第 12 类 RD Sharma 解决方案 - 第 31 章概率 - 练习 31.4 |设置 1

问题 1(i)。一枚硬币被抛三次，所有八种结果的可能性都相同。说明事件 A 和 B 是否独立如果，A = 第一次抛出的结果在头部，B = 最后一次抛出的结果在尾部

解决方案：

According to question:

A coin is tossed thrice 

So, Sample Space = {HTT, HHT, HTH, HHH, THT, THH, TTH, TTT}

Now, 

A = The first throw result in head 

A = {HHT, HTH, HHH, HTT}

Now, B = The last throw result in tail

B = {HHT, HTT, THT, TTT}

A ∩ B = {HHT, HTT}

P(A) = 4 / 8 = 1/2

Similarly,

P(B) = 1/2

Now,

P(A ∩ B) = 2/ 8 = 1/ 4

P(A), P(B) = 1/ 2, 1/ 2 

P(A) × P(B) = 1/4

As we know that P(A ∩ B) = P(A)×P(B)

So, A and B are independent events.

问题 1(ii)。一枚硬币被抛三次，所有八种结果的可能性都相同。说明事件 A 和 B 是否独立如果，A = 正面数为奇数，B = 反面数为奇数

解决方案：

According to question:

A coin is tossed thrice

So, Sample Space = {HTT, HHT, HTH, HHH, THT, THH, TTH, TTT}

A = the number of head is odd, B = the number of tails is odd

So, A = {HTT, THT, TTH, HHH}

B = {THH, HTH, HHT, TTT}

Here, A ∩ B = {} = ∅ 

P(A) = 4/8 = 1/2

P(B) = 4/8 = 1/2

And, P(A ∩ B) = 0/8 = 0

Now, P(A) × P(B) = 1/2 × 1/2 = 1/4

So, we can see that 

P(A) × P(B) ≠ P(A ∩ B)

Hence, A and B are not independent events.

问题 1(iii)。一枚硬币被抛三次，所有八种结果的可能性都相同。说明事件 A 和 B 是否独立如果，A = 正面二的数量，B = 最后一次掷出正面的结果

解决方案：

According to question:

A coin is tossed thrice

So, Sample Space = {HTT, HHT, HTH, HHH, THT, THH, TTH, TTT}

Now,

A = The number of head two

A = {HHT, HTH, THH}

Now, B = The last throw results in head

B = {HHH, HTH, THH, TTH}

A ∩ B = {THH, HTH}

P(A) = 3/8 

P(B) = 4/8 = 1/2

And, P(A ∩ B) = 2/8 = 1/4

Now, P(A) × P(B) = 3/8 ×1/2 = 3/16

So, P(A) × P(B) ≠ P(A ∩ B)

Hence, A and B are not independent events.

问题 2. 证明在掷一对骰子时，第一个骰子上出现的数字 4 与第二个骰子上出现的数字 5 无关。

解决方案：

According to question:

A pair of dice are thrown. So, it has 36 elements in its sample space.

A = Occurrence of number 4 on the first die.

P(A) = {(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)}

B = Occurrence of 5 on the second die.

P(B) = {(1, 5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 5)}

A ∩ B = {(4, 5)}

Now, P(A) = 6/36 = 1/6, P(B) = 6/36 = 1/6 and P(A ∩ B) = 1/36

P(A) × P(B) = 1/6 × 1/6 = 1/36, which is equal to P(A ∩ B).

Hence, A and B are independent events.

问题 3(i)。从一包 52 张牌中抽取一张牌，这样每张牌被选中的可能性相同。说明事件 A 和 B 是否独立，如果，

A = 抽到的牌是国王或王后。

B = 抽出的牌是皇后或杰克。

解决方案：

According to question:

A card is drawn form 52 cards. We know that it has 4 Kings, 4 Queen, 4 Jack.

Now, A = The card drawn is a king or queen.

So, P(A) = (4 + 4)/52 = 8/52 = 2/13

B = The card drawn is a queen or a jack.

So, P(B) = (4 + 4)/52 = 8/52 = 2/13

Now, A ∩ B = The drawn card is queen (queen is common in both)

P(A ∩ B) = 4/52 = 1/ 13

Now, P(A) × P(B) = 2/13 × 2/13 = 4/169

We can see that the above expression is not equal to P(A ∩ B). 

So, A and B are not independent events.

问题 3(ii)。从一包 52 张牌中抽取一张牌，这样每张牌被选中的可能性相同。说明事件 A 和 B 是否独立，如果，

A = 抽到的牌是黑色的。

B = 抽到的牌是国王。

解决方案：

According to question:

A card is drawn form 52 cards. We know that there are 26 black cards 

in which 2 kings are black.

Now, A = The card drawn is black.

So, P(A) = 26/52 = 1/2

B = The card drawn is a king.

So, P(B) = 4/52 = 1/13

Now, A ∩ B = The drawn card is a black king

P(A ∩ B) = 2/52 = 1/ 26

Now, P(A) × P(B) = 1/2 × 1/13 = 1/26

P(A) × P(B) = P(A ∩ B) 

So, A and B are independent events.

问题 3(iii)。从一包 52 张牌中抽取一张牌，这样每张牌被选中的可能性相同。说明事件 A 和 B 是否独立，如果，

A = 抽到的牌是黑桃。

B = 抽出的牌是一张 A。

解决方案：

According to question:

A card is drawn form 52 cards. We know that there are 13 spades and 

4 Ace in which 1 card is ace of spade.

Now, A = The card drawn is a spade.

So, P(A) = 13/52 = 1/4

B = The card drawn is an ace.

So, P(B) = 4/52 = 1/13

Now, A ∩ B = The drawn card is an ace of spade.

P(A ∩ B) = 1/52 = 1/ 52

Now, P(A) × P(B) = 1/4 × 1/13 = 1/52

P(A) × P(B) = P(A ∩ B)

So, A and B are independent events.

问题 4：一枚硬币被抛 3 次。让事件 A、B 和 C 定义如下：

A = 第一次抛是正面，B = 第二次抛是正面，C = 恰好两个正面连续抛。

检查 (i) A 和 B (ii) B 和 C (iii) C 和 A 的独立性

解决方案：

According to question:

A coin is tossed thrice

So, Sample Space = {HTT, HHT, HTH, HHH, THT, THH, TTH, TTT}

A = first toss is head

A = {HHH, HHT, HTH, HTT} 

So, P(A) = 4/8 = 1/2

B = second toss is head

B = {HHH, HHT, THH,THT}

So, P(B) = 4/8 = 1/2

C = exactly two head in a row, i.e., P(C) = {THH, HHT}

So, P(C) = 2/8 = 1/4

Now, A ∩ B = {HHH, HHT}, i.e., P(A ∩ B) = 2/8 = 1/4

Now, B ∩ C = {HHT, THH}, i.e., P(B ∩ C) = 2/8 = 1/4

And, A ∩ C = {HHT}, i.e., P(A ∩ C) = 1/8

(i) P(A) × P(B) = 1/2 × 1/2 = 1/4 

So, P(A) × P(B) = P(A ∩ B)

Hence, A and B are independent events.

(ii) P(B) × P(C) = 1/2 × 1/4 = 1/8

So, P(B)×P(C) ≠ P(B ∩ C)

Hence, B and C are not independent events.

(iii) P(A) × P(C) = 1/2 × 1/4 = 1/8

So, P(A) × P(C) = P(A ∩ C)

Hence, A and C are independent events.

问题 5. 如果 A 和 B 是两个事件，使得 P(A) = 1/4，P(B) = 1/3，并且 P(A ∪ B) = 1/2，证明 A 和 B 是独立事件.

解决方案：

According to question:

It is given that 

P(A) = 1/4, P(B) = 1/3 and P(A ∪ B) = 1/2

As we know that,

P(A ∩ B) = P(A) + P(B) – P(A ∪ B)

= 1/4 + 1/3 – 1/2

= 1/12

Now, P(A) × P(B) = 1/4 × 1/3 = 1/12

So, P(A) × P(B) = P(A ∩ B)

Hence, A and B are independent events.

问题 6. 给定两个独立事件 A 和 B，使得 P(A) = 0.3 和 P(B) = 0.6，求

(i) P(A ∩ B) (ii) P(A ∩ B') (iii) P(A' ∩ B)

(iv) P(A' ∩ B') (v) P(A ∪ B)

(vi) P(A ⁄ B) (vii) P(B ⁄ A)

解决方案：

According to question:

It is given that A and B are independent events and P(A) = 0.3, P(B) = 0.6

(i) Since, A and B are independents events so,

 P(A ∩ B) = P(A) × P(B)

= 0.3 × 0.6 = 0.18

(ii) P(A ∩ B‘)= P(A) – P(A ∩ B)

= 0.3 – 0.18 = 0.12

(iii) P(A‘ ∩ B)= P(B) – P(A ∩ B)

= 0.6 – 0.18 = 0.42

(iv) P(A‘ ∩ B‘) = P(A‘) × P(B‘)

= [1 – P(A)] × [1 – P(B)]

= [1 – 0.3] × [1- 0.6]

= 0.7 × 0.4 = 0.28

(v) P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

= 0.3 + 0.6 – 0.18 = 0.72

(vi) P(A ⁄ B) = P(A ∩ B) / P(B)

= 0.18/ 0.6 = 0.3

(vii) P(B ⁄ A) = P(A ∩ B) / P(A)

= 0.18/ 0.3 = 0.6

问题 7：如果 P(not B) = 0.65，P(A ∪ B) = 0.85，且 A 和 B 是独立事件，则求 P(A)。

解决方案：

According to question:

It is given that P(not B) = 0.65, P(A ∪ B) = 0.85

We know that, P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

Since, A and B are independent

So, P(A ∩ B) = P(A) × P(B)

Also, P(not B) = 0.65,  

So, P(B) = 0.35            -(Since P(not B) = 1 – P(B))

Hence, We have

⇒ 0.85 = P(A) + 0.35 – P(A) × (0.35)

⇒ 0.5 = P(A)[1-0.35]

⇒ 0.5/ 0.65 = P(A)

So, P(A) = 0.77

问题 8. 如果 A 和 B 是两个独立事件，使得 P(A' ∩ B) = 2/15 且 P(A ∩ B') = 1/6，则求 P(B)。

解决方案：

According to question:

It is given that P(A‘ ∩ B) = 2/15 and P(A ∩ B‘) = 1/6

Since, A and B are independent,

So, P(A‘) × P(B) = 2/15 ⇒ [1 – P(A)] P(B) = 2/15         -(1)

and P(A) × P(B‘) = 1/6 ⇒ P(A) [1 – P(B)] = 1/6         -(2)

From eq(1), We get 

P(B) = 2/15 × 1/(1 – P(A))

On substituting the above value of eq(1) in the eq(2) we get,

P(A)[1 – 2/(15(1-P(A)))] = 1/6

  ⇒ P(A)[15(1 – P(A)) – 2] / [15(1 – P(A))] = 1/6

 ⇒ 6P(A)(13 – 15P(A)) = 15(1 – P(A))

 ⇒ 2P(A)(13 – 15P(A)) = 5 – 5P(A)

 ⇒ 26P(A) – 30 [P(A)]² + 5P(A) – 5 = 0

 ⇒ -30 [P(A)]² + 31 P(A) – 5 = 0         -(3)

This is the form of quadratic equation replace P(A) by x in eq(3)

-30x² + 31x – 5 = 0

30x² – 31x + 5 = 0

So, x = -b ± √b² – 4ac / 2a 

Where, a = 30, b= -31 and c = 5

⇒ x = 31 ± √(-31)² – 4(30)(5) / 60

= 31 ± √961 – 960 / 60

= 30 ± 19 / 60

= 50/60, 12/60 = 5/6, 1/5 

So, P(A) = 5/6 or 1/5

Now, P(A) [1-P(B)] = 1/6

On putting, P(A) = 5/6, we get

5/6[1 – P(B)] = 1/6

⇒ 1 – P(B) = 1/5

⇒ P(B) = 1 – 1/5 = 4/5

Now, putting, P(A) = 1/5, we get

1/5[1 – P(B)] = 1/6

⇒ 1 – P(B) = 5/6

⇒ P(B) = 1 – 5/6 = 1/6

Hence, P(B) = 4/5 or 1/6      

问题 9. A 和 B 是两个独立的事件。 A 和 B 发生的概率是 1/6，两者都不发生的概率是 1/3。求两个事件发生的概率。

解决方案：

According to question:

It is given that P(A ∩ B) = 1/6, P(A‘ ∩ B‘) = 1/ 3

We know that

P(A‘ ∩ B‘) = P(A‘) × P(B‘)

1/3 = (1 – P(A))(1 – P(B))

1/3 = 1 – P(B) – P(A) + P(A) P(B)

1/3 = 1 – P(B) – P(A) + P(A ∩ B)

1/3 = 1 – P(B) – P(A) + 1/6

Now, P(A) + P(B) = 1+ 1/6 – 1/3 = 5/6

So, P(A) = 5/6 – P(B)         -(1)

Now, Given that, P(A ∩ B) = 1/6

⇒ P(A) P(B) = 1/6

⇒ [5/6 – P(B)]P(B) = 1/6          -(From eq(1)) 

⇒ 5/6 P(B) – {P(B)}² = 1/6

⇒ {P(B)}² – 5/6 P(B) + 1/6 = 0

⇒ 6 {P(B)}² – 5 P(B) + 1 = 0

After solving this quadratic equation, we get,

⇒ [2P(B) – 1][3P(B) – 1] = 0

⇒ 2 P(B) -1 = 0 or 3P(B) -1 = 0 

⇒ P(B) = 1/2 or P(B) = 1/3

Now,

Using eq(1), 

Putting P(B) = 1/2, ⇒ P(A) = 5/6 – 1/2 = 1/3

Putting P(B) = 1/3, ⇒ P(A) = 5/6 – 1/3 = 1/2

Hence, P(A) = 1/3, P(B) = 1/2 or P(A) = 1/2, P(B) = 1/3.

问题 10. 如果 A 和 B 是两个独立事件，使得 P(A ∪ B) = 0.60 且 P(A) = 0.2，求 P(B)。

解决方案：

According to question:

It is given that, A and B are independent events and P(A ∪ B) = 0.60, P(A) = 0.2, 

where A and B are independent events. 

So, P(A ∩ B) = P(A) × P(B)

Now, we know that, P(A∪ B) = P(A) + P(B) -P(A∩B)

⇒ 0.6 = 0.2 + P(B) – P(A) × P(B)

⇒ 0.6 – 0.2 = P(B) – 0.2×P(B)          -(Since, P(A) = 0.2)

⇒ 0.4 = 0.8 P(B)

⇒ P(B) = 0.4/0.8 = 0.5

问题 11. 掷骰子两次。求每次投掷得到大于 3 的数字的概率。

解决方案：

According to question

A dice is tossed twice 

Let us consider the events:

A = Getting a number greater than 3 on first toss

B = Getting a number greater than 3 on second toss

Since, number greater than3 on die are 4, 5, 6

Now, P(A) = 3/6 = 1/2

And, P(B) = 3/6 = 1/2

Now, P(Getting a number greater than 3 on each toss)

= P(A ∩ B) 

= P(A) P(B)

= 1/2 × 1/2 = 1/4

Hence, the required probability = 1/4

问题 12. 假设 A 可以解决问题的概率是 2/3，B 可以解决相同问题的概率是 3/5。找出两者都不能解决问题的概率。

解决方案：

According to question,

It is given that, Probability that A can solve a problem = 2/3

⇒ P(A) = 2/3

⇒ P(A‘) = 1 – 2/3 = 1/3

Now, Probability that B can solve the same problem = 3/5

 ⇒ P(B) = 3/5

⇒ P(B‘) = 1 – 3/5 = 2/5

Now we have the find the probability that none of them solve the problem

Now, P(None of them solve the problem)

= P(A‘ ∩ B‘) = P(A‘) × P(B‘)

⇒ 1/3 × 2/5 = 2/15

Hence, The required probability = 2/15