Class 12 RD Sharma Solutions – Chapter 32 Mean and Variance of a Random Variable – Exercise 32.2 | Set 2

Question 13: Two cards are selected at random from a box which contains five cards numbered 1,1,2,2, and 3. Let X denote the sum and Y the maximum of the two numbers drawn. Find the probability distribution, mean and variance of X and Y.

Solution:

As box contains cards numbered as 1,1,2,2 and 3

∴ possible sums of card numbers are 2,3,4 and 5

Hence, X can take values 2,3,4 and 5

X=2 [ when drawn cards are (1,1)]

X=3 [when drawn cards are (1,2) or (2,1)]

X=4 [when drawn cards are (2,2) or (3,1) or (1,3)]

X=5 [when drawn cards are (2,3) or (3,2)]

As Y is a random variable representing maximum of the two numbers drawn

∴ Y can take values 1,2 and 3.

Y=1 [when drawn cards are 1 and 1]

Y=2 [when drawn cards are (1,2) or (2,2) or (2,1)]

Y=3 [when drawn cards are (1,3) or (3,1) or (2,3) or (3,2)]

Note : P(1) represents probability of drawing card numbered as 1, similarly P(2) and P(3)

∴ P(X=2) = P(1)P(1) = 2/5 x 1/4 = 0.1

[For drawing first card we had 2 favourable outcomes as 1,1 out of total 5 ,in second time of drawing ,as we drew a card numbered as 1 we are having 1 favourable outcome out of total remaining of 4]

Similarly,

P(X=3) = P(2)P(1) + P(1)P(2) = 2/5 x 2/4 + 2/5 x 2/4 = 0.4

P(X=4) = P(2)P(2)+P(3)P(1)+P(1)P(3) = 2/5 x 1/4 + 2/5 x 1/4 + 1/5 x 2/4 = 0.3

P(X=5) = P(2)P(3)+P(3)P(2) = 2/5 x 1/4 + 2/5 x 1/4 = 0.2

Similarly,

P(Y=1) = P(1)P(1) = 2/5 x 1/4 = 0.1

P(Y=2) = P(1)P(2)+P(2)P(1)+P(2)P(2) = 2/5 x 2/4 + 2/5 x 2/4 + 2/5 x 1/4 = 0.5

P(Y=3) = P(2)P(3)+P(3)P(2)+ P(1)P(3)+P(3)P(1)

= 2/5 x 1/4 + 2/5 x 1/4 + 2/5 x 1/4 +2/5 x 1/5 + 1/5 x 2/4 = 0.4

Now we have pi and xi.

Let’s proceed to find mean and variance.

Mean of any probability distribution is given by Mean = ∑xipi

Variance is given by:

Variance = ∑ xi²pi – (∑xipi)²

∴ first we need to find the products i.e. pixi and pixi² and add them to get mean and apply the above formula to get the variance.

Following table representing probability distribution gives the required products : 

xi	pi	xipi	xi2pi
0	1/16	0	0
1	7/16	7/16	7/16
2	5/16	10/16	20/16
3	2/16	6/16	18/16

∴ Mean for(X) = 0.2+1.2+1.2+1 = 3.6

Variance for(X) = 0.4+3.6+4.8+5.0-3.6² = 13.8-3.6² = 0.84

Similarly probability distribution for Y is given below:

yi	pi	yipi	yi2pi
1	0.1	0.1	0.1
2	0.5	1.0	2.0
3	0.4	1.2	3.6

∴ Mean for(Y) = 0.1+1.0+1.2 = 2.3

Variance for(Y) = 0.1+2.0+3.6-2.3² = 5.7-2.3² = 0.41

Question 14: A die is tossed twice. A ‘success’ is getting an odd number on a toss. Find the variance of the number of successes.

Solution:

As success is considered when we get an odd number when we roll a die.

As die is rolled twice , so we can get no success or a single success or we can get odd both the times an odd number.

If X is the random variable denoting the success then X can take value 0,1 or 2

∵ P(getting an odd number in a single rolling of die) = 3/6 = 1/2

As rolling a die is an independent event:

∴ P(getting an odd on first roll and probability of getting odd on second roll)=P(getting an odd on first roll) x P(getting an odd on second roll)

Note: P(AՈB) = P(A)P(B) where A and B are independent events.

∴ P(X=0) = P(even number on first throw) x P(even on second throw) = 1/2 x 1/2 = 1/4

P(X=1) = P(even number on first throw) x P(odd on second throw) +

P(odd number on first throw) x P(even on second throw) = 1/2 x 1/2 + 1/2 x 1/2 = 1/2

P(X=2) = P(odd number on first throw) x P(odd on second throw) = 1/2 x 1/2 = 1/4

Now we have pi and xi.

Let’s proceed to find mean and variance.

Mean of any probability distribution is given by Mean = ∑xipi

Variance is given by:

Variance = ∑ xi²pi – (∑xipi)²

∴ first we need to find the products i.e. pixi and pixi² and add them to get mean and apply the above formula to get the variance.

Following table representing probability distribution gives the required products :

xi	pi	xipi	xi2pi
0	1/4	0	0
1	1/2	1/2	1/2
2	1/4	1/2	1

∵ Variance = ∑ xi²pi – (∑xipi)²

∴ Variance = 0 + 1/2 + 1 – (0 + 1/2 + 1/2)² = 0.5

Question 15: A box contains 14 bulbs, out of which 5 are defective. 3 bulbs are randomly drawn, one by one without replacement, from the box. Find the probability distribution of the number of defective bulbs.

Solution:

Let X be the random variable denoting the number of defective bulbs drawn in each draw. Since we are drawing a maximum of 3 bulbs at a time, So we can get at max 3 defective bulbs as total defective bulbs are 5.

∴ X can take values 0,1,2 and 3

P(X=0) = P(drawing no defective bulbs)

As we are finding probability for 0 defective bulbs ,so we will select all 3 bulbs

from 9 good bulbs.

n(s) = total possible ways = 14C₃

∴ P(X=0) = 9C₃/14C₃ = 3/13

P(X=1) = P(drawing 1 defective bulbs and 2 good bulbs)

As we are finding probability for 1 defective bulbs ,so we will select 2 bulbs

from 9 good bulbs and 1 from 5 defective ones

∴ P(X=1) = (9C₂ x 5C₁)/ 14C₃ = 45/91

Similarly,

P(X=2) = (9C₁ x 5C₂)/14C₃ = 45/182

P(X=3) =  5C₃ /14C₃ = 5/182

So, Probability distribution is given below: 

xi	pi
0	3/13
1	45/91
2	45/182
3	5/182

Question 16: In roulette, Fig. 32.2, the wheel has 13 numbers 0,1,2,….,12 marked on equally spaced slots. A player sets ₹10 on a given number. He receives ₹100 from the organizer of the game if the ball comes to rest in this slot; otherwise, he gets nothing. If X denotes the player’s net gain/loss, find E (X). 

Solution:

As player sets Rs 10 on a number ,if he wins he get Rs 100

∴ his profit is Rs 90.

If he loses, he suffers a loss of Rs 10

He gets a profit when ball comes to rest in his selected slot.

Total possible outcome = 13

Favourable outcomes = 1

∴ probability of getting profit = 1/13

And probability of loss = 12/13

If X is the random variable denoting gain and loss of player

∴ X can take values 90 and -10

P(X=90) = 1/13

And P(X=-10) = 12/13

Now we have pi and xi.

Let’s proceed to find mean

Mean of any probability distribution is given by Mean = ∑xipi

∴ first we need to find the products i.e. pixi and add them to get mean

Following table representing probability distribution gives the required products :

xi	pi	xipi
90	1/13	90/13
-10	12/13	-120/13

E(X) = Mean = 90/13 + (-120/13) = 90/13 – 120/13 = -30/13

Question 17: Three cards are drawn at random (without replacement) from a well shuffled pack of 52 cards. Find the probability distribution of number of red cards. Hence find the mean of the distribution.

Solution:

We have total 26 red cards in a deck of 52 cards.

As we are drawing maximum 3 cards at a time so we can get maximum 3 red cards.

If X denotes the number of red cards ,then X can take values from 0,1,2 and 3

P(X=0) = probability of drawing no red cards

We need to select all 3 cards from remaining 26 cards

Total possible ways of selecting 3 cards = 52C₃

∴ P(X=0) = 26C₃ / 52C₃ = 2/17

P(X=1) = P(selecting one red and 2 black cards) = (26C₁ x 26C₂ ) / 52C₃ = 13/34

P(X=2) = P(selecting 2 red and 1 black cards) = (26C₂ x 26C₁) / 52C₃ = 13/34

P(X=3) = 26C₃/52C₃ = 2/17

So, Probability distribution is given below:

xi	pi	xipi
0	2/17	0
1	13/34	13/34
2	13/34	26/34
3	2/17	6/17

Mean = 0 + 13/34 + 26/34 + 6/17 = 1.5

Question 18: An urn contains 5 red 2 black balls. Two balls are randomly drawn, without replacement. Let X represent the number of black balls drawn. What are the possible values of X? Is X a random variable? If yes, find the mean and variance of X.

Solution:

X represents the number of black balls drawn.

∴ X can take values 0,1 and 2

∵ there are total 7 balls

n(S) = total possible ways of selecting 2 balls = 7C₂

P(X=0) = P(selecting no black balls) = 5C₂/7C₂ = 10/21

P(X=1) = P(selecting 1 black ball and 1 red ball)

= (5C₁ x 2C₁) / 7C₂ = 10/21

P(X=2) = P(selecting 2 black ball and 0 red ball) = (5C₀ x 2C₂) / 7C₂ = 1/21

X is said to be a random variable if some of the probabilities associated with each value of X is 1

Here,

P(X=0) + P(X=1) + P(X=2) = 20/42 + 20/42 + 2/42 = 1

∴ X is a random variable.

Now we have pi and xi.

Let’s proceed to find mean and variance.

Mean of any probability distribution is given by Mean = ∑xipi

Variance is given by:

Variance = ∑ xi²pi – (∑xipi)²

∴ first we need to find the products i.e. pixi and pixi² and add them to get mean and apply the above formula to get the variance.

Following table representing probability distribution gives the required products :

xi	pi	xipi	xi2pi
0	10/21	0	0
1	10/21	10/21	10/21
2	1/21	2/21	4/21

∴ Mean = 10/21 + 2/21  = 4/7

∴ Variance = 0 + 10/21 + 4/21 – (4/7)² = 50/147

Question 19: Two numbers are selected at random (without replacement) from positive integers 2,3,4,5,6 and 7. Let X denote the larger of the two numbers obtained. Find the mean and variance of the probability distribution of X.

Solution:

∵ two numbers are selected at random like {(2,3) or (5,4) or (4,5)..etc}

Total ways of selecting two numbers without replacement = 6 x 5 =30

As X denote the larger of two numbers selected

∴ X can take values 3,4,5,6 and 7

P(X=3) = P(larger number is 3) =  (2/30)[{2,3},{3,2}]

P(X=4) = P(larger number is 4) =  (4/30)[{2,4},{4,2},{3,4},{4,3}]

P(X=5) = P(larger number is 5) =  (6/30)[{2,5},{3,5},{4,5} and their reverse order]

P(X=6) = P(larger number is 6) =  (8/30)[{2,6},{3,6},{4,6},{5,6} and their reverse order]

P(X=7) = P(larger number is 7) =  (10/30)[{2,7},{3,7},{4,7},{5,7},{6,7} and their reverse order]

Now we have pi and xi.

Let’s proceed to find mean and variance.

Mean of any probability distribution is given by Mean = ∑xipi

Variance is given by:

Variance = ∑ xi²pi – (∑xipi)²

∴ first we need to find the products i.e. pixi and pixi² and add them to get mean and apply the above formula to get the variance.

Following table representing probability distribution gives the required products:

xi	pi	xipi	xi2pi
3	2/30	6/30	18/30
4	4/30	16/30	64/30
5	6/30	30/30	150/30
6	8/30	48/30	288/30
7	10/30	70/30	490/30

∴ Mean = 6/30 + 16/30 + 30/30 + 48/30 + 70/30 = 17/3

∴ Variance = 18/30 + 64/30 + 150/30 + 288/30 + 490/30 – (17/3)² = 14/9

Question 20: In a game, a man wins ₹5 for getting a number greater than 4 and loses ₹1 otherwise, when a fair die is thrown. The man decided to throw a die thrice but to quits as and when he gets a number than 4. Find the expected value of amount he wins/lose.

Solution:

We are asked to find the expected amount he wins or lose i.e we have to find the mean of probability distribution of random variable X denoting the win/loss.

As he decided to throw the dice thrice but to quit at the instant he loses

∴ if he wins in all throw he can make earning of Rs 15

If he wins in first two throw and lose in last, he earns Rs (10-1) = Rs 9

If he wins in first throw and then loses ,he earns = Rs 4

If he loses in first throw itself, he earns Rs = -1

Thus X can take values -1,4,9 and 15

P(getting a number greater than 4 in a throw of die) = 2/6 = 1/3

P(getting a number not greater than 4 in a throw of die) = 4/6 = 2/3

P(X=-1) = P(getting number less than or equal to 4) = 2/3

P(X=4) = P(getting > 4) x P(getting ≤ 4) = 1/3 x 2/3 = 2/9 

P(X=9) = P(getting > 4) x P(getting > 4) x P(getting ≤ 4) = 1/3 x 1/3 x 2/3 = 2/27  

P(X=15) = P(getting > 4) x P(getting > 4) x P(getting > 4) =  1/3 x 1/3 x 1/3 = 1/27

So, Probability distribution is given below:

xi	pi	xipi
-1	2/3	-2/3
4	2/9	8/9
9	2/27	18/27
15	1/27	15/27

∵ Mean = ∑xipi

Mean = -2/3 + 8/9 + 18/27 + 15/27 = 39/27 = 1.44

He can win around Rs 1.45