RD Sharma Class 12 Ex 32.1 Solutions Chapter 32 Mean and Variance of a Random Variable

Here we provide Sharma Class 12 Ex 32.1 Solutions Chapter 32 Mean and Variance of a Random Variable for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 12 Ex 32.1 Solutions Chapter 32 Mean and Variance of a Random Variable book pdf download. Now you will get step-by-step solutions to each question.

TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter32
Exercise32.1
CategoryRD Sharma Solutions

Here we provide RD Sharma Class 12 Ex 19.15 Solutions Chapter 19 Indefinite Integrals for English medium students, Which will very helpful for every student in their exams. Students can download the latest Sharma Class 12 Ex 19.15 Solutions Chapter 19 Indefinite Integrals book pdf download. Now you will get step-by-step solutions to each question.

TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter19
Exercise19.15
CategoryRD Sharma Solutions

Table of Contents

RD Sharma Class 12 Ex 32.1 Solutions Chapter 32 Mean and Variance of a Random Variable

Calculate the mean deviation from the median of the following frequency distribution :

Question 1(i): Find the mean and standard deviation of each of the following probability distributions:

xi: 2 3 4

pi: 0.3 0.5 0.3

Solution:

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

Standard Deviation is given by SD = √Variance where 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 gives the required products:

xipixipixi2pi
20.20.40.8
30.31.54.5
40.51.24.8

∴ mean = 0.4 + 1.5 + 1.2 = 3.1

And variance = 0.8 + 4.5 + 4.8 – (3.1)2 = 0.49

∴ Standard deviation = √ 0.49 = 0.7

Question 1(ii): Find the mean and standard deviation of each of the following probability distributions:

xi: 1 3 4 5

pi: 0.4 0.1 0.2 0.3

Solution:

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

Standard Deviation is given by SD = √ Variance where 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 gives the required products:

xipixipixi2pi
10.40.40.4
30.10.30.9
40.20.83.2
50.31.57.5

∴ mean = 0.4 + 0.3+0.8+1.5= 3.0

And variance =0.4+0.9+3.2+7.5 – (3.0)2 = 3

∴ Standard deviation = √ 3= 1.732

Question 1(iii): Find the mean and standard deviation of each of the following probability distributions:

xi: -5 -4 1 2

pi: 1/4 1/8 1/2 1/8

Solution:

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

Standard Deviation is given by SD = √ Variance where 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 gives the required products:

xipixipixi2pi
-51/4-1.256.25
-41/8-0.52
11/20.50.5
21/80.250.5

∴ mean = -1.25-0.5+0.5+0.25 = -1

And variance = 6.25+2+0.5+0.5 – (-1)2 = 8.25

∴ Standard deviation = √8.25= 2.9

Question 1(iv): Find the mean and standard deviation of each of the following probability distributions:

xi: -1 0 1 2 3

pi: 0.3 0.1 0.1 0.3 0.2

Solution:

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

Standard Deviation is given by SD = √ Variance where 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 gives the required products:

xipixipixi2pi
-10.3-0.30.3
00.100
10.10.10.1
20.30.61.2
30.20.61.8

∴ mean = -0.3 + 0 + 0.1 + 0.6 + 0.6 = 1.0

And variance =0.3 + 0 + 0.1 + 1.2 + 1.8 – (1)2 = 2.4

∴ Standard deviation = √2.4 = 1.5

Question 1(v): Find the mean and standard deviation of each of the following probability distributions:

xi: 1 2 3 4

pi: 0.4 0.3 0.2 0.1

Solution:

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

Standard Deviation is given by SD = √Variance where 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 gives the required products :

xipixipixi2pi
10.40.40.4
20.30.61.2
30.20.61.8
40.10.41.6

∴ mean = 0.4+0.6+0.6+0.4 = 2.0

And variance = 0.4 +1.2 + 1.8 + 1.6– (2)2 = 1.0

∴ Standard deviation = √1 = 1

Question 1(vi): Find the mean and standard deviation of each of the following probability distributions:

xi: 0 1 3 5

pi: 0.2 0.5 0.2 0.1

Solution:

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

Standard Deviation is given by SD = √ Variance where 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 gives the required products:

xipixipixi2pi
00.200
10.50.50.5
30.20.61.8
50.10.52.5

∴ mean = 0+0.5+0.6+0.5 = 1.6

And variance = 0 +0.5 + 1.8 + 2.5– (1.6)2 = 2.24

∴ Standard deviation = √2.24 = 1.497

Question 1(vii): Find the mean and standard deviation of each of the following probability distributions:

xi: -2 -1 0 1 2

pi: 0.1 0.2 0.4 0.2 0.1 

Solution:

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

Standard Deviation is given by SD = √ Variance where 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 gives the required products:

xipixipixi2pi
-20.1-0.20.4
-10.2-0.20.2
00.400
10.20.20.2
20.10.20.4

∴ mean = -0.2-0.2+0+0.2+0.2 = 0

And variance = 0 +0.4+0.2+0.2+0.4– (0)2 = 1.2

∴ Standard deviation = √1.2 = 1.095

Question 1(viii): Find the mean and standard deviation of each of the following probability distributions:

xi: -3 -1 0 1 3

pi: 0.05 0.45 0.20 0.25 0.05

Solution:

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

Standard Deviation is given by SD = √ Variance where 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 gives the required products:

xipi xipixi2pi
-30.05-0.150.45
-10.45-0.450.45
00.2000
10.250.250.25
30.050.150.45

∴ mean = -0.15-0.45+0+0.25+0.15 = -0.2

And variance = 0 +0.45+0.25+0.45+0.45– (-0.2)2 = 1.56

∴ Standard deviation = √1.56 = 1.248

Question 1(ix): Find the mean and standard deviation of each of the following probability distributions:

xi: 0 1 2 3 4 5

pi: 1/6 5/18 2/9 1/6 1/9 1/18

Solution:

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

Standard Deviation is given by SD = √ Variance where 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 gives the required products :

xipixipixi2pi
01/600
15/185/185/18
22/94/98/9
31/61/23/2
41/94/916/9
51/185/1825/18

∴ Mean =  0+5/18+4/9+1/2+4/9+5/18 = 35/18

Variance =  0+5/18+8/9+3/2+16/9+25/18 -{35/18)= 665/324  

∴ standard deviation = √ (665/324) = √665/18

Question 2: A discrete random variable X has the probability distribution given below:

X: 0.5 1 1.5 2

P(X): k k2 2k2 k

(i) Find the value of k. (ii) Determine the mean of the distribution.

Solution:

To find the value of k we will be using the very basic idea of probability.

Note: We know that the sum of the probabilities of all random variables taken from a given sample space is equal to 1.

∴ P(X=0.5) + P(X=1) + P(X=1.5) + P(X=2) = 1

∴ k + k2 + 2k2 + k = 1

⇒ 3k+ 2k – 1 = 0

⇒ 3k2 + 3k – k – 1 = 0

⇒ 3k(k+1) – (k+1) = 0

⇒ (3k-1)(k+1) = 0

∴ k = 1/3 or k = -1

∵ k represents probability of an event. Hence 0≤P(X)≤1

∴ k = 1/3

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

Now we have,

X: 0.5 1 1.5 2

P(X): 1/3 1/9 2/9 1/3

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

∴ Mean = 0.5 x (1/3) + 1 x (1/9) + 1.5 x (2/9) +2 x (1/3) = 23/18.

Question 3: Find the mean-variance and standard deviation of the following probability distribution

Xi: a b

Pi: p q

Where p+q=1.

Solution:

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

Standard Deviation is given by SD = √ Variance where 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.

∴ p1x1 = ap and p2x2 = bq Similarly p1x1= a2p and p2x22= b2q

∴ Mean = ap + bq

Variance = a2p + b2q – (ap + bq)2

=a2pq + b2pq + 2abpq [p + q=1]

=pq(a-b)2

∴ SD = √{pq(a-b)2 } = |a-b|√pq

Question 4: Find the mean and variance of the number of tails in three tosses of a coin.

Solution:

When we toss a coin three times we have the following possibilities:

{HHH,HHT,HTH,THH,HTT,THT,TTH,TTT}

Let X be a random variable representing number of tails in 3 tosses of a coin.

∵ probability of getting a head or probability of getting a tail are independent events and P(GETTING A HEAD) = P(GETTING A TAIL) = 1/2

∴ P(Head in first toss) and P(Head in second toss) and P(head in third toss) can be given by their individual products.

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

Thus,

P(X=0) = P(HHH) = P(H)P(H)P(H) = 1/2 x 1/2 x 1/2 = 1/8

P(X=1) = P(HHT or HTH or THH) = P(HHT)+P(HTH)+P(THH)

= P(H)P(H)P(T)+ P(H)P(T)P(H)+ P(T)P(H)P(H)

= 1/2 x 1/2 x 1/2 + 1/2 x 1/2 x 1/2 + 1/2 x 1/2 x 1/2

= 3/8

P(X=2) = P(HTT or THT or TTH) = P(HTT)+P(THT)+P(TTH)

= P(H)P(T)P(T)+ P(T)P(H)P(T)+ P(T)P(T)P(H)

= 1/2 x 1/2 x 1/2 + 1/2 x 1/2 x 1/2 + 1/2 x 1/2 x 1/2

= 3/8

P(X=3) = P(TTT) = P(T)P(T)P(T) = 1/2 x 1/2 x 1/2 = 1/8

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 gives the required products: 

xipixipixi2pi
01/800
13/83/83/8
23/83/43/2
31/83/89/8

∴ Mean = 0 + 3/8 + 3/8 + 1/8 = 3/2

Variance = 0 + 3/8 + 3/4 + 3/8 – (3/2)2 = 3/4

Question 5: Two cards are drawn simultaneously from a pack of 52 cards. Compute the mean and standard deviation of the number of kings.

Solution:

In a deck of 52 cards there are 4 kings each of one suit respectively.

Let X be the random variable denoting the number of kings for an event when two cards are drawn simultaneously.

∴ X can take values 0 , 1 or 2. 

P(X=0) = 48C2/52C2    = 48×47/52×51 = 188/221

[For selecting 0 kings, we removed all 4 kings from deck and selected out of 48]

P(X=1) = 4C1 x 48C1/52C2 = 48 x 4 x 2/52 x 51 = 32/221

[For selecting 1 king, we need to select and 1 out of 4 and not any other]

P(X=2) = 4C2/52C2 = 4 x 3/52 x 51 = 1/221

[For selecting 2 king, we need to select and 2 out of 4]

Now we have pi and xi.

Let’s proceed to find mean and standard deviation.

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

Standard Deviation is given by SD = √ Variance where 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 gives the required products:

xipixipixi2pi
0188/22100
132/22132/22132/221
21/2212/2214/221

∴ mean = 0 + 32/221 + 2/221 = 34/221

Variance = 0 + 32/221 + 4/221 – (34/221) = 400/2873

∴ Standard deviation = √variance = √(400/2873) = 20/√2873

Question 6: Find the mean, variance and standard deviation of the number of tails in three tosses of a coin.

Solution:

When we toss a coin three times we have the following possibilities:

{HHH,HHT,HTH,THH,HTT,THT,TTH,TTT}

Let X be a random variable representing number of tails in 3 tosses of a coin.

∵ probability of getting a head or probability of getting a tail are independent events and P(GETTING A HEAD) = P(GETTING A TAIL) = 1/2

∴ P(Head in first toss) and P(Head in second toss) and P(head in third toss) can be given by their individual products.

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

Thus,

P(X=0) = P(HHH) = P(H)P(H)P(H) = 1/2 x 1/2 x 1/2 = 1/8

P(X=1) = P(HHT or HTH or THH) = P(HHT)+P(HTH)+P(THH)

= P(H)P(H)P(T)+ P(H)P(T)P(H)+ P(T)P(H)P(H)

= 1/2 x 1/2 x 1/2 + 1/2 x 1/2 x 1/2 + 1/2 x 1/2 x 1/2

= 3/8

P(X=2) = P(HTT or THT or TTH) = P(HTT)+P(THT)+P(TTH)

= P(H)P(T)P(T)+ P(T)P(H)P(T)+ P(T)P(T)P(H)

= 1/2 x 1/2 x 1/2 + 1/2 x 1/2 x 1/2 + 1/2 x 1/2 x 1/2

= 3/8

P(X=3) = P(TTT) = P(T)P(T)P(T) = 1/2 x 1/2 x 1/2 = 1/8

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

Standard Deviation is given by SD = √Variance

∴ 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 gives the required products :

xipixipixi2pi
1/800
13/83/83/8
23/83/43/2
31/83/89/8

∴ Mean = 0 + 3/8 + 3/4 + 3/8 = 3/2

Variance = 0 + 3/8 + 3/2 + 9/8  – (3/2) = 3/4

Standard Deviation = √(3/4) = 0.87

Question 7: Two bad eggs are accidentally mixed up with ten good ones. Three eggs are drawn at random with replacement from this lot. Compute the mean for the number of bad eggs drawn.

Solution:

As there are total of two bad eggs. Therefore while drawing 3 eggs we can draw 1 bad egg or 2 or 0 bad eggs.

Let X be the random variable denoting number of bad eggs that can be drawn in each draw.

Clearly X can take values 0,1 or 2

P(X=0) = P(all 3 are good eggs) = 2Cx 10C/12C3 = 120/220 = 6/11

[Since there are 10 good eggs so for selecting all good we took all three from 10 and 0 eggs from 2 bad ones. Total sample points are no of ways of selecting 3 eggs from total of 12 eggs]

Similarly,

P(X=1) = P(1 bad and 2 good eggs) =  2C1 x 10C2/12C3  = 9/22 

P(X=2) = P(2 Bad eggs and 1 good egg) =2C2 x 10C1 /12C3 = 1/22

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 gives the required products :

xipixipi
06/110
19/229/22
21/221/11

∴ mean = 0 + 9/22 + 1/11 = 1/2

Question 8: A pair of fair dice is thrown. Let X be the random variable which denotes the minimum of the two numbers which appear. Find the probability distribution, mean and variance of X.

Solution:

When a pair of fair dice is thrown there are total 36 possible outcomes.

X denotes the minimum of two numbers which appear

∴ X can take values 1,2,3,4,5 and 6

P(X=1) = 11/36

[Possible Pairs: (1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(3,1),(4,1),(5,1),(6,1)]

P(X=2) = 9/36

[Possible Pairs: (2,2),(3,2),(4,2),(5,2),(6,2),(2,6),(2,5),(2,4),(2,3)]

P(X=3) = 7/36

[Possible Pairs: (3,3),(3,4),(4,3),(5,3),(3,5),(3,6),(6,3)]

P(X=4) = 5/36

[Possible Pairs: (4,4),(5,4),(4,5),(4,6),(6,4)]

P(X=5) = 3/36

[Possible Pairs (5,5),(5,6),(6,5)]

P(X=6) = 1/36

[Possible Pairs: (6,6)]

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

Standard Deviation is given by SD = √Variance

∴ 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 gives the required products:

Required Probability distribution table:

xipixipixi2pi
111/3611/3611/36
29/3618/361
37/3621/3663/36
45/3620/3680/36
53/3615/3675/36
61/366/361

∴ Mean = 11/36 + 18/36 + 21/ 36 + 20/36 + 15/36 + 6/36 = 91/36

Variance = 11/36 + 1 + 63/36 + 80/36 + 75/36 + 1 – (91/36)2 =2555/1296

Standard deviation = √variance = 1.403

Question 9: A fair coin is tossed four times. Let X denote the number of heads occurring. Find the probability distribution, mean and variance of X.

Solution:

Say, H represents event of getting a head and T represents getting a tail.

When we toss a coin 4 times we have the following possibilities:

{HHHH,HHHT,HHTH,THHH,HTHH,THHT,TTHH,HHTT,THTH…………,TTTT}

A total of 24 = 16 possibilities.

Let X be a random variable representing number of heads occurring in 4 tosses of a coin.

∵ probability of getting a head or probability of getting a tail are independent events and P(GETTING A HEAD) = P(GETTING A TAIL) = 1/2

∴ P(Head in first toss) and P(Head in second toss) and P(head in third toss) and P(tail in 4th toss) can be given by their individual products.

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

Thus,

P(X=0) = P(TTTT) = P(T)P(T)P(T)P(T) = 1/2 x 1/2 x 1/2 x 1/2 = 1/16

Selecting a coin out of 4 which will show head rest all showing tail

= 4C1 x P(HHHT) = 4C1 x(1/2) x (1/2) x (1/2) x (1/2) = 1/4

similarly ,

P(X=2) = 4C2 x(1/2)4 = 3/8

P(X=3) = 4C3 x (1/2)4 = 1/4

P(X=4) = P(HHHH) = 1/16

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 :

xipi xipi xi2pi
01/1600
11/41/41/4
23/83/43/2
31/43/49/4
41/161/41

∴ Mean = 0+ 1/4 + 3/4 + 3/4 + 1/4 = 2

Variance = 0 + 1/4 + 3/2 + 9/4 + 1 – (2) = 1

Question 10: A fair die is tossed. Let X denote twice the number appearing. Find probability distribution, mean and variance of X.

Solution:

When a fair dice is thrown there are total 6 possible outcomes.

∵ X denote twice the number appearing on die

∴ X can take values 2,4,6,8,10 and 12

As appearance of a number on a fair die is equally likely

i.e. P(appearing of 1) = P(appearing of 2) = P(appearing of 3) = P(appearing of 4) = P(appearing of 5) = P(appearing of 6) = 1/6

∴ appearance of twice of the number is also equally likely with a probability of 1/6.

P(X=2)=P(X=4)=P(X=6)=P(X=8)=P(X=10)=P(X=12)=1/6

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 pixiand add them to get mean and apply the above formula to get the variance.

Following table gives the required products:

Required Probability distribution table:

xipixipixi2pi
21/62/64/6
41/64/616/6
61/6136/6
81/64/364/6
101/65/3100/6
121/62144/6

∴ Mean = 2/6 + 4/6 + 1 + 4/3 + 5/3 + 2 = 7

Variance = 4/6 + 16/6 + 36/6 + 64/6 + 100/6 + 144/6 – 72 = 70/6.

Question 11: A fair die is tossed. Let X denote 1 or 3 according as an odd or an even number appears. Find the probability distribution, mean and variance of X.

Solution: 

When a fair dice is thrown there are total 6 possible outcomes.

∵ X denote 1 or 3 according as an odd or an even number appears.

P(appearing of even number on a die) = 3/6 [favourable outcomes {2,4,6}]

P(appearing of an odd number on a die) = 3/6 [favourable outcomes {1,4,3}]

P(X=1) = 3/6 = 1/2

P(X=3) = 3/6 = 1/2

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 gives the required products :

Required Probability distribution table:-

xipixipixi2pi
11/21/21/2
31/23/29/2

∴ Mean = 1/2 + 3/2 = 2

Variance = 1/2 + 9/2 – (2)2 = 1.

Question 12: A fair coin is tossed four times. Let X denote the longest string of heads occurring. Find the probability distribution, mean and variance of X.

Solution:

Say, H represents event of getting a head and T represents getting a tail.

When we toss a coin 4 times we have the following possibilities:

{HHHH,HHHT,HHTH,THHH,HTHH,THHT,TTHH,HHTT,THTH…………,TTTT}

A total of 24 = 16 possibilities.

∵ probability of getting a head or probability of getting a tail are independent events and P(GETTING A HEAD) = P(GETTING A TAIL) = 1/2

∴ P(Head in first toss) and P(Head in second toss) and P(head in third toss) and P(tail in 4th toss) can be given by their individual products.

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

As X is a random variable representing longest string of head occurring in 4 tosses.

∴ X can take following values:

X = 0 [ all tails (TTTT) ]

X = 1 [Longest string contains only 1 head e.g. (HTTT),(TTTH),(HTHT)..]

X = 2 [ Longest string contain only 2 head e.g. (HHTT),(HHTH),(THHT)…]

X = 3 [Longest string contain only 3 head e.g. ( HHHT) And (THHH)]

X = 4 [ Longest string contain 4 heads i.e. (HHHH) ]

Thus,

P(X=0) = 1/16

P(X=1) = 7/16 [by counting number of favourable outcomes as explained]

P(X=2) = 5/16

P(X=3) = 2/16

P(X=4) = 1/16

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 :

xipixipixi2pi
01/1600
17/167/167/16
25/1610/1620/16
32/166/1618/16
41/161/41

∴ Mean = 0 + 7/16 + 10/16 + 6/16 + 1/4 = 1.7

Variance = 0 + 7/16 + 20/16 + 18/16 + 1 – (1.7)2 = 0.935

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 : 

xipi xipixi2pi
01/1600
17/167/167/16
25/1610/1620/16
32/166/1618/16

∴ 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:

yipiyipiyi2pi
10.10.10.1
20.51.02.0
30.41.23.6

∴ 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 :

xipixipixi2pi
01/400
11/21/21/2
21/41/21

∵ 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) =  5C/14C3 = 5/182

So, Probability distribution is given below: 

xipi
03/13
145/91
245/182
35/182

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 :

xipixipi
901/1390/13
-1012/13-120/13

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) = 26C/ 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:

xipixipi
02/170
113/3413/34
213/3426/34
32/176/17

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 :

xipixipixi2pi
010/2100
110/2110/2110/21
21/212/214/21

∴ 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:

xipixipixi2pi
32/306/3018/30
44/3016/3064/30
56/3030/30150/30
68/3048/30288/30
710/3070/30490/30

∴ 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:

xipixipi
-12/3-2/3
42/98/9
92/2718/27
151/2715/27

∵ Mean = ∑xipi

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

He can win around Rs 1.45

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment in the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share rdsharmasolutions.in to your friends.

Leave a Comment

Your email address will not be published.