第12类RD Sharma解–第32章随机变量的均值和方差–练习32.1 |套装2

问题16.连续抽出两张纸牌，并从洗净的52张纸牌中替换。求出国王数的概率分布。

解决方案：

Given that two cards are drawn with replacement from well shuffled pack of 52 cards.

Then the values of random variable for the probability distribution could be,

i. No king

ii. One king

iii. Two kings

i. No king:

P(X=0)=(48/52)x(48/52)

           =144/169=0.85

ii. One king:

P(X=1)=(48/52)x(4/52)+(48/52)x(4/52)

          =24/169=0.14

iii. Two kings:

P(X=2)=(4/52)x(4/52)

          =1/169=0.005

问题17：连续抽出两张纸牌，而不是从52张纸牌中随机抽取。找到王牌数的概率分布。

解决方案：

Given that two cards are drawn successively without replacement from a deck.

Then the values of random variable for the probability distribution for the number of aces could be,

i. No ace

ii. One ace

iii. Two aces

i. No ace:

P(X=0)=⁴⁸C₂/⁵²C₂

           =48×47/52×51

           =188/221=0.85

ii. One ace:

P(X=1)=⁴⁸C₁x⁴C₁/⁵²C₂

           =48x4x2/52×51

           =32/221=0.144

iii. Two aces:

P(X=2)=⁴C₂/⁵²C₂

           =4×3/52×51

           =1/221=0.0045

问题18：从一个装有4个白色和6个红色球的袋子中，随机抽取3个球，而无需替换，得出白球数目的概率分布。

解决方案：

Given that 3 balls are drawn in a random from a bag containing 4 white and 6 red balls.

Then the values of random variable for the probability distribution for number of white balls would be:

i. No white balls

ii. One white ball

iii. Two white balls

iv. Three white balls

i. No white balls:

P(X=0)=⁶C₃/¹⁰C₃

           =6x5x4/10x9x8

           =1/6=0.16

P(X=1)=⁶C₂x⁴C₁/¹⁰C₃

           =6x5x4x3/10x9x8

          =1/2=0.5

P(X=2)=⁶C₁x⁴C₂/¹⁰C₃

           =6x4x3x3/10x9x8

           =3/10=0.3

P(X=3)=⁴C₃/¹⁰C₃

           =4x3x2/10x9x8

           =1/30=0.03

问题19：在两次掷出的两个骰子中找出Y的概率，其中Y表示总共出现9次的次数。

解决方案：

Given that 2 dice are thrown two times and Y represents the number of times a total of 9 appears.

A total of 9 appears when the dice outcomes are: (3,6) , (4,5) , (5,4) , (3,6)

Probability of getting a total of 9 = 4/36=1/9=0.11

Then the values of random variable for the probability distribution of Y would be: 0, 1, 2

P(X=0)=(32/36)x(32/36)

               =64/81=0.79

P(X=1)=(32/36)x(4/36)+(4/36)x(32/36)

           =16/81=0.19

P(X=2)=(4/36)x(4/36)

           =1/81=0.012

问题20.从包含25个物品的批次中，其中5个是有缺陷的，随机选择4个。令X为发现的缺陷数量。如果选择的项目没有替换，则获得X的概率分布。

解决方案：

Given that in 25 items 5 are defective and 4 are chosen at random.

Then the values for random variable for the probability distribution for number of defective ones could be:

i. No defective

ii. One defective

iii. Two defective

iv. Three defective

v. Four defective

i. No defective:

P(X=0)=²⁰C₄/²⁵C₄

               =20x19x18x17/25x24x23x22

            =969/2530=0.38

P(X=1)=²⁰C₃x⁵C₁/²⁵C₄

           =20x19x18x5x4/25x24x23x22

           =114/253=0.45

P(X=2)=²⁰C₂x⁵C₂/²⁵C₄

           =20x19x5x4x3x2/25x24x23x22

           =38/253=0.15

P(X=3)=²⁰C₁x⁵C₃/²⁵C₄

           =20x5x4x3x4/25x24x23x22

           =4/253=0.01

P(X=4)=⁵C₄/²⁵C₄

           =5x4x3x2/25x24x23x22

           =1/2530=0.0003

问题21：连续抽出三张牌，并从洗净的52张牌中替换。随机变量X表示抽出的三张牌中的心数。确定X的概率分布。

解决方案：

Given that three cards are drawn successively with replacement from well-shuffled deck.

Then the values of random variable for the probability distribution of number of hearts would be:

i. No hearts

ii. One heart

iii. Two hearts

iv. Three hearts

i. No hearts:

P(X=0)=(39/52)x(39/52)x(39/52)

           =27/64=0.42

P(X=1)=(39/52)x(39/52)x(13/52)x3

           =27/64=0.42

P(X=2)=(39/52)x(13/52)x(13/52)x3

           =9/64=0.14

P(X=3)=(13/52)x(13/52)x(13/52)

           =1/64

问题22.骨灰盒包含4个红色和3个蓝色的球。在随机替换的3个球中找到蓝色球数的概率分布。

解决方案：

Given that an urn contains 4 red and 3 blue balls and 3 balls are drawn with replacement.

Then the values of random variable for the probability distribution of number of blue balls drawn would be:

i. No blue balls

ii. One blue ball

iii. Two blue balls

iv. Three blue balls

i. No blue balls:

P(X=0)=(4/7)x(4/7)x(4/7)

           =64/343=0.18

P(X=1)=(4/7)x(4/7)x(3/7)x3

           =144/343=0.41

P(X=2)=(4/7)x(3/7)x(3/7)x3

           =108/343=0.31

P(X=3)=(3/7)x(3/7)x(3/7)

           =27/343=0.07

问题23：从一张洗净的52张纸牌中同时抽出两张纸牌。当获得黑桃被认为是成功时，找到成功次数的概率分布。

解决方案：

Given that two cards are drawn from a deck.

Then the values of the random variable for the probability distribution of number of spades would be:

i. No spade

ii. One spade

iii. Two spades

i. No spade:

P(X=0)=³⁹C₂/⁵²C₂

           =39×38/52×51=19/34=0.55

P(X=1)=³⁹C₁x¹³C₁/⁵²C₂

           =39x13x2/52×51

           =13/34=0.38

P(X=2)=¹³C₂/⁵²C₂

           =13×12/52×51

           =1/17=0.05

问题24.一个公平的死者被扔了两次。如果出现在顶部的数字小于3，则表示成功。查找成功次数的概率分布。

解决方案：

Given that a fair dice is tossed twice and when a number less than 3 occurs it is a success.

The probability that the number on the top is less than 3=2/6

Then the value of random variable for the probability distribution would be: 0 , 1 , 2

P(X=0)=(4/6)x(4/6)

           =16/36=0.4

P(X=1)=(4/6)x(2/6)x2

           =16/36=0.4

P(X=2)=(2/6)x(2/6)

           =4/36=0.11

问题25.包含5个红色和2个黑色的球。随机选择两个球。令X代表黑球的数量。 X的可能值是什么。X是随机变量吗?

解决方案：

Given that an urn contains 5 red and 2 black balls and two balls are selected randomly.

Then the values of random variable for the probability distribution of the number of black balls would be:

i. No black balls

ii. One black ball

iii. Two black balls

These are the possible values of X.

Yes X is a random variable.

问题26.令X代表抛硬币6次后的正面数和反面数之差。 X的可能值是什么?

解决方案：

Given that X is the difference between the number of heads and the number of tails when a coin is tossed 6 times.

The possible outcomes are(T,H): (6,0), (5,1), (4,2), (3,3), (2,4), (1,5), (0,6)

The possible values of random variable X would be:

X= 6, 4, 2, 0

问题27.从包括3个缺陷的10个灯泡中随机抽取2个灯泡的样本。查找缺陷灯泡数量的概率分布。

解决方案：

Given that a lot of 10 bulbs contains 3 defective ones.

Then the values of random variables for the probability distribution of number of defective bulbs would be:

i. No defective bulb

ii. One defective bulb

iii. Two defective bulbs

i. No defective bulbs

P(X=0)=⁷C₂/¹⁰C₂

           =7×6/10×9

           =7/15=0.4

P(X=1)=⁷C₁x³C₁/¹⁰C₂

           =7X3X2/10X9

           =7/15=0.4

P(X=2)=³C₂/¹⁰C₂

           =3×2/10×9

           =1/15=0.06

问题28.从装有8个红色和4个白色小球的盒子中抽出四个小球，不得更换。如果X表示绘制的红色球的数量，则找到X的概率分布。

解决方案：

Given that 4 balls are drawn without replacement from a box containing 8 red and 4 white balls.

Then the values of random variable for the probability distribution of number of red balls drawn would be:

i. No red ball

ii. One red ball

iii. Two red balls

iv. Three red balls

v. Four red balls

i. No red ball:

P(X=0)=⁴C₄/¹²C₄

           =4x3x2/12x11x10x9

           =1/495=0.002

P(X=1)=⁴C₃x⁸C₁/¹²C₄

           =4x3x2x8x4/12x11x10x9

           =32/495=0.06

P(X=2)=⁴C₂x⁸C₂/¹²C₄

           =4x3x8x7x3x2/12x11x10x9

           =56/165=0.33

P(X=3)=⁴C₁x⁸C₃/¹²C₄

           =4x8x7x6x4/12x11x10x9

           =224/495=0.45

P(X=4)=⁸C₄/¹²C₄

           =8x7x6x5/12x11x10x9

           =14/99=0.14

问题29.随机变量X的概率分布如下：

X	0	1	2	3
P(X)	k	k/2	k/4	k/8

i)确定k的值。

所以lution：

We know that the sum of probability distributions is equal to 1.

=>k + k/2 + k/4 + k/8 = 1

=>15k/8=1

=>k=8/15

ii)确定P(X ＜＝ 2)和P(X＞ 2)。

解决方案：

P(X<=2)=P(X=0)+P(X=1)+P(X=2)

                    = k + k/2 + k/4

                     =7k/4=7×8/4×15

                     =14/15=0.93

P(X>2)=P(X=3)

           =k/8=8/15×8=1/15

           =0.06

iii)求P(X <= 2)+ P(X> 2)

解决方案：

P(X<=2)+P(X>2)=8X15/15X8=1

问题30.让X代表您在结果之后申请的大学数量，P(X = x)代表您获得x所大学数量的录取概率。鉴于

kx，如果x = 0或1

如果x = 2，则为2kx

如果x = 3或4，则P(X = x)= k(5-x)

如果x> 4，则为0

其中k是一个正常数。找出k的值。此外，找到您将被(i)一所大学(ii)最多两所大学(iii)至少两所大学录取的概率。

解决方案：

When x=0, P(X)=k(0)=0

                 x=1, P(X)=k(1)=k

                 x=2, P(X)=2k(2)=4k

                 x=3, P(X)=k(5-3)=2k

                 x=4, P(X)=k(5-4)=k

The probability distribution of X would be:

X	0	1	2	3	4
P(X)	0	k	4k	2k	k

We know that the sum of probability distribution is equal to 1.

=>0+k+4k+2k+k=1

=>8k=1

=>k=1/8

i. exactly one college:

P(X=1)=k=1/8=0.125

ii. at most 2 colleges:

P(X<=2)=P(X=0)+P(X=1)+P(X=2)

              =0+k+4k=5k=5/8=0.625

iii. at least 2 colleges:

P(X>=2)=P(X=2)+P(X=3)+P(X=4)

              =4k+2k+k=7k=7/8=0.875