第 11 类 RD Sharma 解决方案 - 第 33 章概率 - 练习 33.4 |设置 2

问题 15. 从一包 52 张牌中，随机抽取 4 张牌。找出抽出的牌颜色相同的概率？

解决方案：

From a pack of 52 cards, 4 cards are drawn

Hence, Sample space, n(S) = ⁵²C₄          -(1)

Let A be the event of getting cards of same color,

Since there are two sets of same color,

n(A) = 2 * ²⁶C₄           -(2)

P(A) = 2 * ²⁶C₄/⁵²C₄

= 92/833 

Note: The factorial of the respective cases will give a large number, 

           so in such cases simplify the factorial in final step       

问题16 ：两次考试共有100名学生参加，60人通过了第一次，50人通过了第二次考试，30人通过了两次考试。找出随机选择的学生至少通过一次考试的概率？

解决方案：

There are 100 students. Hence, sample space will be –

n(S) = 100          -(1)

Let A be the event that 60 students passed in first exam,

n(A) = 60

= 60/100           -(2)

Let B be the event that 50 students passed in first exam,

n(A) = 50

= 50/100          -(3)

30 passed both of the examinations,

P(A ∩ B) = 30/100          -(4)

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)           

= 60/100 + 50/100 – 30/100          -(From 2, 3, 4)

= 4/5 

问题 17. 一个盒子里有 10 个白球、6 个红球和 10 个黑球。从盒子里随机抽出一个球，球是红色还是白色的概率是多少？

解决方案：

There are 10 white, 6 red and 10 black balls, hence the sample space will be,

n(S) = 10 + 6 + 10

= 26

Let W be the event of drawing the White balls,

n(W) = 10

P(W) = 10/26          -(1)

Let R be the event of drawing the Red balls,

n(R) = 6

P(R) = 6/26          -(2)

E & R are mutually exclusive events –

n(R ∩ E) = 0           -(red and white balls can’t be drawn to be together)

P(E ∪ R) = P(E) + P(R) – P(E ∩ R)

= 10/26 + 6/26

= 16/26

= 8/13

问题 18. 在一场比赛中，马 A、B、C、D 的胜率分别为 1:3、1:4、1:5 和 1:6。找出其中一个赢得比赛的概率？

解决方案：

We have P(A) : $P(\bar{A})$ = 1 : 3

 = P(A) / 1 – P(A)          -($P(\bar{A})$ = 1 – P(A))

 = 1/4          -(1)

Similarly P(B) = 1/5          -(2)

P(C) = 1/6          -(3)

P(D) = 1/7          -(4)

Probability that at least one of the horse wins is P(A ∪ B ∪ C  ∪ D)

= 1/4 + 1/5 + 1/6 + 1/7          -(From 1, 2, 3, 4)

= 319/420

问题 19. 一个人乘火车旅行的概率是 3/5，他乘飞机旅行的概率是 1/4。求他乘火车或飞机旅行的概率？

解决方案：

Let T be the event that persons travels by train –

P(T) = 3/5          -(1)

Let A be the event that persons travels by PLANE –

P(A) = 1/4          -(2)

P(A ∪ T) = P(A) + P(T)          -((A ∩  T) = 0)

= 3/5 + 1/4           -(From 1 , 2)

= 17/20 

问题 20. 从一副洗好的 52 张牌中抽出两张牌。找出两张牌都是黑色或国王的概率？

解决方案：

Two cards are drawn from well shuffled deck of 52 cards,

n(S) = ⁵²C₂          -(1)

Let A be the event of getting black cards,

n(A) = ²⁶C₂

P(A) =²⁶C₂/⁵²C₂           -(2)

Let B be the event of getting KING cards,            

n(B) = ⁴C₂

P(B) = ⁴C₂/⁵²C₂           -(3)

Also, n(A ∩ B) = ²C₂

            P(A ∩ B) = 2/⁵²C₂          -(4)

Now,

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

= 26 * 25/52 * 51 + 4 * 3 / 52 * 51 – 2/52 * 51

= 55/221

问题 21. 在两次考试评分的入学考试中，随机选择的学生通过考试 1 的概率为 0.8，通过第二次考试的概率为 0.7。其中一项考试通过的概率为0.95。两者都通过的概率是多少？

解决方案：

Let A be the event of selecting a random student who passed in examination 1,

P(A) = 0.8          -(1)

Let A be the event of selecting a random student who passed in examination 2,

P(B) = 0.7          -(2)

Now,

The probability of selecting a random student passing in at least one of the examination is –

P(A ∪ B) = 0.95          -(3)

Probability of passing both of the examination is P(A ∩ B),

P(A ∩ B) = P(A) + P(B) -P(A ∪ B)

= 0.8 + 0.7 – 0.95

=1.5-0.95

=0.55

问题 22。一个盒子包含 40 个螺母和 30 个螺栓。一半的螺栓和螺母生锈了。如果随机抽取两件物品，它们生锈并且都是螺栓的概率是多少？

解决方案：

A box contains 40 nuts and 30 bolts,

half of them are rusted –

=> 40/2 = 20 nuts are rusted

=> 30/2 = 15 bolts are rusted

Since two items are drawn,

Sample space n(S) = ⁷⁰C₂          -(1)

Let A be the event of choosing the rusted item –

n(A) = ³⁵C_2.         -(20 nuts + 15 bolt = 35)

P(A) = ³⁵C₂/⁷⁰C₂

35 * 34/ 70 * 69          -(2)

Let B be the event of choosing two rusted bolts-

n(B) = ³⁰C₂

P(B) = ³⁰C₂/⁷⁰C₂

= 30 * 29 / 70 * 69          -(3)

Also, n(A n B) = 15          -(bolts are rusted)

P(A n B) = 15*14/70* 69          -(4)

Now,

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

= (35*34/70*69) + (30*29/70*69) – (15*14/70*69)          -(From 2, 3, 4)

= 185/483

问题 23. 从前 200 个整数中随机选择一个整数，求该整数能被 6 或 8 整除的概率？

解决方案：

A integer is chosen at random from 200 integers,

SAMPLE SPACE = n(S) = 200          -(1)

Let A be the event of choosing a number divisible by 6,

n(A) ={6,12,18. . . 198}

n(A) = 33          (Using T n formula)

P(A) = 33/200          -(2)

Let B be the event of choosing a number divisible by 8,

n(B) = {8,16,24. . . 200}

n(B) = 25

P(B) = 25/200

= 1/4          -(3)

Also, n(A n B)= {24,48. . 192}

= 8

P(A n B) = 8/200          -(4)

Now,

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

= 1/4          -(from 2, 3, 4)

问题 24. 当硬币被抛 4 次时，找到 2 或 3 个反面的概率？

解决方案：

A coin is tossed 4 times,

Hence, Sample space – n(S) = 2 ⁴ =16

Let A be the event of getting 2 tails,

A = {HHTT, HTHT, TTHH, THTH, THHT, HTTH}

n(A) = 6          

P(A) = 6/16         -(1)

Let B the event of getting 3 tails,

B = {HTTT, THTT, TTHT, TTTH}

n(B) = 4  

P(B) = 4/16          -(2)

A & B are the mutually exclusive events-

Hence, P(A ∩ B) = 0

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

= 6/16 + 4/16-0

= 10/16

= 5/8

问题 25. 假设从 1 到 1000 中选择一个整数，求它是 2 或 9 的倍数的概率？

解决方案：

A number is chosen from 1 to 1000,

Hence, sample space, n(S) = 1000

Number of multiples of 2 from 1 to 1000 are – 500

Number of multiples of 9 from 1 to 1000 are – 111

Out of 111, 55 are even numbers

Hence, total number multiple of 2 or 9 – 500 + 56

Probability of number multiple of 2 or 9 – 

= 556/1000

= 0.556

问题 26。在大都市地区，一个家庭拥有彩色电视机、黑白电视机或两者兼有的概率为 0.87、0.36 和 0.30。家庭拥有其中一套的概率是多少？

解决方案：

Let A be the probability of a family having a color TV set,

P(A) = 0.87          -(1)

Let B be the probability of a family having a black and white TV set,

P(B) = 0.36          -(2)

Also, P(A ∩ B) = 0.3

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

= 0.87 + 0.36 – 0.30

= 0.93

问题 27. 如果 A 和 B 是互斥事件，使得 P(A) = 0.35 和 P(B) = 0.45 找到，

(i) P(A∪B)

(ii) P(A ∩ B)

㈢ $P(A ∩ \bar{B})$ [特克斯] [/特克斯]

(四) $P(\bar{A} ∩ \bar{B})$ 

*** QuickLaTeX cannot compile formula:
 

*** Error message:
Error: Nothing to show, formula is empty

解决方案：

(i) P(A ∪ B) = P(A) + P(B)          (A and B are mutually exclusive events
                                                     i.e P( A∩ B) = 0)                   

= 0.80

(ii) P(A ∩ B) 

P(A ∩ B) = 0          (A and B are mutually exclusive events)

(iii) $P(A ∩ \bar{B})$ = P(A)

= 0.35

(iv) $P(\bar{A}∩\bar{B})$ = 1 – P(A ∪ B)

= 1 – 0.80

= 0.20

问题 28。样本空间由 9 个基本事件 E1、E2、E3 组成。 . . E9，其概率为 P(E1) = P(E2) = 0.08, P(E3) = P(E4) = 0.1,P(E6) = P(E7) = 0.2,P(E8) = P(E9) = 0.07，假设 A = {E1，E5，E8}，B ={E2，E5，E8，E9}

(i) 计算 P(A), P(B) & P(A ∩ B)

(ii) 求 P(A ∪ B)

(iii) 列出事件 A ∪ B 的燃烧，并找到 P(A ∪ B)

(iv) 计算

*** QuickLaTeX cannot compile formula:
 

*** Error message:
Error: Nothing to show, formula is empty

 

解决方案：

Given: P(E1) = P(E2) = 0.08, P(E3) = P(E4) = 0.1, P(E6) = P(E7) = 0.2, P(E8) = P(E9) = 0.07

Suppose: A = {E1, E5, E8}, B ={E2, E5, E8, E9}          -(I)

B^c = {E1,E3,E4,E6.E7}

P(E5) = 1 – (0.08 + 0.1 + 0.2 + 0.07}

=0.1

(i) P(A) = 0.08 + 0.1 + 0.07          (From Given)

= 0.25

P(B) = 0.08 + 0.1 + 0.07 + 0.07

= 0.32

P(A ∩ B) = 0.1 + 0.07

= 0.17

(ii) P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

= 0.57-0.17

= 0.40

(iii) $P(\bar{B})$ = 1 – P(B)

= 1 -0.32

= 0.68