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

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

Table of Contents

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

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:

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.

=>3c+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)=1C1x1C1x1C12C12C12C1

           =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

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:

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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