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Class 12 RD Sharma Solutions – Chapter 32 Mean and Variance of a Random Variable – Exercise 32.1 | Set 2

  • Last Updated : 25 Jan, 2021

Question 16. Two cards are drawn successively with replacement from well-shuffled pack of 52 cards. Find the probability distribution of number of kings.

Solution: 

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

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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

Question 17. Two cards are drawn successively without replacement from a well-shuffled pack of 52 cards. Find the probability distribution of the number of aces.

Solution: 

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

Question 18. Find the probability distribution of the number of white balls drawn in a random draw of 3 balls without replacement, from a bag containing 4 white and 6 red balls.

Solution: 

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

Question 19. Find the probability of Y in two throws of two dice, where Y represents the number of times a total of 9 appears.

Solution: 

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

Question 20. From a lot containing 25 items, 5 of which are defective, 4 are chosen at random. Let X be the number of defectives found. Obtain the probability distribution of X if the items are chosen without replacement.

Solution: 

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

Question 21. Three cards are drawn successively with the replacement from well-shuffled deck of 52 cards. A random variable X denotes the number of hearts in the three cards drawn. Determine the probability distribution of X.

Solution: 

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

Question 22. An urn contains 4 red and 3 blue balls. Find the probability distribution of the number of blue balls in a random draw of 3 balls with replacement.

Solution: 

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

Question 23. Two cards are drawn simultaneously from a well-shuffled deck of 52 cards. Find the probability distribution of the number of successes, when getting a spade is considered a success.

Solution: 

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

Question 24. A fair die is tossed twice. If the number appearing on the top is less than 3, it is a success. Find the probability distribution of number of successes.

Solution: 

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

Question 25. An urn contains 5 red and 2 black balls. Two balls are randomly selected. Let X represent the number of black balls. What are the possible values of X. Is X a random variable?

Solution: 

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.

Question 26. Let X represent the difference between the number of heads and the number of tails when a coin is tossed 6 times. What are the possible values of X?

Solution: 

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

Question 27. From a lot of 10 bulbs, which includes 3 defectives, a sample of 2 bulbs is drawn at random. Find probability distribution of the number of defective bulbs.

Solution: 

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

Question 28. Four balls are to be drawn without replacement from a box containing 8 red and 4 white balls. If X denotes the number of red balls drawn, then find the probability distribution of X.

Solution: 

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

Question 29. The probability distribution of a random variable X is given below:

X0123
P(X)kk/2k/4k/8

i) Determine the value of k.

Solution: 

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) Determine P(X<=2) and P(X>2).

Solution: 

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) Find P(X<=2)+P(X>2)

Solution: 

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

Question 30. Let X, denote the number of colleges where you apply after your results and P(X=x) denotes your probability of getting admission in x number of colleges. It is given that

kx, if x=0 or 1

2kx, if x=2

P(X=x)= k(5-x), if x=3 or 4

0, if x>4

where k is a positive constant. Find the value of k. Also, find the probability that you will get admission in (i) exactly one college (ii) at most 2 colleges (iii) at least 2 colleges.

Solution: 

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:

X01234
P(X)0k4k2kk

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




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