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Class 12 RD Sharma Solutions- Chapter 32 Mean and Variance of a Random Variable – Exercise 32.1 | Set 1

  • Last Updated : 25 Jan, 2021

Question 1. Which of the following distributions of probabilities of random variables are their probability distributions?

i.

X3210-1
P(X)0.30.20.40.10.05

Solution:

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We know that the sum of probability distribution is always 1.

Sum of probabilities (P(X))=P(X=3)+P(X=2)+P(X=1)+P(X=0)+P(X=-1)

                                          =0.3+0.2+0.4+0.1+0.05=1.05>1

The sum of probability distribution is not equal to 1. Hence, it is not the probability distribution of the given random variables.

ii.

X012
P(X)0.60.40.2

Solution:

Sum of probabilities (P(X))=P(X=0)+P(X=1)+P(X=2)

                                          =0.6+0.4+0.2=1.2>1



The sum of probability distribution is not equal to 1. Hence, it is not the probability distribution of the given random variables.

iii.

X01234
P(X)0.10.50.20.10.1

Solution:

Sum of probabilities (P(X))=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)

                                          =0.1+0.5+0.2+0.1+0.1=1

The sum of probability distribution is equal to 1. Hence, it is the probability distribution of the given random variables.

iv.

X0123
P(X)0.30.20.40.1

Solution:

Sum of probabilities (P(X))=P(X=0)+P(X=1)+P(X=2)+P(X=3)

                                          =0.3+0.2+0.4+0.1=1



The sum of probability distribution is equal to 1. Hence, it is the probability distribution of the given random variables.

Question 2. A random variable X has the following probability distribution:

X-2-10123
P(X)0.1k0.22k0.3k

Find the value of k.

Solution: 

We know that the sum of probability distribution is always 1.

Sum of probability distribution (P(X))=P(X=-2)+P(X=-1)+P(X=0)+P(X=1)+P(X=2)+P(X=3)=1

            =>0.1+k+0.2+2k+0.3+k=1

            =>0.6+4k=1

            =>4k=1-0.6

            =>k=0.1

Question 3. A random variable X has the following probability distribution:

X012345678
P(X)a3a5a7a9a11a13a15a17a

i. Find the value of a.



Solution: 

We know that the sum of probability distribution is always 1.

Sum of probability distribution (P(X))=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)+P(X=6)+P(X=7)+P(X=8)=1

         =>a+3a+5a+7a+9a+11a+13a+15a+17a=1

         =>81a=1

         =>a=1/81

ii. Find P(X<3).

Solution:

P(X<3)=P(X=0)+P(X=1)+P(X=2)

           =1/81+3/81+5/81

           =9/81=1/9

iii. Find P(X>=3).

Solution:

P(X>=3)=P(X=3)+P(X=4)+P(X=5)+P(X=6)+P(X=7)+P(X=8)

              =7/81+9/81+11/81+13/81+15/81+17/81

               =72/81=8/9

iv. Find P(0<X<5).

P(0<X<5)=P(X=1)+P(X=2)+P(X=3)+P(X=4)

                =3/81+5/81+7/81+9/81

                =24/81=8/27=0.296

Question 4.1. The probability distribution function of a random variable X is given by

X012
P(X)3c34c-10c25c-1

where c>0. Find c.

Solution: 

We know that the sum of probability distributions of a random variable is always 1.

=>3c3 +4c-10c2+5c-1=1

=>3c3-10c2+9c-2=0

Let c=1

3(1)-10(1)+9(1)-2=12-12=0

Therefore c=1.

By horn’s method :



We get a quadratic equation : 3c2-7c+2=0

From this quadratic equation we get,

=>3c2-6c-c+2=0

=>3c(c-2)-1(c-2)=0

=>(3c-1)(c-2)=0

=>3c-1=0;    c-2=0

=>3c=1;     c=2

=>c=1/3;    c=2;    c=1

We know that a single probability distribution cannot be 1 or more than one. So we take c=1/3.

Therefore, c=1/3.

Question 4.2. Find P(X<2).

Solution: 

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

            =3(1/3)3+4(1/3)-10(1/3)2

            =1/9+4/3-10/9

            =1/3=0.33

Question 4.3. Find P(1<X<=2).

Solution: 

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

                  =5(1/3)-1

                  =5/3-1

                 =2/3=0.66

Question 5. Let X be a random variable which assumes values x1, x2,x3,x4  such that 2P(X=x1)=3P(X=x2)=P(X=x3)=5P(X=x4). Find the probability distribution of X.

Solution: 

Sum of probability distributions= P(X=x1)+P(X=x2)+P(X=x3)+P(X=x4)=1

Given,

      2P(X=x1)=3P(X=x2)=P(X=x3)=5P(X=x4)

=>P(X=x2)=2/3P(X=x1) ; P(X=x3)=2/1(P(X=x1) ; P(X=x4)=2/5(P(X=x1)

=>P(X=x1)+2/3(P(X=x1)+2/1(P(X=x1)+2/5(P(X=x1)=1

=>61/15(P(X=x1)=1

=>P(X=x1)=15/61=0.24

=>P(X=x2)=2/3(P(X=x1)

                 =2/3(15/61)

                 =10/61=0.16

=>P(X=x3)=2/1(P(X=x1)

                 =2(15/61)

                 =30/61=0.49

=>P(X=x4)=2/5(P(X=x1))

                 =2/5(15/61)

                 =6/61=0.09

Question 6. A random variable X takes the values 0,1,2 and 3 such that: P(X=0)=P(X>0)=P(X<0) ; P(X=-3)=P(X=-2)=P(X=-1) ; P(X=1)=P(X=2)=P(X=3). Obtain the probability distribution of X.

Solution: 

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

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



Given,

P(X=0)=P(X>0)=P(X<0)

=>P(X=0)+P(X=0)+P(X=0)=1

=>3P(X=0)=1

=>P(X=0)=1/3

=>P(X>0)=1/3

=>P(X=1)+P(X=2)+P(X=3)=1/3

Given,

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

=>3P(X=1)=1/3

=>P(X=1)=1/9 ; P(X=2)=1/9 ; P(X=3)=1/9

=>P(X<0)=1/3

=>P(X=-1)+P(X=-2)+P(X=-3)=1/3

Given,

P(X=-3)=P(X=-2)=P(X=-1)

=>3P(X=-1)=1/3

=>P(X=-1)=1/9 ; P(X=-2)=1/9 ; P(X=-3)=1/9

Question 7. Two cards are drawn from a well shuffled deck of 52 cards. Find the probability distribution of the number of aces.

Solution: 

Given that two cards are drawn from a well shuffled deck of 52 cards.

Then the random variables for the probability distribution of the number of aces could be 

i. No ace is drawn

ii. One ace is drawn

iii. Two aces are drawn

i. No ace is drawn:

P(X=0)=52-4C2/52C2

              =48C2/52C2

          =48!/2!x46!/52!/2!x50!

          =48×47/52×51

          =188/221

          =0.85

ii. One ace is drawn:

P(X=1)=4C1x48C1/52C2

           =4x48x2/52×51

           =32/221

           =0.14

iii. Two aces are drawn:

P(X=2)=4C2/52C2

           =6×2/52×51

           =1/221

           =0.004

Question 8. Find the probability distribution of number of heads, when three coins are tossed.

Solution: 

Given that three coins are tossed simultaneously.

Then the random variables for the probability distribution of the number of heads could be,

i. No heads

ii. One head

iii. Two heads

iv. Three heads

i. No heads:

P(X=0)=1C1x1C1x1C1/ 2C1x 2C1x 2C1

           =1x1x1/2x2x2

           =1/8

           =0.125

ii. One head:

P(X=1)=1C1+1C1+1C1/8

          =1+1+1/8

          =3/8

          =0.37

iii. Two heads:

P(X=2)=1C1+1C1+1C1/8

           =3/8



           =0.37

iv. Three heads:

P(X=3)=1/8

           =0.125

Question 9. Four cards are drawn simultaneously from a well shuffled pack of 52 playing cards. Find the probability distribution of the number of aces.

Solution: 

Given that four cards are drawn simultaneously from a well shuffled pack of 52 cards.

Then the random variables for the probability distribution of the number of aces drawn could be,

i. No aces

ii. One ace

iii. Two aces

iv. Three aces

v. Four aces

i. No aces

P(X=0)=48C4/ 52C4

           =48x47x46x45/49x50x51x52

           =0.71

ii. One ace

P(X=1)=4C1x48C3/52C4

           =4x48x47x46x4/49x50x51x52

           =0.25

iii. Two aces

P(X=2)= 4C2x48C2/ 52C4

           =6x48x47x12/49x50x51x52

           =0.024

iv. Three aces

P(X=3)= 4C3x48C1/52C4

           = 4x48x24/49x50x51x52

           = 0.0007

v. Four aces

P(X=4)=4C4/ 52C4

           =1/ 270725

           =0.000003694

Question 10. A bag contains 4 red and 6 black balls. Three balls are drawn at random. Find the probability distribution of the number of red balls.

Solution: 

Given that three balls are drawn at random from a bag.

Then the value of random variable for the probability distribution of number of red balls could be,

i. No red ball

ii. One red ball

iii. Two red balls

iv. Three red balls

i. No red balls:

P(X=0)=6C3/10C3

           =6x5x4/10x9x8

           =1/6=0.16

ii. One red ball:

P(X=1)=6C2x4C1/10C3

           =6x5x4x3/10x9x8

           =1/2=0.5

iii. Two red balls:

P(X=2)=6C1x4C2/10C3

           =6x4x3x3/10x9x8

           =3/10=0.3

iv. Three red balls:

P(X=3)=4C3/10C3

           =4x3x2/10x9x8

           =1/30=0.03

Question 11. Five defective mangoes are accidentally mixed with 15 good ones. Four mangoes are drawn at random from this lot. Find the probability distribution of the number of defective mangoes.

Solution: 

Given that five defective mangoes are mixed with 15 good ones.

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

i. No defective

ii. One defective



iii. Two defective

iv. Three defective

v. Four defective

i. No defective:

P(X=0)=15C4/20C4

           =15x14x13x12/20x19x18x17

           =91/323=0.28

ii. One defective:

P(X=1)=15C3x5C1/20C4

           =15x14x13x5x4/20x19x18x17

           =455/969=0.469

iii. Two defective:

P(X=2)=15C2x5C2/20C4

           =15x14x5x4x6/20x19x18x17

           =70/323=0.21

iv. Three defective:

P(X=3)=15C1x5C3/20C4

           =15x5x4x3x4/20x19x18x17

           =10/323=0.03

v. Four defective:

P(X=4)=5C4/20C4

           =5x4x3x2/20x19x18x17

           =1/969=0.001

Question 12. Two dice are thrown together and the number appearing on them is noted. X denotes the sum of the two numbers. Assuming that all the 36 outcomes are equally likely, what is the probability distribution of X?

Solution: 

Given that two dice are thrown simultaneously.

Then the outcomes would be as follows:

(1,1) ; (1,2) ; (1,3) ; (1,4) ; (1,5) ; (1,6) ;

(2,1) ; (2,2) ; (2,3) ; (2,4) ; (2,5) ; (2,6) ;

(3,1) ; (3,2) ; (3,3) ; (3,4) ; (3,5) ; (3,6) ;

(4,1) ; (4,2) ; (4,3) ; (4,4) ; (4,5) ; (4,6) ;

(5,1) ; (5,2) ; (5,3) ; (5,4) ; (5,5) ; (5,6) ;

(6,1) ; (6,2) ; (6,3) ; (6,4) ; (6,5) ; (6,6) 

The values of the random variable could be: 2,3,4,5,6,7,8,9,10,11,12

P(X=2)=1/36=0.02

P(X=3)=2/36=1/18=0.05

P(X=4)=3/36=1/12=0.08

P(X=5)=4/36=1/9=0.11

P(X=6)=5/36=0.13

P(X=7)=6/36=1/6=0.16

P(X=8)=5/36=0.13

P(X=9)=4/36=1/9=0.11

P(X=10)=3/36=1/12=0.08

P(X=11)=2/36=1/18=0.05

P(X=12)=1/36=0.02

Question 13. A class has 15 students whose ages are 14,17,15,14,21,19,20,16,18,17,20,17,16,19 and 20 years respectively. One student is selected in such a manner that each has the same chance of being selected and the age X of the selected student is recorded. What is the probability distribution of the random variable X?

Solution: 

Given that the students are selected without any bias.

Then the values of the random variable X could be: 14,15,16,17,18,19,20,21

P(X=14)=2/15=0.13

P(X=15)=1/15=0.06

P(X=16)=2/15=0.13

P(X=17)=3/15=0.2

P(X=18)=1/15=0.06

P(X=19)=2/15=0.13

P(X=20)=3/15=0.2

P(X=21)=1/15=0.06

Question 14. Five defective bolts are accidentally mixed with twenty good ones. If four bolts are drawn at random from this lot, find the probability distribution of the number of defective bolts.

Solution: 

Given that five defective bolts are mixed with 20 good ones.

Then the values of random variable for the probability distribution would 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

ii. One defective:

P(X=1)=20C3x5C1/25C4

           =20x19x18x5x4/25x24x23x22



           =114/253=0.45

iii. Two defective:

P(X=2)=20C2x5C2/25C4

           =20x19x5x4x6/25x24x23x22

           =38/253=0.15

iv. Three defective:

P(x=3)=20C1x5C2/25C4

           =20x5x4x4x3/25x24x23x22

           =4/253=0.015

v. Four defective:

P(X=4)=5C4/25C4

           =5x4x3x2/25x24x23x22

           =1/2530=0.0004

Question 15. Two cards are drawn successively with replacement from well shuffled pack of 52 cards. Find the probability distribution of number of aces.

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 ace

ii. One ace

iii. Two aces

i. No ace:

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

           =144/169=0.85

ii. One ace:

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

           =24/169=0.14

iii. Two aces:

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

           =1/169=0.005




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