌中随机抽取的，一张一张接一张，不用更换。找出其中一个是红牌而另一个是黑牌的概率吗?
解决方案：
According to question:
It is given that,
Two cards are drawn from a well-shuffled pack of 52 cards,
one after another without replacement.
There are 26 red and 26 black cards.
Now, We have to find that,
P(one red and other black card)
= P[(R ∩ B) ∪ (B ∩ R)]
= P(R ∩ B) + P(B ∩ R)
= P(R) × P(B/R) + P(B) × P(R/B)
= 26/52 × 26/51 + 26/52 × 26/51
= 26/51
Hence, the required probability 26/51.
问题12.彩票的编号从1到10。随机抽取两张彩票。求出一张票证上的数字是5的倍数，而另一张票证上的4的倍数的概率。
解决方案：
According to question:
It is given that,
Tickets are numbered from 1 to 10. And, Two tickets are drawn at random.
Now, Let us consider,
A = Ticket is a multiple of 5.
B = Ticket is a multiple of 4.
Since 5 and 10 are multiple of 5 . So, P(A) = 2/10 = 1/5.
and , 4, 8 are the multiple of 4. So, P(B) = 2/10 = 1/5.
Now, we have to find that,
P(one number is multiple of 5 and other is the multiple of 4)
= P[(A∩B) ∪ (B ∩ A)]
= P(A∩B) + P(B ∩ A)
= P(A)×P(B/A) + P(B) × P(A/B)
= 1/5 × 2/9 + 1/5 × 2/9
= 4/45
Hence, the required probability 4/45.