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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.

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:

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

### 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:

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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