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.
| Textbook | NCERT |
| Class | Class 12th |
| Subject | Maths |
| Chapter | 32 |
| Exercise | 32.1 |
| Category | RD Sharma Solutions |
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.
| X | 3 | 2 | 1 | 0 | -1 |
| P(X) | 0.3 | 0.2 | 0.4 | 0.1 | 0.05 |
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.
| X | 0 | 1 | 2 |
| P(X) | 0.6 | 0.4 | 0.2 |
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.
| X | 0 | 1 | 2 | 3 | 4 |
| P(X) | 0.1 | 0.5 | 0.2 | 0.1 | 0.1 |
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.
| X | 0 | 1 | 2 | 3 |
| P(X) | 0.3 | 0.2 | 0.4 | 0.1 |
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 | -1 | 0 | 1 | 2 | 3 |
| P(X) | 0.1 | k | 0.2 | 2k | 0.3 | k |
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:
| X | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| P(X) | a | 3a | 5a | 7a | 9a | 11a | 13a | 15a | 17a |
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
| X | 0 | 1 | 2 |
| P(X) | 3c3 | 4c-10c2 | 5c-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
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:
| X | 0 | 1 | 2 | 3 |
| P(X) | k | k/2 | k/4 | k/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:
| X | 0 | 1 | 2 | 3 | 4 |
| P(X) | 0 | k | 4k | 2k | k |
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
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.