Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

NCERT Exemplar for Class 12 Maths Chapter-13 (Book Solutions)

ffImage
banner

NCERT Exemplar for Class 12 Maths - Probability - Free PDF Download

Free PDF download of NCERT Exemplar for Class 12 Maths Chapter 13 - Probability solved by expert Maths teachers on Vedantu.com as per NCERT (CBSE) Book guidelines. All Chapter 13 - Probability Exercise questions with solutions to help you to revise complete syllabus and score more marks in your Examinations.

Popular Vedantu Learning Centres Near You
centre-image
Sharjah, Sharjah
location-imgKing Abdul Aziz St - Al Mahatta - Al Qasimia - Sharjah - United Arab Emirates
Visit Centre
centre-image
Abu Dhabi, Abu-Dhabi
location-imgMohammed Al Otaiba Tower - 1401, 14th Floor - opposite to Nissan Showroom West Zone building - Al Danah - Zone 1 - Abu Dhabi - United Arab Emirates
Visit Centre
centre-image
22 No Phatak, Patiala
location-img#2, Guhman Road, Near Punjabi Bagh, 22 No Phatak-Patiala
Visit Centre
centre-image
Chhoti Baradari, Patiala
location-imgVedantu Learning Centre, SCO-144 -145, 1st & 2nd Floor, Chotti Baradari Scheme Improvement Trust, Patiala-147001
Visit Centre
centre-image
Janakpuri, Delhi
location-imgVedantu Learning Centre, A-1/173A, Najafgarh Road, Opposite Metro Pillar 613, Block A1, Janakpuri, New Delhi 110058
Visit Centre
centre-image
Tagore School, Gudha-Gorji
location-imgTagore Public School Todi, Gudha Gorji, Jhunjhunu, Rajasthan 333022
Visit Centre
View More
Courses
Competitive Exams after 12th Science

Access NCERT Exemplar Solutions for Class 12 Mathematics Unit 13 – Probability

Solved Examples

Short Answer Questions

1. A and B are two candidates seeking admission in a college. The probability that A is selected is 0.7 and the probability that exactly one of them is selected is 0.6. Find the probability that B is selected.

Ans: Let p be the probability that B gets selected.

P(A is selected )=0.7 

Than, P(A)=1P(A)=10.7=0.3

And P(Exactly one of A,B is selected)=0.6

P(A is selected, B is not selected; B is selected, A is not selected )=0.6

P(AB)+P(AB)=0.6 

P(A)P(B)+P(A)P(B)=0.6

(0.7)(1p)+(0.3)p=0.6

p=0.25

Thus, the probability that B gets selected is 0.25.


2. The probability of simultaneous occurrence of at least one of two events A and B is p. If the probability that exactly one of A,B occurs is q, then prove that

P(A)+P(B)=22p+q .

Ans: Given, P (exactly one of A, B occurs) = q, we get

P(AB)P(AB)=q

pP(AB)=q

P(AB)=pq

1P(AB)=pq

P(AB)=1p+q

P(A)+P(B)P(AB)=1p+q

P(A)+P(B)=(1p+q)+P(AB)

P(A)+P(B)=(1p+q)+(1P(AB))

P(A)+P(B)=(1p+q)+(1p)

P(A)+P(B)=22p+q


3. 10% of the bulbs produced in a factory are of red colour and 2% are red and defective. If one bulb is picked up at random, determine the probability of its being defective if it is red.

Ans: Let A and B be the events that the bulb is red and defective respectively.

P(A)=10100=110

P(AB)=2100=150

P(BA)=P(AB)P(A)=150×101=15

So, the probability of its being defective, if it is red, is 15.


4. Two dice are thrown together. Let A be the event getting 6 on the first die and B be the event getting 2 on the second die. Are events A and B independent?

Ans: A={(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}

B={(1,2),(2,2),(3,2),(4,2),(5,2),(6,2)}

AB={(6,2)}

P(A)=636=16,P(B)=16

P(AB)=136

A and B will be independent if, P(AB)=P(A)P(B)

LHS=P(AB)=136

RHS=P(A)P(B)=16×16=136

LHS=RHS

So, A and B are independent events.


5. A committee of 4 students is selected at random from a group consisting 8 boys and 4 girls. Given that there is at least one girl in the committee, calculate the probability that there are exactly 2 girls in the committee.

Ans:  Let A denote the event that at least one girl will be chosen, and B the event that exactly 2 girls will be chosen. We require P(BA).

Because A denotes the event that at least one girl will be chosen, A' denotes that no girl is chosen, i.e., 4 boys are chosen. Then

P(A)=8C412C4

=70495=1499[nCr=n!r!(nr)!]

Than, P(A)=11499=8599

Now P(AB)=P(2 boys and 2girls)=8C24C212C4

=6×28495=56165

So, P(BA)=P(AB)P(A)=56165×9985=168425 


6. Three machines E1,E2,E3 in a certain factory produce 50%,25% and 25%, respectively, of the total daily output of electric tubes. It is known that 4% of the tubes produced one each of machines E1 and E2 are defective, and that 5% of those produced on E3 are defective. If one tube is picked up at random from a day's production, calculate the probability that it is defective.

Ans: Let D be the event that the picked up tube is defective.

Let A1,A2 and A3 be the events that the tube is produced on machines E1,E2 and E3, respectively.

P(D)=P(A1)P(DA1)+P(A2)P(DA2)+P(A3)P(DA3)..(1)

Given, P(A1)=50100=12,P(A2)=14,P(A3)=14 

Also P(DA1)=P(DA2)=4100=125

And,P(DA3)=5100=120

Putting these values in (1), we get

P(D)=12×125+14×125+14×120

=150+1100+180=17400=.0425


7. Find the probability that in 10 throws of a fair die a score which is a multiple of 3 will be obtained in at least 8 of the throws.

Ans: Given that, obtaining multiple of 3 in a throw of a die is a success.

There are total 10 throws of a fair die.

Let p be the probability of getting multiple of 3 in a throw of a die.

Since there can be a total 6 outcomes in a throw of a die. That is, {1,2,3,4,5,6}.

And a multiple of 3 in a die is {3},{6}.

p=26

p=13

Then, let q be the probability of not getting a multiple of 3 in a throw of a die.

And we know, p+q=1

q=1p

q=113

q=313

q=23

Let X be a random variable representing a number of successes (getting multiple of 3 in die) out of 10 throws of a die.

Then, the probability of getting r successes out of n throws of die is given by,

P(X=r)=nCrprqnr

Here n=10.

Now, put values of n,p, and q in the above equation.

P(X=r)=10Cr(13)r(23)10r (i) 

We need to find the probability of getting a multiple of 3 in at least 8 of the throws out of 10 throws of a fair die.

It is given,

Probability =P(X8)

This can be written as, P(X8)=P(X=8)+P(X=9)+P(X=10)

Just out r=8,9,10 in equation (i) to find the value of P(X=8),P(X=9),P(X=10) respectively, then substitute in the above equation.

P(X8)=10C8(13)8(23)108+10C9(13)9(23)109+10C10(13)10(23)1010

P(X8)=10C8(13)8(23)2+10C9(13)9(23)1+10C10(13)10(23)0

P(X8)=[(10!(108)!8!)×(22310)]+[(10!(109)!9!)×(2310)]+[(10!(1010)!10!)×(13)10]

P(X8)=[(10!2!8!)×4×(13)10]+[(10!9!)×2×(13)10]+[(10!10!)×(13)10]

P(X8)=(13)10[(10×9×8!2×8!×4)+(10×9!9!×2)+1]

P(X8)=(13)10[180+20+1]

P(X8)=201×(13)10

P(X8)=201310


8. A discrete random variable X has the following probability distribution:

X

1

2

3

4

5

6

7

P(X)

C

2C

2C

3C

C2

2C2

7C2+C


Find the value of C. Also find the mean of the distribution.

Ans: Since pi=1, we have

C+2C+2C+3C+C2+2C2+7C2+C=1

or, 10C2+9C1=0

or ,(10C1)(C+1)=0

Either (10C1)=0 or (C+1)=0

C=110,C=1 

C=1 cannot be possible.probability cannot be negative.

So, the permissible value of C=110

Mean =i=1nxipi=i=17xipi

i=17xipi=1×110+2×210+3×210+4×310+5(110)2+6×2(110)2+7(7(110)2+110)

i=17xipi=110+410+610+1210+5100+12100+49100+710

i=17xipi=3.66.


Long Answer

9. 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, find the probability distribution of X.

Ans: Since 4 balls have to be drawn, therefore, X can take the values 0,1,2,3,4.

P(X=0)=P( no red ball )=P(4 white balls )

=4C412C4=1495

P(X=1)=P(1 red ball and 3 white balls )

=8C1×4C312C4=32495

P(X=2)=P(2 red balls and 2 white balls )

=8C2×4C212C4=168495

P(X=3)=P(3 red balls and 1 white ball ) 

=8C3×4C112C4=224495

P(X=4)=P(4 red balls )=8C412C4=70495

So, the following is the required probability distribution of X


X

0

1

2

3

4

P(X)

1495

32495 

168495

224495

70495


10. Determine variance and standard deviation of the number of heads in three tosses of a coin.

Ans:  Let X denote the number of heads tossed. So, X can take the values 0,1,2,3. When a coin is tossed three times, we get

Sample space S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT }

P(X=0)=P( no head )=P(TTT)=18

P(X=1)=P( one head )=P(HTT,THT,TTH)=38

P(X=2)=P( two heads )=P(HHT,HTH,THH)=38

P(X=3)=P( three heads )=P(HHH)=18

So, the probability distribution of X is: 

X

0

1

2

3

P(X)

18

38

38

18


Variance of X=σ2=xi2piu2...(1)

where μ is the mean of X given by

μ=xipi=0×18+1×38+2×38+3×18

=32......(2)

Now xi2pi=02×18+12×38+22×38+32×18=3....(3)

From (1), (2) and (3), we get

σ2=3(32)2=34

And ,standard deviation =σ2=34=32.


11. Refer to Example 6. Calculate the probability that the defective tube was produced} on machine E1.

Ans: Now, we have to find P(A1/D)

P(A1/D)=P(A1D)P(D)=P(A1)P(D/A1)P(D)

P(A1/D)=12×12517400=817


12. A car manufacturing factory has two plants, X and Y. Plant X manufactures 70% of cars and plant Y manufactures 30%. 80% of the cars at plant X and 90% of the cars at plant Y are rated of standard quality. A car is chosen at random and is found to be of standard quality. What is the probability that it has come from plant X?

Ans: Let E be the event that the car is of standard quality. Let B1 and B2 be the events that the car is manufactured in plants X and Y, respectively. 

P(B1)=70100=710,P(B2)=30100=310

P(EB1)= Probability that a standard quality car is manufactured in plant X

P(EB1)=80100=810

And, P(EB2)=90100=910

P(B1E)= Probability that a standard quality car has come from plant X

P(B1E)=P(B1)×P(EB1)P(B1)P(EB1)+P(B2)P(EB2)

P(B1E)=710×810710×810+310×910=5683

Hence, the required probability is 5683.


Objective Type Questions

Choose the correct answer from the given four options in each of the Examples 13 to 17.

13. Let A and B be two events. If P(A)=0.2,P(B)=0.4,P(AB)=0.6, then P(AB) is equal to

(A) 0.8

(B) 0.5

(C) 0.3

(D) 0

Ans: Correct answer - D 

From the given data P(A)+P(B)=P(AB).

This means, P(AB)=0. Thus, P(AB)=P(AB)P(B)=0


14. Let A and B be two events such that P(A)=0.6,P(B)=0.2, and P(AB)=0.5.

Then P(AB) equals

(A) 110

(B) 310

(C) 38

(D) 67

Ans: Correct answer - C

P(AB)=P(AB)P(B)

P(AB)=0.5×0.2=0.1

P(AB)=P(AB)P(B)=1P(AB)1P(B)

P(AB)=1P(A)P(B)+P(AB)10.2=38


15. If A and B are independent events such that 0<P(A)<1 and 0<P(B)<1, then which of the following is not correct?

(A) A and B are mutually exclusive

(B) A and B are independent

(C) A and B are independent

(D) A and B are independent

Ans: Correct answer - A

We have A and B are independent events,

Hence P(AB)=P(A)

Hence P(AB)P(B)=P(A)

Multiplying both sides by P(B), we get

P(AB)=P(A)P(B)

Two events A and B are said to be mutually exclusive if the sets A and B are disjoint.

Consider two events A and B such that A has non-zero probability and B=S.

Since AS, we have AB=Aϕ. Hence the events are not mutually exclusive.

However, P(AB)=P(A) and P(A)P(B)=P(A)

Hence the events are independent but not mutually exclusive. Hence option a is incorrect.

Now if A and B are independent then, P(AB)=P(A)P(B)

Now we know that A=A(BB)=(AB)(AB)

Hence P(A)=P((AB)(AB))

Since the events (AB) and (AB) are disjoint, these events are mutually exclusive Hence P(A)=P((AB)(AB))=P(AB)+P(AB)

Hence, we have

P(A)=P(A)P(B)+P(AB) P(AB)=P(A)(1P(B)) We know that P(B)=1P(B)

Hence we have P(AB)=P(A)P(B)

Hence the events A and B' are independent,

Since A and B are independent, we have A ' and B ' are also independent.

Hence options b and c are correct.

Now we know that P(A/B)=P(AB)P(B)

Hence, we have

P(A/B)+P(A/B)=P(AB)P(B)+P(AB)P(B)=P(AB)+P(AB)P(B)

Since AB and AB are mutually exclusive events, we have

P(AB)+P(AB)=P((AB)(AB))=P((AA)B)=P(B)

Hence, we have

P(A/B)+P(A/B)=P(B)P(B)=1

Hence option d is correct.

Therefore, the option that is incorrect is option A.


16. Let X be a discrete random variable. The probability distribution of X is given below:

X

30

10

-10

P(X)

15

310

12


Then E(X) is equal to

(A) 6

(B) 4 

(C) 3

(D) 5

Ans: Correct answer - B

Because we know that E(X)=i=1nxipi

So, E(X)=30×15+10×31010×12=4


17. Let X be a discrete random variable assuming values x1,x2,,xn with probabilities P1,P2,.Pn, respectively. Then variance of X is given by

(A) E(X2)

(B) E(X2)+E(X)

(C) E(X2)[E(X)]2

(D) E(X2)[E(X)]2

Ans: Correct answer - C

As we know that probability of X is equal to mean (μ) of X.

The probability of X and X2 is

μ=E(X)=i=1nxipi

E(X2)=i=1nxi2pi

The variance of X(σ) when it is discrete is given by the addition of square of the mean of X (μ) and the probability of X2.

σ2=i=1n(xiμ)2pi=i=1nxi2piμ2 or σ=E(Xμ)2

The standard deviation of the random variable X is defined as σ2= variance (X)

variance(X)=i=1nxi2pi+[i=1nxipi]2 

variance(X)=E(X2)+[E(X)]2


Fill in the blanks in Examples 18 and 19

18. If A and B are independent events such that P(A)=p,P(B)=2p and P (Exactly one of A,B)=59, then p=.....

p=13=512[(1p)(2p)+p(12p)=3p4p2=59]


19. If A and B are independent events then P(AB)=1......

Ans: P(AB)=1P(AB)=1P(A)P(B)

(since A and B are independent).


State whether each of the statement in Examples 20 to 22 is True or False 

20. Let A and B be two independent events. Then P(AB)=P(A)+P(B).

Ans: False, because P(AB)=P(A)P(B) when events A and B are independent.


21. Three events A,B and C are said to be independent if

P(ABC)=P(A)P(B)P(C).

Ans: False. Reason is that A,B,C will be independent if they are pairwise independent and P(ABC)=P(A)P(B)P(C).


22. One of the conditions of Bernoulli trials is that the trials are independent of each other.

Ans: True. 


Exercise

Short Answer Questions

1. For a loaded die, the probabilities of outcomes are given as under:  

P(1)=P(2)=0.2,P(3)=P(5)=P(6)=0.1 and P(4)=0.3.

The die is thrown two times. Let A and B be the events,'same number each time' and 'a total score is 10 or more',respectivly. Determine whether or not A and B are independent.

Ans: Given 

P(1)=P(2)=0.2,

P(3)=P(5)=P(6)=0.1 

and P(4)=0.3

Die is thrown two times

A = Same number each time and B= Total score is 10 or more

So, P(A)=[P(1,1)+P(2,2)+P(3,3)+P(4,4)+P(5,5)+P(6,6)]

=[P(1)P(1)+P(2)P(2)+P(3)P(3)+P(4)P(4)+P(5)P(5)+P(6)P(6)]

=[0.2×0.2+0.2×0.2+0.1×0.1+0.3×0.3+0.1×0.1+0.1×0.1]

=0.04+0.04+0.01+0.09+0.01+0.01=0.20 

Now B={(4,6),(6,4),(5,5),(5,6),(6,5),(6,6)}

P(B)=P(4,6)+P(6,4)+P(5,5),P(5,6),(6,5),(6,6)}

=P(4)P(6)+P(6)P(4)+P(5)P(5)+P(5)P(6)+P(6)P(5)+P(6)P(6)

=0.3×0.1+0.1×0.3+0.1×0.1+0.1×0.1+0.1×0.1+0.1×0.1

=0.03+0.03+0.01+0.01+0.01+0.01=0.10

And, AB={(5,5),(6,6)}

P(AB)=P(5,5)+P(6,6)=P(5)P(5)+P(6)P(6)

=0.1×0.1+0.1×0.1=0.01+0.01=0.02

 A and B both will be  independent events, if 

P(AB)=P(A)P(B).

Here, P(AB)=0.02 and P(A)P(B)=0.20×0.10=0.02

So, P(AB)=P(A)P(B)=0.02

Hence, A and B are independent events.


2. Refer to Exercise 1 above. If the die were fair, determine whether or not the events A and B are independent.

Ans:  We have

A={(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)}

n(A)=6 and n(S)=62=36

So, P(A)=n(A)n(B)=636=16

and B={(4,6),(6,4),(5,5),(6,5),(5,6),(6,6)}

n(B)=6 and n(S)=62=36 

So, P(B)=n(B)n(S)=636=16

AB={(5,5),(6,6)}

n(AB)=2 and n(S)=36

So, (AB)=236=118

and, P(A)P(B)=136

So,P(AB)P(A)P(B)

Hence, A and B are not independent events.


3. The probability that at least one of the two events A and B occurs is 0.6. If A and B occur simultaneously with probability 0.3, evaluate P(A)+P(B).

Ans: We know that, AB denotes the occurrence of at least one of A and B and AB denotes the occurrence of both A and B, simultaneously.

Thus, P(AB)=0.6 and P(AB)=0.3

Also, P(AB)=P(A)+P(B)P(AB)

0.6=P(A)+P(B)0.3

P(A)+P(B)=0.9

[1P(A¯)]+[1P(B¯)]=0.9[P(A)=1P(A¯) and P(B)=1P(B¯)]

P(A¯)+P(B¯)=20.9=1.1


4. A bag contains 5 red marbles and 3 black marbles. Three marbles are drawn one by one without replacement. What is the probability that at least one of the three marbles drawn be black, if the first marble is red? 

Ans: Let R stand for red marbles and B is for black.

If at least one of the three marbles drawn be black, if the first marble is red.

(i) E1=Second ball is black and third ball is red .

(ii)E2= Second ball is black and third ball is also black .

(iii)E3= Second ball is red and third ball is black .

P(E1)=P(R1)P(B1R1)P(R2R1B1)=583746=60336=528

P(E2)=P(R1)P(B1R1)P(B2R1B1)=583726=30336=556

And P(E3)=P(R1)P(R2R1)P(B1R1R2)=584736=60336=528

P(E)=P(E1)+P(E2)+P(E3)=528+556+528

P(E)=10+5+1056=2556

So, the probability=2556


5. Two dice are thrown together and the total score is noted. The events E,F and G are 'a total of 4', 'a total of 9 or more', and 'a total divisible by 5', respectively. Calculate P(E),P(F) and P(G) and decide which pairs of events, if any, are independent.

Ans: Two dice are thrown together 

SO,n(S)=36

E= A total of 4={(2,2),(3,1),(1,3)}

so, n(E)=3

and F=A total of 9 or more 

={(3,6),(6,3),(4,5),(4,6),(5,4),(6,4),(5,5),(5,6),(6,5),(6,6)} 

so, n(F)=10

and G= a total divisible by 5={(1,4),(4,1),(2,3),(3,2),(4,6),(6,4),(5,5)}

so, n(G)=7

Here,we see that  (EF)=ϕ and (EG)=ϕ

and (FG)={(4,6),(6,4),(5,5)}

n(FG)=3 and (EFG)=ϕ

P(E)=n(F)n(S)=336=112

P(F)=n(F)n(S)=1036=518

P(G)=n(G)n(S)=736

P(FG)=336=112

and P(F)P(G)=518736=35648

Since,  P(FG)P(F)P(G)

Hence, there is no pair which is independent.


6. Explain why the experiment of tossing a coin three times is said to have binomial distribution.

Ans: We know that, a random variable X takes values 0,1,2,,n is said to be binomial distribution having parameters n and p, if its probability is given by

P(X=r)=nCrprqnr

Where q=1p,and r=0,1,2,,n

Similarly, in case tossing a coin 3 times, 

n=3 and X has values r=0,1,2 and 3 with p=12 and q=12.

Hence, it is said to have a binomial distribution.


7. If A and B are two events such that P(A)=12,P(B)=13 and P(AB)=14. Find:

(i) P(AB)

Ans: We have, P(A)=12,P(B)=13 and P(AB)=14

P(AB)=P(AB)P(B)=1413=34


(ii) P(BA)

Ans: We have, P(A)=12,P(B)=13 and P(AB)=14

P(BA)=P(AB)P(A)=1412=12 


(iii) P(AB)

Ans: P(AB)=1P(AB)=134=14


(iv) P(AB)

Ans: P(AB)=P(AB)P(B)=1P(AB)1P(B)

=1[P(A)+P(B)P(AB)]1P(B)

=1[12+1314]113=1142423=102423=58


8. Three events A, B and C have probabilities 25,13 and 12 respectively. Given that P(AB)=15 and P(BC)=14 find the value of P(CB) and P(AC).

Ans: We have,P(A)=25,P(B)=13,P(C)=12

P(AC)=15 and P(BC)=14

P(C/B)=P(BC)P(B)=1413=34

P(AC)=1P(AC)=1[P(A)+P(C)P(AC)]

P(AC)=1[25+1215]=1[4+5210]=1710=310


9. Let E1 and E2 be two independent events such that P(E1)=p1 and P(E2)=P2.

Describe in words of the events whose probabilities are:

(i) P1P2

Ans: Here, P(E1)=p1 and P(E2)=p2

So,P1P2P(E1)P(E2)=P(E1E2)

So, E1 and E2 occur.


(ii) (1p1)p2

Ans: (1P1)P2=P(E1)P(E2)=P(E1E2)

So, E1 does not occur but E2 occurs.


(iii) 1(1p1)(1p2)

Ans: 1(1P1)(1P2)=1P(E1)P(E2)=1P(E1E2)

=1[1P(E1E2)]=P(E1E2)

So, either E1 or E2 or both E1 and E2 occurs.


(iv) p1+p22p1p2

Ans: P1+P22P1P2=P(E1)+P(E2)2P(E1)P(E2)

=P(E1)+P(E2)2P(E1E2)

=P(E1E2)P(E1E2)

So, either E1 or E2 occurs but not both.


10. A discrete random variable X has the probability distribution given as below:

X

0.5

1

1.5

2

P(X)

k

k2

2k2

k


(i) Find the value of k

Ans: =1nPi=1, where Pi0 

P1+P2+P3+P4=1 

k+k2+2k2+k=1 

3k2+2k1=0 

3k2+3kk1=0 

3k(k+1)1(k+1)=0 

(3k1)(k+1)=0 

k=13 and k=1  

but k is 0k=13


(ii) Determine the mean of the distribution.

Ans: Mean of the distribution 

E(X)=i=1nx,pi=0.5(k)+1(k2)+1.5(2k2)+2k=4k2+2.5k =419+2.513[k=13] =4+7.59=2318


11. Prove that

(i) P(A)=P(AB)+P(AB)

Ans: To prove P(A)=P(AB)+P(AB¯)

RHS=P(AB)+P(AB¯)

=P(A)P(B)+P(A)P(B¯)

=P(A)[P(B)+P(B¯)]

=P(A)[P(B)+1P(B)][P(B¯)=1P(B)]

=P(A)=LHS Hence proved.


(ii) P(AB)=P(AB)+P(AB¯)+P(A¯B)

Ans: To prove, P(AB)=P(AB)+P(AB¯)+P(A¯B)

RHS=P(A)P(B)+P(A)P(B¯)+P(A¯)P(B)

=P(A)P(B)+P(A)[1P(B)]+[1P(A)]P(B) 

=P(A)P(B)+P(A)P(A)P(B)+P(B)P(A)P(B)

=P(A)+P(B)P(A)P(B)

=P(A)+P(B)P(AB)

=P(AB)=LHS 


12. If X is the number of tails in three tosses of a coin ,determine the standard deviation of X.

Ans: Given that,X=0,1,2,3

P(Xx)=nCz(p)xqnx,P(X=r)=nCr(p)r(q)nr

where n=3,p=12,q=12 

and x=0,1,2,3,n=3,p=12,q=12,r=0,1,2,3

X

0

1

2

3

P(X)

18

38

38

18

XP(X)

0

34

38

38

X2P(X)

0

38

32

98


We know that, Var(X)=E(X2)[E(X)]2 Eq...(i)

Where, E(X2)=i=1nxi2P(xi) and E(X)=i=1nxPP(x)

E(X2)=i=1nxi2P(Xi)=0+38+32+98=248=3

And [E(X)]2=[i=1nx2P(x)]2=[0+38+34+38]2=[128]2=94

Var(X)=394=34[ using Eq. ( i) )

Standard deviation of X=Var(X)=34=32


13. In a dice game, a player pays a stake of Re1 for each throw of a die. She receives Rs 5 if the die shows a3,Rs2 if the die shows a1 or 6 , and nothing otherwise. What is the player's expected profit per throw over a long series of throws?

Ans: Let X be the random variable of profit per throw.

X

-1

1

4

P(X)

12

13

16


Since, she loses Rs1 for getting any of 2,4 or 5 .

so, P(X=1)=16+16+16=36=12

P(X=1)=16+16=13 [die shows1 or 6 ]

P(X=4)=16 [ die shows a 3]

Player's expected profit =XP(X)

=1×12+1×13+4×16

=3+2+46=36=12=0.50


14. Three dice are thrown at the same time. Find the probability of getting three two's, if it is known that the sum of the numbers on the dice was six.

Ans: The dice is thrown three times 

sample space [n(S)]=63=216

Let E1 =event of sum of numbers on the dice was six 

and E2 =event of three two's.

n(E1)=10

E1={(1,1,4),(1,2,3),(1,3,2),(1,4,1),(2,1,3),(2,2,2),(2,3,1),(3,1,2),(3,2,1),(4,1,1)}

and  E2={2,2,2}

n(E2)=1

Also, (E1E2)=1

P(E2E1)=P(E1E2)P(E1)=121610216=110


15. suppose 10,000 tickets are sold in a lottery each for Re 1 . First prize is of Rs 3000 and the second prize is of Rs. 2000 . There are three third prizes of Rs. 500 each. If you buy one ticket, what is your expectation?

Ans: Let X be   the random variable where X=0,500,2000and3000

X

0

500

2000

3000

P(X)

999510000

310000

110000

110000


E(X)=XP(X)

=0×999510000+150010000+200010000+300010000

=1500+2000+300010000

=650010000=1320=0.65


16. A bag contains 4 white and 5 black balls. Another bag contains 9 white and 7 black balls. A ball is transferred from the first bag to the second and then a ball is drawn at random from the second bag. Find the probability that the ball drawn is white.

Ans: Let, W1={4 white balls }

B1={5 black balls }

And W2={9 white balls } , B2={7 black balls }

Let E1 = event that ball transferred from the first bag is white 

and E2 = event that the ball transferred from the first bag is black.

and E = event that the ball drawn from the second bag is white.

P(EE1)=1017P(EE2)=917

P(E1)=49 

and P(E2)=59

P(E)=P(E1)P(EE1)+P(E2)P(EE2)

=491017+59917 =40+45153=85153=59


17. Bag I contains 3 black and 2 white balls, Bag II contains 2 black and 4 white balls. A bag and a ball is selected at random. Determine the probability of selecting a black ball.

Ans: Given, Bag I ={3B,2W}

and  Bag II ={2B,4W}

Let E1 = event that bag I is selected 

E2 = event that bag II is selected

And E = event that a black ball is selected 

P(E1)=12,P(E2)=12,P(EE1)=35,P(EE2)=26=13

P(E)=P(E1)P(EE1)+P(E2)P(EE2)

=1235+1226=310+212

=18+1060=2860=715


18. A box has 5 blue and 4 red balls. one ball is drawn at random and not replaced. Its colour is also not noted. Then another ball is drawn at random. What is the probability of the second ball being blue?

Ans:Given, A box = (5 blue, 4 red) 

Let E1 = event that first ball drawn is blue 

E2 = event that first ball drawn is red 

and E = event that the second ball drawn is blue. 

P(E)=P(E1)P(EE1)+P(E2)P(EE2) =5948+4958=2072+2072=4072=59


19. Four cards are successively drawn without replacement from a deck of 52 playing cards. What is the probability that all the four cards are king?

Ans: Let's assume E1E2E3 and E4 be the events respectively the first, second, third and fourth card is king .

P(E1E2E3E4)=P(E1)P(E2E1)P(E3E1E2)P[E4(E1E2E3)E4]

=452351250149=2452515049

=113172549=1270725


20. A die is thrown 5 times. Find the probability that an odd number will come up exactly three times.

Ans: Here,p=(16+16+16)=12 and q=1p=112=12,n=5, r=3

P(X=r)=nCr(p)r(q)nr=5C3(12)3(12)53 =5!3!2!1814=1032=516


21. Ten coins are tossed. What is the probability of getting at least 8 heads?

Ans: Here, n=10, p=12,q=12

P(X=r)=P(r=8)+P(r=9)+P(r=10)

=10C8(12)8(12)108+10C9(12)9(12)109+10C10(12)10(12)1010

=10!8!2!(12)10+10!9!1!(12)10+10!0!10!(12)10

=(12)10[10×92+10+1]

=(12)1056=1272356=7128


22. The probability of a man hitting a target is 0.25. He shoots 7 times. What is the probability of his hitting at least twice?

Ans: Here, n=7,p=0.25=14,q=114=34 

P(X2)=1[P(r=0)+P(r=1)]

=1[7C0(14)0(34)70+7C1(14)1(34)71]

=1[7!0!7!(34)7+7!116!(14)(34)6]

=1[(34)6(341+147)]

=1[3545(104)]=1[35×1047]=1[27271064256]

=1[729016384]=136458192=45478192


23. A lot of 100 watches is known to have 10 defective watches. If 8 watches are selected (one by one with replacement) at random, what is the probability that there will be at least one defective watch?

Ans: Probability (defective watch out of 100 watches) =10100=110

Here, p=110, q=910,n=8 and r1

P(X1)=1P(x=0)=18C0(110)0(910)80

=18!0!8!(910)8=1(910)8


24. Consider the probability distribution of a random variable X:

X

0

1

2

3

4

P(X)

0.1

0.25

0.3

0.2

0.15

Calculate (i) V(X2)

Ans: We have

X

0

1

2

3

4

P(X)

0.1

0.25

0.3

0.2

0.15

XP(X)

0

0.25

0.6

0.6

0.60

X2P(X)

0

0.25

1.2

1.8

2.40


Var(X)=E(X2)[E(X)]2

here, E(X)=i=1nxPi(x) 

E(X)=0+0.25+0.6+0.6+0.60=2.05

and E(X2)=i=1nx2P(xi)

E(X2)=0+0.25+1.2+1.8+2.40=5.65

V(X2)=14V(X)=14[5.65(2.05)2]

14[5.654.2025]=14×1.4475=0.361875


(ii) Variance of X.

Ans: V(x)=1.44475


25. The probability distribution of a random variable X is given below:

X

0

1

2

3

P(X)

k

k2

k4

k8


(i) Determine the value of k.

Ans: We know, i=1nPi=1,i=1,2,,n and pi0

k+k2+k4+k8=1

8k+4k+2k+k=8

k=815


(ii) Determine P(X2) and P(X>2)

Ans: P(X2)=P(0)+P(1)+P(2)=k+k2+k4

=(4k+2k+k)4=7k4=74815=1415

And P(X>2)=P(3)=k8=18815=115


(iii) Find P(X2)+P(X>2).

Ans: P(X2)+P(X>2)=1415+115=1


26. For the following probability distribution, determine standard deviation of the random variable X.

X

2

3

4

P(X)

0.2

0.5

0.3

Ans: 

X

2

3

4

P(X)

0.2

0.5

0.3

XP(X)

0.4

1.5

1.2

X2P(X)

0.8

4.5

4.8


We know that, standard deviation of X=VarX

where, VarX=E(X2)[E(X)]2

=i=1nxi2P(x1)[i=1nxiPi]2

VarX=[0.8+4.5+4.8][0.4+1.5+1.2]2

=10.1(3.1)2=10.19.61=0.49

Standard deviation of X=VarX=0.49=0.7


27. A biased die is such that P(4)=110 and other scores being equally likely. The die is tossed twice. If X is the 'number of four seen', find the variance of the random variable x

Ans: Here, X=0,1,2

P(f)=110 and Pret=910

So, P(X=0)=P(ret ) P(ret) )=910910=81100

P(X=1)=P(ret)P(f)+P(f)P(ret )=910110+110910=18100

P(X=2)=P(f)P(f)=110110=1100

X

0

1

2

P(X)

81100

18100

1100

XP(X)

0

18100

2100

X2P(X)

0

1100

4100


Var(X)=E(X2)[E(X)]2=X2P(X)[XP(X)]2

=[0+18100+4100][0+18100+2100]2

=22100(20100)2=1150125

=11250=950=18100=0.18


28. A die is thrown three times. Let X be 'the number of two seen'. Find the expectation of x.

Ans: We have, X=0,1,2,3.

Ans: Here, we have X=0,1,2,3

[ die is thrown 3 times ]

and p=16,q=56

P(X=0)=P( not 2)P( not 2)P(not2)=565656=125216

P(X=1)=P(2)P(not2)P(not2)+P(not2)P(2)P(not2)

+P(not2)P(not2)P(2)

=165656+561656+565616=25216+25216+25216=75216

P(X=2)=P(2)P(2)P( not 2)+P(2)P(not2)P(2)+P(not2)P(not2)P(2)

=161656+165616+561616=5216+5216+5216=15216

P(X=3)=P(2)P(2)P(2)=161616=1216

Now E(X)=i=1npixi

=0×125216+1×75216+2×15216+3×1216

=0+75216+30216+3216=75+30+3216=108216=12

Hence, the required expectation is 12.


29. Two biased dice are thrown together. For the first die P(6)=12, the other scores being equally likely while for the second die, P(1)=25 and the other scores are equally likely. Find the probability distribution of 'the number of one seen'.

Ans: Given,for first die, P(6)=12 and P(6)=12

P(1)+P(2)+P(3)+P(4)+P(5)=12 P(1)=110 

P(1¯)=910

[\because P(1)=P(2)=P(3)=P(4)=P(5)]$ 

For second die, P(1)=25 and P(1¯)=125=35 

Let X= Number of one's seen 

X=0,P(X=0)=P(1)P(1¯)=91035=2750=0.54 P(X=1)=P(1¯)P(1)+P(1)P(1¯)=91025+11035 

=1850+350=2150=0.42 

P(X=2)=P(1)P(1)=11025=250=0.04

Hence, the required probability distribution is as below.

X

0

1

2

P(X)

0.54

0.42

0.04


30. Two probability distributions of the discrete random variable X and Y are given below:

X

0

1

2

3

P(X)

15

25

15

15


Y

0

1

2

3

P(Y)

15

310

25

110


Prove that, E(Y2)=2E(X)

Ans: We have to prove that, E(Y2)=2E(X)

E(X)=XP(X)

=015+125+215+315=75

2E(X)=145(i)

E(Y)2=Y2P(Y)

=015+1310+425+9110

=310+85+910=2810=145

E(Y)2=145( ii )

Hence, E(Y2)=2E(X).


31. A factory produces bulbs. The probability that any one bulb is defective is 150 and they are packed in boxes of 10 . From the single box, find the probability that 

(i) none of the bulbs is defective 

Ans: Let X is denoting that a bulb is defective.

Given, n=10,p=150,q=1150=4950

P(X=r)=nC,prqnr

None of the bulbs is defective, i.e., r=0

P(x=0)=10C0(150)0(4950)100=(4950)10


(ii) exactly two bulbs are defective 

Ans: If exactly two bulbs are defective

P(x=2)=10C2(150)2(4950)102 =45(49)8(50)10=45×(150)10×(49)8 


(iii) more than 8 bulbs work properly.

Ans: More than 8 bulbs work properly,so, We can say that less than 2 bulbs are only defective P(x<2)=P(x=0)+P(x=1) =10C0(150)0(4950)10+10C1(150)1(4950)9=(4950)10+15(4950)9

=(4950)9(4950+15)=(4950)9(5950)=59(49)9(50)10 .


32. Suppose you have two coins which appear identical in your pocket. You know that one is fair and one is a 2-headed coin. If you take one out, toss it and get a head, what is the probability that it was a fair coin?

Ans: Let E1= event that coin coin is fair

E2= event that one coin is 2 -headed

and H= event that the tossed coin gets a head.

P(E1)=12,P(E2)=12,P(HE1)=12,P(HE2)=1

Using Bayes' Theorem, we get

P(E1/H)=P(E1)P(HE1)P(E1)P(HE1)+P(E2)P(HE2)

=12121212+121=1414+12=1434=13

The probability of a fair coin=13


33. Suppose that 6% of the people with blood group O are left handed and 10% of those with other blood groups are left handed, 30% of the people have blood group O. If a left handed person is selected at random, what is the probability that he/she will have blood group O ?

Ans: Let E1= event that a person selected is of blood group O

E2=  event that the person selected is of other group and H=  event that selected person is left handed 

P(E1)=0.30 and P(E2)=0.70

P(HE1)=0.06P(HE2)=0.10

So, from Bayes' Theorem

P(E1H)=P(E1)P(HE1)P(E1)P(HE1)+P(E2)P(HE2)

=0.30×0.060.30×0.06+0.70×0.10=0.0180.018+0.070=0.0180.088=944


34. Two natural numbers r and s are drawn one at a time, without replacement from the set S={1,2,3,,n}. Find P(rp/sp), where pS.

Ans: Given, S={1,2,3,,n}

P(rpsp)=P(PS)P(S)=p1n×nn1=p1n1

Hence, the required probability is p1n1.


35. Find the probability distribution of the maximum of the two scores obtained when a die is thrown twice. Determine also the mean of the distribution.

Ans: Let X =random variable scores when a die is thrown twice.

X=1,2,3,4,5,6,S={(1,1),(1,2),(3,3),(3,4),(3,5),,(6,6)}

 So, P(X=1)=1616=136 

P(X=2)=1616+1616+1616=336

P(X=3)=1616+1616+1616+1616+1616=536

Similarly

P(X=4)=736

P(X=5)=936 

and P(X=6)=1136

So, the required distribution:

X

1

2

3

4

5

6

P(X)

136

336

536

736

936

1136


Mean E(X)=i=1nxipi

=1×136+2×336+3×536+4×736+5×936+6×1136

=136+636+1536+2836+4536+6636=16136


36. The random variable X can take only the values 0,1,2. If P(X=0)=P(X=1)=p and E(X2)=E(X), then find the value of p.

Ans: Given , X=0,1,2

and P(X) at X=0 and 1 is p

Let P(X) at X=2 is x

p+p+x=1x=12p

Now we have the following distributions:

X

0

1

2

P(X)

p

p

12p


E(X)=0.p+1.p+2(12p)=p+24p=23p 

and E(X2)=0.p+1.p+4(12p)=p+48p=4p 

Given that: E(X2)=E(X)

47p=23p4p=2p=12

Hence, the required value of p = 12


37. Find the variance of the distribution:

X

0

1

2

3

4

5

P(X)

16

518

29

16

19

118


Ans: Variance (X)=E(X2)[E(X)]2

E(X)=i=1npixi

=0×16+1×518+2×29+3×16+4×19+5×118 

=0+518+49+36+49+518=5+8+9+8+518=3518 

And, E(X2)=0×16+1×518+4×29+9×16+16×19+25×118 

=518+89+96+169+2518=5+16+27+32+2518=10518

Var(X)=105183518×3518=18901225324=665324


38. A and B throw a pair of dice alternately. A wins the game if he gets a total of 6 and B wins if she gets a total of 7 . If A starts the game, find the probability of winning the game by A in the third throw of the pair of dice.

Ans: Let A1 = event of getting a total of 6

={(2,4),(4,2),(1,5),(5,1),(3,3)}

and B1 = event of getting a total of 7

={(2,5),(5,2),(1,6),(6,1),(3,4),(4,3)}

Let P(A1) is the probability, if A wins in a throw =536

and P(B1) is the probability, if B wins in a throw =16

The required probability of winning A in third throw

=P(A¯1)P(B¯1)P(A1)=313656536=7757776.


39. Two dice are tossed. Find whether the following two events A and B are independent. A={(x,y):x+y=11} and B={(x,y):x5} where (x,y) denotes a typical sample point.

Ans: We have ,

A={(x,y):x+y=11}  and B={(x,y):x5}

A={(5,6),(6,5)}

B={(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),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}

n(A)=2,n(B)=30 and n(AB)=1

P(A)=236=118

and P(B)=3036=56

P(A)P(B)=11856=5108 and P(AB)=136

P(A)P(B)P(AB)

So,events A and B are not independent.


40. An urn contains m white and n black balls. A ball is drawn at random and is put back into the urn along with k additional balls of the same colour as that of the ball drawn. A ball is again drawn at random. Show that the probability of drawing a white ball now does not depend on k.

Ans: Let A = event of having m white and n black balls

E1= {first ball drawn of white colour}

E2= {first ball drawn of black colour}

And, E3= {second ball drawn of white colour}

P(E1)=mm+n and P(E2)=nm+n

P(E3E1)=m+km+n+k and P(E3E2)=mm+n+k

Now P(E3)=P(E1)P(E3E1)+P(E2)P(E3E2)

=mm+n×m+km+n+k+nm+n×mm+n+k

=mm+n+k[m+km+n+nm+n]

=mm+n+k[m+n+km+n]=mm+n

So, the probability of drawing a white ball does not depend upon k.


Long Answer Type Questions

41. Three bags contain a number of red and white balls as follows, Bag I: 3 red balls, Bag II: 2 red balls and 1 white ball and Bag III: 3 white balls. The probability that bag i will be chosen and a ball is selected from it is i6, where i=1,2,3.

What is the probability that 

(i) a red ball will be selected 

Ans: Given,

Bag I=three red balls and no white ball

Bag II=two red balls and one white ball

Bag III=no red ball and three white balls

Let E1,E2 and E3 be the events of choosing Bag I, Bag II and Bag III respectively and a ball is drawn from it.

P(E1)=16,P(E2)=26 and P(E3)=36

Let E be the event that red ball is selected

P(E)=P(E1)P(EE1)+P(E2)P(EE2)+P(E3)P(EE3)

=1633+2623+360=318+418=718


(ii) a white ball is selected?

Ans: Let F be the event that  ball is selected is white

P(F)=1P(E)[P(E)+P(F)=1]

=1718=1118


42. Refer to Exercise Q.41 above. If a white ball is selected, what is the probability that it came from

(i) Bag II

Ans: We will use here Bayes' Theorem

P(E2/F)=P(E2)P(F/E2)P(E1)P(F/E1)+P(E2)P(F/E2)+P(E3)P(F/E3) =2613160+2613+361=218218+36=211


(ii) Bag III?

Ans: P(E3/F)=P(E3)P(F/E3)P(E1)P(F/E1)+P(E2)P(F/E2)+P(E3)P(F/E3) =361160+2613+361=36218+36=36×1811=911


43. A shopkeeper sells three types of flower seeds A1,A2 and A3. They are sold as a mixture, where the proportions are 4:4:2, respectively. The germination rates of the three types of seeds are 45%,60% and 35%. Calculate the probability

(i) of a randomly chosen seed to germinate

Ans: Given, A1:A2:A3=4:4:2

P(A1)=410,P(A2)=410 and P(A3)=210

where A1,A2 and A3 are the 3 types of seeds.

Let E = event that a seed germinates and 

E¯ = event that a seed does not germinate

So, P(EA1)=45100,P(EA2)=60100 and P(EA3)=35100 and P(EA1)=55100,P(EA2)=40100 and P(EA3)=65100

P(E)=P(A1)P(EA1)+P(A2)P(EA2)+P(A3)

=41045100+41060100+21035100

=1801000+2401000+701000=4901000=0.49


(ii) that it will not germinate given that the seed is of type A3

Ans: From sol.(i)

P(EA3)=1P(EA3)=135100=65100=0.65


(iii) that it is of the type A2 given that a randomly chosen seed does not germinate.

Ans: Using Bayes' Theorem, 

P(A2E)=P(A2)P(EA2)P(A1)P(EA1)+P(A2)P(EA2)+P(A3)P(EA3)

=4104010040100+41040100+21065100

=1602201000+1601000+1301000=160220+160+130=160510=1651=0.314


44. A letter is known to have come either from TATA NAGAR or from CALCUTTA. On the envelope, just two consecutive letters TA are visible. What is the probability that the letter came from TATA NAGAR?

Ans: Let E1 =event that the letter comes from TATA NAGAR and E2 = event that the letter comes from CALCUTTA Also 

E3 = event that on the letter, two consecutive letters TA are visible

P(E1)=12 and P(E2)=12 and P(E3E1)=28  (because For TATA NAGAR, the two consecutive letters visible are TA,AT,TA,AN,NA,AG,GA,AR)

and P(E3E2)=17

(because CALCUTA, the two consecutive letters visible are CA, AL, LC, CU, UT, TT and TA)

Now using Bayes' Theorem, 

P(E1E3)=P(E1)P(E3E1)P(E1)P(E3E1)+P(E2)P(E3E2)

=12281228+1217=1818+114=187+456=711


45. There are two bags, one of which contains 3 black and 4 white balls while the other contains 4 black and 3 white balls. A die is thrown. If it shows up 1 or 3,a ball is taken from the first bag but if it shows up any other number, a ball is chosen from the second bag. Find the probability of choosing a black ball.

Ans: Let E1 = event of selecting Bag I

E2 = event of selecting Bag II

E3 = event that black ball is selected

P(E1)=26=13 and P(E2)=113=23

P(E3E1)=37 and P(E3E2)=47

P(E3)=P(E1)P(E3E1)+P(E2)P(E3E2)

=1337+2347=3+821=1121


46. There are three urns containing 2 white and 3 black balls, 3 white and 2 black balls and 4 white and 1 black balls, respectively. There is an equal probability of each urn being chosen. A ball is drawn at random from the chosen urn and it is found to be white. Find the probability that the ball drawn was from the second urn.

Ans: Let U1={two white,three black balls}

U2={three white,two black balls}

U3={four white,one black balls}

P(U1)=P(U2)=P(U3)=13

Let H = event of drawing white ball chosen.

So, P(HU1)=25,P(HU2)=35 and P(HU3)=45

P(U2H)=P(U2)P(HU2)P(U1)P(HU1)+P(U2)P(HU2)+P(U3)P(HU3)

=13351325+1335+1345=3525+35+45=39=13


47. By examining the chest X-ray, the probability that TB is detected when a person is actually suffering is 0.99. The probability of a healthy person diagnosed to have TB is 0.001. In a certain city, 1 in 1000 people suffer from TB. A person is selected at random and is diagnosed to have TB. What is the probability that he actually has TB?

Ans: Let E1 = event that  person has TB

E2 = event that  person does not have TB

and H = event that the person is diagnosed to have TB.

P(E1)=11000=0.001,P(E2)=111000=9991000=0.999

P(HE1)=0.99,P(HE2)=0.001

P(E1H)=P(E1)P(HE1)P(E1)P(HE1)+P(E2)P(HE2)

=0.001×0.990.001×0.99+0.999×0.001=0.990.99+0.999

=0.9900.990+0.999=9901989=110221


48. An item is manufactured by three machines A,B and C. Out of the total number of items manufactured during a specified period, 50% are manufactured on machine A, 30% on B and 20% on C. 2% of items produced on A and 2% of items produced on B are defective and 3% of these produced on machine C are defective. All the items are stored at one godown. One item is drawn at random and is found to be defective. What is the probability that it was manufactured on machine A?

Ans: Let E1 = event that the item is manufactured on machine A

E2 = event that the item is manufactured on machine B

E3 = event that the item is manufactured on machine C

Let H= event that the selected item is defective.

Using Bayes' Theorem,

P(E1)=50100,P(E2)=30100,P(E3)=20100 

P(HE1)=2100,P(HE2)=2100 and P(HE3)=3100

P(E1H)=P(E1)P(HE1)P(E1)P(HE1)+P(E2)P(HE2)+P(E3)P(HE3)

=50100×210050100×2100+30100×2100+20100×3100

=100100+60+60=100220=1022=511


49. Let X be a discrete random variable whose probability distribution is defined as follows

P(X=x)={k(x+1) for x=1,2,3,42kx for x=5,6,70 otherwise 

where k is a constant-Calculate:

(i) the value of k

Ans: Here, P(X=x)=k(x+1) for x=1,2,3,4

Than, P(X=1)=k(1+1)=2k;P(X=2)=k(2+1)=3k

P(X=3)=k(3+1)=4k;P(X=4)=k(4+1)=5k

Also, P(X=x)=2kx for x=5,6,7

P(X=5)=2(5)k=10k;P(X=6)=2(6)k=12k

P(X=7)=2(7)k=14k

and for otherwise = 0 .

The probability distribution:

X

1

2

3

4

5

6

7

otherwise

P(X)

2k

3k

4k

5k

10k

12k

14k

0


We know , i=1nP(Xi)=1

So, 2k+3k+4k+5k+10k+12k+14k=1

50k=1k=150

So, the value of k is 150


(ii) E(X) 

Ans: Now the probability distribution:

X

1

2

3

4

5

6

7

P(X)

250

350

450

550

1050

1250

1450


E(X)=1×250+2×350+3×450+4×550+5×1050+6×1250 +7×1450

=250+650+1250+2050+5050+7250+9850=26050=265=5.2 


(iii) Standard deviation of X.

Ans: Standard deviation = Variance 

Variance =E(X2)[E(X)]2 E(X2)=1×250+4×350+9×450+16×550+25×1050 36×1250+49×1450

36×1250+49×1450 3250+68650=149850

=250+1250+3650+8050+25050+43250+68650=149850

Than, Variance (X)=149850(265)2

=14985067625=1498135250=14650=2.92

S.D =2.92=1.7 


50. The probability distribution of a discrete random variable X is given as under:

X

1

2

4

2A

3A

5A

P(X)

12

15

325

110

125

125


Calculate: (i) The value of A if E(X)=2.94

Ans: We know,E(X)=i=1nPiXi

So, E(X)=1×12+2×15+4×325+2A×110+3A×125+5A×125

2.94=12+25+1225+A5+3A25+A5

2.94=0.5+0.4+0.48+13A25=1.38+13A25

2.941.38=13A25

1.56=13A25

A=1.56×2513=0.12×25

So, A=3


(ii) Variance of X

Ans: Now the distribution:

X

1

2

4

6

9

15

P(X)

12

15

325

110

125

125


E(X2)=1×12+4×15+16×325+36×110+81×125+225×125

=12+45+4825+3610+8125+22525

=0.5+0.8+1.92+3.6+3.24+9.00=19.06

So,Variance (X)=E(X2)[E(X)]2

=19.06(2.94)2=19.068.64=10.42


51. The probability distribution of a random variable as under: P(X=x)={kx2 for x=1,2,32kx for x=4,5,60 otherwise 

where k is a constant. 

Calculate:

(i) E(X) 

(ii) E(3X2)

(iii) P(X4)

Ans:  

X

1

2

3

4

5

6

otherwise

P(X)

k

4k

9k

8k

10k

12k

0


We know,i=1nP(Xi)=1

k+4k+9k+8k+10k+12k=1

44k=1k=144

E(X)=i=1nPiXi=1×k+2×4k+3×9k+4×8k+5×10k+6×12k =k+8k+27k+32k+50k+72k=190k

=190×144=9522=4.32 (approx.)


(ii) E(3X2)

Ans: We know,i=1nP(Xi)=1

k+4k+9k+8k+10k+12k=144k=1k=144

E(3X2)=3[k+4×4k+9×9k+16×8k+25×10k+36×12k]

3[k+16k+81k+128k+250k+432k]=3[908k]

=3×908×144=272444=61.9 (approx.) 


(iii) P(X4)

Ans: We know,i=1nP(Xi)=1

k+4k+9k+8k+10k+12k=1

44k=1k=144

P(X4)=P(X=4)+P(X=5)+P(X=6)

=8k+10k+12k=30k

=30×144=1522


52. A bag contains (2n+1) coins. It is known that n of these coins have a head on both sides whereas the rest of the coins are fair. A coin is picked up at random from the bag and is tossed. If the probability that the toss results in a head is 3142, determine the value of n.

Ans: Given ,n coins are two headed coins and the remaining (n+1) coins are fair.

Let E1 = event that unfair coin is selected

E2 = event that the fair coin is selected

E = event that the toss results in a head

P(E1)=n2n+1 and P(E2)=n+12n+1

P(EE1)=1 (sure event) and P(EE2)=12

P(E)=P(E1)P(EE1)+P(E2)P(EE2)

=n2n+11+n+12n+112=12n+1(n+n+12)

=12n+1(2n+n+12)=3n+12(2n+1)

But P(E)=3142( given )

3n+12(2n+1)=3142  

3n+12n+1=3121

63n+21=62n+31

n=10


53. Two cards are drawn successively without replacement from a well shuffled deck of cards. Find the mean and standard deviation of the random variable X where X is the number of aces.

Ans: Let X be the random variable , X=0,1,2

and E=  event of drawing an ace

and F=  event of drawing non-ace.

So, P(E)=452 and P(E¯)=4852 

P(X=0)=P(E¯)P(E¯)=48524751=188221 

P(X=1)=P(E)P(E¯)+P(E¯)P(E)=452×4851+4852×451=32221 

P(X=2)=P(E)P(E)=452351=1221

Distribution Table:

X

0

1

2

P(X)

188221

32221

1221


Mean E(X)=0×188221+1×32221+2×1221=32221+2221=34221=213 

E(X2)=0×188221+1×32221+4×1221=32221+4221=36221

 Variance =E(X2)[E(X)]2

=36221(213)2=362214169=4686813×221=4002873 

Standard deviation =4002873=0.377 (approx.)


54. A die is tossed twice. A success' is getting an even number on a toss. Find the variance of the number of successes.

Ans: Let E = event of getting an even number on tossing a die.

P(E)=36=12 and P(E)=112=12 

Here X=0,1,2 

P(X=0)=P(E)P(E)=1212=14 P(X=1)=P(E)P(E)+P(E)P(E)=1212+1212=14+14=24 P(X=2)=P(E)P(E)=1212=14

X

0

1

2

P(X)

14

24

14


E(X)=0×14+1×24+2×14=24+24=1

E(X2)=0×14+1×24+4×14=32

Variance (X)=E(X2)[E(X)]2=321=12=0.5


55. There are 5 cards numbered 1 to 5 , one number on one card. Two cards are drawn at random without replacement. Let X denote the sum of the numbers on two cards drawn. Find the mean and variance of X.

Ans: Sample space 

S={(1,2),(2,1),(1,3),(3,1),(2,3),(3,2),(1,4),(4,1),(1,5),(5,1),(2,4),(4,2),(2,5),(5,2),(3,4),(4,3),(3,5),(5,3),(5,4),(4,5)}

So, n(S)=20

Let X be the random variable denoting the sum of the numbers on two cards drawn.

X=3,4,5,6,7,8,9

So, P(X=3)=220

P(X=4)=220

P(X=5)=420 

P(X=6)=420

P(X=7)=420  

P(X=8)=220

P(X=9)=220

Mean,E(X)=3×220+4×220+5×420+6×420+7×420+8×220+9×220

=620+820+2020+2420+2820+1620+1820=12020=6

E(X2)=9×220+16×220+25×420+36×420+49×420+64×220+81×220

=1820+3220+10020+14420+19620+12820+16220=78020=39

Variance (X)=E(X2)[E(X)]2=39(6)2=3936=3


Objective Type Questions

Choose the correct answer from the given four options in each of the Exercises from Q. 56 to 82.

56. If P(A)=45 and P(AB)=710, then P(BA) is equal to

(A) 110

(B) 18

(C) 78

(D) 1720

Ans: Correct option is (c) 

Because P(A)=45,P(AB)=710

P(BA)=P(AB)P(A)=7/104/5=78


57. If P(AB)=710 and P(B)=1720, then P(AB) equals

(A) 1417

(B) 1720

(C) 78

(D) 18

Ans:  Correct option is (a) 

Because P(AB)=710 and P(B)=1720

P(AB)=P(AB)P(B)=7/1017/20=1417


58. If P(A)=310,P(B)=25 and P(AB)=35, then P(BA)+P(AB) equals

(A) 14

(B) 13

(C) 512

(D) 72

Ans:  Correct option is (d) 

 Because, P(A)=310,P(B)=25 and P(AB)=35

P(BA)+P(AB)=P(BA)P(A)+P(AB)P(B)

=P(A)+P(B)P(AB)P(A)+P(A)+P(B)P(AB)P(B)

[P(AB)=P(A)+P(B)P(AB) i.e., P(AB)=P(A)+P(B)P(AB)] 

=310+2535310+310+253525

=110310+11025=13+14=712


59. If P(A)=25,P(B)=310 and P(AB)=15, then P(AB).P(BA) P(AB)P(BA) is equal to

(A) 56

(B) 57

(C) 2542

(D) 1

Ans:  Correct option is (c) 

Because, P(A)=25,P(B)=310 and P(AB)=15

P(AB)=P(AB)P(B)=1P(AB)1P(B)

=1[P(A)+P(B)P(AB)]1P(B)

=1(25+31015)1310

=1(4+3210)710=112710=57

And P(BA)=P(BA)P(A)=1P(AB)1P(A)

=112125=1/23/5=56[P(AB)=12]

P(AB)P(BA)=5756=2542


60. If A and B are two events such that P(A)=12,P(B)=13,P(A/B)=14, then P(AB) equals

(A) 112

(B) 34

(C) 14

(D) 316

Ans: Correct option is (c) 

Because, P(A)=12,P(B)=13 and P(AB)=14 P(AB)=P(AB)P(B)

P(AB)=P(AB)P(B)=1413=112

Now, P(AB)=1P(AB)

=1[P(A)+P(B)P(AB)]

=1[12+13112]=1[6+4112]

=1912=312=14


61. If P(A)=0.4,P(B)=0.8 and P(BA)=0.6, then P(AB) is equal to

(A) 0.24

(B) 0.3

(C) 0.48

(D) 0.96

Ans: Correct option is (d) 

Because, P(A)=0.4,P(B)=0.8 and P(BA)=0.6,

P(BA)=P(BA)P(A)

P(BA)=P(BA)P(A)

=0.6×0.4=0.24

P(AB)=P(A)+P(B)P(AB)

=0.4+0.80.24

=1.20.24=0.96


62. If A and B are two events and Aϕ,Bϕ, then

(A) P(AB)=P(A)P(B) 

(B) P(A/B)=P(AB)P(B)

(C) P(AB)P(BA)=1

(D) P(AB)=P(A)/P(B)

Ans: Correct option is (b)

Because,if Aϕ and Bϕ, then P(AB)=P(AB)P(B)


63. A and B are events such that P(A)=0.4,P(B)=0.3 and P(AB)=0.5, Then P(BA) equals

(A) 23

(B) 12

(C) 310

(D) 15

Ans: Correct option is (d)

Because, P(A)=0.4,P(B)=0.3 and P(AB)=0.5

P(AB)=P(A)+P(B)P(AB)

P(AB)=0.4+0.30.5=0.2

P(BA)=P(A)P(AB)

=0.40.2=0.2=15


64. If it is given that A and B are two events such that P(B)=35,P(AB)=12 and P(AB)=45, then P(A) equals

(A) 310

(B) 15

(C) 12

(D) 35

Ans: Correct option is (c)

Because, P(B)=35,P(AB)=12 and P(AB)=45

P(AB)=P(AB)P(B)

12=P(AB)3/5

P(AB)=35×12=310

And P(AB)=P(A)+P(B)P(AB)

45P(A)+35310

45=P(A)+35310

P(A)=4535+310=86+310=12P(A)=4535+310=86+310=12


65. In Exercise 64 above, P(BA) is equal to

(A) 15 

(B) 310

(C) 12

(D) 35

Ans: Correct option is (d) 

Because,P(BA)=P(BA)P(A)=P(B)P(BA)1P(A) =35310112=631012=610=35


66. If P(B)=35,P(A/B)=12 and P(AB)=45, then P(AB)+P(AB)= 

(A) 15

(B) 45

(C) 12

(D) 1

Ans: Correct option is (d) 

Because, P(B)=35,P(AB)=12

And P(AB)=45

Since, P(AB)=P(AB)P(B) 

P(AB)=P(AB)P(B)

=12×35=310

Also, P(AB)=P(A)+P(B)P(AB)

P(A)=4535+310=12

P(AB)=1P(AB)=145=15

And P(AB)=1P(AB)=1P(AB)

=1P(A)P(B)

=11225=45

P(AB)+P(AB)=15+45=55=1


67. Let P(A)=713,P(B)=913 and P(AB)=413. Then P(A/B)P(AB) is equal to

(A) 613

(B) 413

(C) 49

(D) 59

Ans: Correct option is (d) 

Because, P(A)=713,P(B)=913 and P(AB)=413 P(AB)=P(AB)P(B)=P(B)P(AB)P(B)

=913413913=513913=59


68. If A and B are such events that P(A)>0 and P(B)1, then P(A/B)P(AB) equals.

(A) 1P(AB)

(B) 1P(AB)

(C) 1P(AB)P(B)

(D) P(A)P(B)

Ans: Correct option is (c)

Because,P(A)>0 and P(B)1

P(AB)=P(AB)P(B)=1P(AB)P(B)


69. If A and B are two independent events with P(A)=35 and P(B)=49, then P(AB) equals

(A) 415

(B) 845

(C) 13 

(D) 29

Ans: Correct option is (d) 

Because, P(AB)=1P(AB)

=1[P(A)+P(B)P(AB)]

=1[35+4935×49][P(AB)=P(A)P(B)]

=1[27+201245]=13545=1045=29


70. If two events are independent, then

(A) they must be mutually exclusive

(B) the sum of their probabilities must be equal to 1

(C) (A) and (B) are both are correct

(D) None of the above is correct

Ans: Correct option is (d)

Because,for independent events A and B are 

P(AB)=P(A)P(B),P(A)0,P(B)0

So, they will not be mutually exclusive events. 

In other words, two independent events having non-zero probabilities of occurrence cannot be mutually exclusive and conversely, two mutually exclusive events having non-zero probabilities of outcome cannot be independent.


71. Let A and B be two events such that P(A)=38,P(B)=58 and P(AB)=34. Then P(AB).P(AB) is equal to

(A) 25 

(B) 38

(C) 310

(D) 625

Ans: Correct option is (d)

Because, P(A)=38,P(B)=58 and P(AB)=34

P(AB)=P(A)+P(B)P(AB)

P(AB)=38+5834=3+568=28=14

P(AB)=P(AB)P(B)=1/45/8=820=25

And P(AB)=P(AB)P(B)=P(B)P(AB)P(B)

=581458=52858=35

P(AB)P(AB)=2535=625


72. If the events A and B are independent, then P(AB) is equal to

(A) P(A)+P(B)

(B) P(A)P(B)

(C) P(A)P(B) 

(D) P(A)/P(B)

Ans: Correct option is (c) 

Because,if A and B are independent, then P(AB)=P(A)P(B)


73. Two events E and F are independent. If P(E)=0.3,P(EF)=0.5, then P(EF)P(FE) equals

(A) 27

(B) 335

(C) 170

(D) 17

Ans: Correct option is (c)

Because, P(E)=0.3,P(EF)=0.5,

Let P(F)=x

P(EF)=P(E)+P(F)P(EF)

=P(E)+P(F)P(E)P(F)

0.5=0.3+x0.3x

x=0.50.30.7=27=P(F)

P(E/F)P(F/E)=P(EF)P(F)P(FE)P(E)

=P(EF)P(E)P(FE)P(F)P(E)P(F)

=P(EF)[P(E)P(F)]P(EF)=P(E)P(F)

=31027=212070=170


74. A bag contains 5 red and 3 blue balls. If 3 balls are drawn at random without replacement the probability of getting exactly one red ball is

(A) 45196

(B) 135392

(C) 1556

(D) 1529

Ans: Correct option is (c)

Probability of getting exactly one red (R) ball

=PRPR¯PR+PRPRPR+PRPR¯PR

=583726+385726+382756

=15476+15476+15476

=556+556+556=1556


75. Refer to above Question 74 . The probability that exactly two of the three balls were red, the first ball being red, is 

(A) 13

(B) 47

(C) 1528

(D) 528

Ans: Correct option is (b)

Let E1= Event that first ball being red

And E2= Event that exactly two of three balls being red

P(E1)=PRPRPR+PRPRPR¯+PRPR¯PR+PRPR¯PR¯

=584736+584736+583746+583726 

=60+60+60+30336=210336

P(E1E2)=PRPR¯PR+PRPRPR¯

=583746+584736=120336

P(E2E1)=P(E1E2)P(E1)=120/336210/336=47


76. Three persons, A, B and C, fire at a target in turn, starting with A. Their probability of hitting the target are 0.4,0.3 and 0.2 respectively. The probability of two hits is

(A) 0.024

(B) 0.188

(C) 0.336 

(D) 0.452

Ans: Correct option is (b) 

Because, P(A)=0.4,P(A¯)=0.6,P(B)=0.3,P(B¯)=0.7, P(C)=0.2 and P(C¯)=0.8

So,Probability of two hits =PAPBPC+PAPBPC+PAPBPC

=0.4×0.3×0.8+0.4×0.7×0.2+0.6×0.3×0.2

=0.096+0.056+0.036=0.188


77. Assume that in a family, each child is equally likely to be a boy or a girl. A family with three children is chosen at random. The probability that the eldest child is a girl given that the family has at least one girl is

(A) 12

(B) 13

(C) 23

(D) 47

Ans: Correct option is (d)

Because, S={(B,B,B,(G,G,G),(B,G,G),(G,B,G),(G,G,B),(G,B,B),(B,G,B),(B,B,G)\}$

E1= Event that a family has at least one girl, then

E1={(G,B,B),(B,G,B),(B,B,G),(G,G,B),(B,G,G),(G,B,G),(G,G,G)}

E2= Event that the eldest child is a girl, then

E2={(G,B,B),(G,G,B),(G,B,G),(G,G,G)} E1E2={(G,B,B),(G,G,B),(G,B,G),(G,G,G)}

P(E2E1)=P(E1E2)P(E1)=4/87/8=47


78. A die is thrown and a card is selected at random from a deck of 52 playing cards. The probability of getting an even number on the die and a spade card is

(A) 12

(B) 14

(C) 18

(D) 34

Ans: Correct option is (c)

Because,let E1= Event for getting an even number on the die

And E2= Event that a spade card is selected

P(E1)=36=12 and P(E2)=1352=14

Then, P(E1E2)=P(E1)P(E2)=1214=18


79. A box contains 3 orange balls, 3 green balls and 2 blue balls. Three balls are drawn at random from the box without replacement. The probability of drawing 2 green balls and one blue ball is

(A) 328

(B) 221 

(C) 128

(D) 167168 

Ans: Correct option is (a)

Because,Probability of drawing 2 green balls and one blue ball=PGPG+PGPBPG

=PGPGPB+PBPGPG+PGPBPG

=382727+283726+382726

=128+128+128=328


80. A flashlight has 8 batteries out of which 3 are dead. If two batteries are selected without replacement and tested, the probability that both are dead is

(A) 3356

(B) 964

(C) 114

(D) 328

Ans: Correct option is (d)

Because, probability =PDPD=3827=328


81. Eight coins are tossed together. The probability of getting exactly 3 heads is

(A) 1256

(B) 732 

(C) 532

(D) 332

Ans: Correct option is (b)

Because, Probability distribution P(X=r)=nCr(p)rqnr

Here, n=8,r=3,p=12 and q=12

Required probability =8C3(12)3(12)83=8!5!3!(12)8 =8763211616=732


82. Two dice are thrown. If it is known that the sum of numbers on the dice was less than 6 , the probability of getting a sum 3 , is

(A) 118

(B) 518

(C) 15

(D) 25

Ans: Correct option is (c)

Because,if let E1= Event that the sum of numbers on the dice was less than 6

And E2= Event that the sum of numbers on the dice is 3

E1={(1,4),(4,1),(2,3),(3,2),(2,2),(1,3),(3,1),(1,2),(2,1),(1,1)} n(E1)=10 

And E2={(1,2),(2,1)}

n(E2)=2

Required probability =210=15


83. Which one is not a requirement of a binomial distribution?

(A) There are 2 outcomes for each trial

(B) There is a fixed number of trials

(C) The outcomes must be dependent on each other

(D) The probability of success must be the same for all the trials

Ans: Correct option is (c)


84. Two cards are drawn from a well shuffled deck of 52 playing cards with replacement. The probability, that both cards are queens, is

(A) 113×113

(B) 113+113

(C) 113×117

(D) 113×451

Ans: Correct option is (a)

Because, probability =452452=113×113 (with replacement)


85. The probability of guessing correctly at least 8 out of 10 answers on a true-false type examination is

(A) 764

(B) 7128

(C) 451024

(D) 741

Ans: Correct option is (b)

Because,P(X=r)=nCr(p)r(q)nr

n=10,p=12,q=12

and r8 i.e., r=8,9,10

P(X=r)=P(r=8)+P(r=9)=P(r=10)

=10C8(12)8(12)108+10C9(12)9(12)+10C10(12)10

=10!8!2!(12)10+10!9!1!(12)10+(12)10 

=(12)10[45+10+1]=(12)1056

=1161656=7128


86. The probability that a person is not a swimmer is 0.3. The

probability that out of 5 persons 4 are swimmers is

(A) 5C4(0.7)4(0.3)

(B) 5C1(0.7)(0.3)4

(C) 5C4(0.7)(0.3)4

(D) (0.7)4(0.3)

Ans: Correct option is (a)

Because, p¯=0.3p=0.7 and q=0.3,n=5 and r=4

Than probability =5C4(0.7)4(0.3)


87. The probability distribution of a discrete random variable X is given below:

X

2

3

4

5

P(X)

5k

7k

9k

11k


The value of k is

(A) 8

(B) 16

(C) 32

(D) 48

Ans: Correct option is (c)

Because,we know, P(X)=1

5k+7k+9k+11k=1 

32k=1

k=32


88. For the following probability distribution

X

-4

-3

-2

-1

0

P(X)

0.1

0.2

0.3

0.2

0.2


E(X) is equal to:

(A) 0

(B) 1

(C) 2

(D) 1.8

Ans: Correct option is (d)

Because, E(X)=XP(X)

=4×(0.1)+(3×0.2)+(2×0.3)+(1×0.2)+(0×0.2)

=0.40.60.60.2=1.8


89. For the following probability distribution:

X

1

2

3

4

P(X)

110

15

310

25


E(X2) is equal to

(A) 3

(B) 5

(C) 7

(D) 10

Ans: Correct option is (d) 

Because, E(X2)=X2P(X)=1110+415+9310+1625 

=110+45+2710+325

=1+8+27+6410=10


90. Suppose a random variable X follows the binomial distribution with parameters n and p, where 0<p<1. If P(x=r)P(x=nr) is independent of n and r, then p equals

(A) 12

(B) 13

(C) 15

(D) 17

Ans: Correct option is (a)

Because, P(X=r)nCr(p)r(q)nr=n!(nr)!r!(p)r(1p)nr[q=1p] (i)

P(X=0)=(1p)n

and P(X=nr)=nCnr(p)nr(q)n(nr)

=n!(nr)!r!(p)nr(1p)+r [q=1p]  [nCr=nCnr]...(ii)

P(x=r)P(x=nr)=n!(nr)!r!pr(1p)nrn!(nr)!r!pnr(1p)+r[ by Eqs. (i) and (ii)] 

=(1pp)nr×1(1pp)r

This is independent of n and r,

if 1pp=11p=2p=12


91. In a college 30% students fail in physics, 25% fail in mathematics and 10% fail in both. One student is chosen at random. The probability that she fails in physics if she has failed in mathematics is

(A) 110

(B) 25

(C) 920

(D) 13

Ans: Correct option is (b) 

Because, P(Ph)=30100=310,P(M)=25100=14 and P(MPh)=10100=110

P(PhM)=P(PhM)P(M)=1/101/4=25


92. A and B are two students. Their chances of solving a problem correctly are 13 and 14 , respectively. If the probability of their making a common error is, 120 and they obtain the same answer, then the probability of their answer to be correct is 

(A) 112

(B) 140

(C) 13120

(D) 1013

Ans: Correct option is (d)

If,we let E1= Event that both A and B solve in the problem

P(E1)=13×14=112

Let E2= Event that both A and B got incorrect solution of the problem

P(E2)=23×34=12

Let E= Event that they got same answer

Here, P(E/E1)=1,P(E/E2)=120

P(E1/E)=P(E1E)P(E)=P(E1)P(E/E1)P(E1)P(E/E1)+P(E2)P(E/E2)

=112×1112×1+12×120=1/210+3120=12012×3=1013


93. A box has 100 pens of which 10 are defective. What is the probability that out of a sample of 5 pens drawn one by one with replacement at most one is defective? 

(A) (910)5

(B) 12(910)4

(C) 12(910)5

(D) (910)5+12(910)4

Ans: Correct option is (d)

Because, n=5,p=10100=110 and q=910,r1

r=0,1

Also, P(X=r)=nCrpγqnr

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

=5C0(110)0(910)5+5C1(110)1(910)4

=(910)5+5110(910)4

=(910)5+12(910)4


State True or False for the statements in each of the Exercises 94 to 103.

94. Let P(A)>0 and P(B)>0. Then A and B can be both mutually exclusive and independent.

Ans: False 

For events to be mutually exclusive -

P(AB)=P(A)+P(B)

But as per the conditions in question, it is not necessary that they will meet the condition because it might be possible that

P(AB)0

For events to be independent-

P(AB)=P(A)P(B)

Again P(A)>0 and P(B)>0 are not sufficient conditions to validate them.


95. If A and B are independent events, then A and B are also independent.

Ans: True

Since if A and B are independent then, P(AB)=P(A)P(B)

Now we know that A=A(BB)=(AB)(AB)

Hence P(A)=P((AB)(AB))

Since the events (AB) and (AB) are disjoint, these events are mutually exclusive

Hence P(A)=P((AB)(AB))=P(AB)+P(AB)

Hence, we have

P(A)=P(A)P(B)+P(AB)

P(AB)=P(A)(1P(B))

We know that P(B)=1P(B)

Hence we have P(AB)=P(A)P(B)

Hence the events A and B ' are independent,

Since A and B are independent, we have A and B are also independent.


96. If A and B are mutually exclusive events, then they will be independent also.

Ans: False

If A and B are mutually exclusive, then if A occuss, then B cannot occur. Then if two events are mutually exclusive, they caunot be independent So, the correct oplion is B.


97. Two independent events are always mutually exclusive.

Ans: False

Two events A and B are said to be independent if the occurrence of one does not affect the probability of the probability of the occurrence of the other. Thus if A and B are two independent events, the P(AB)=P(A).P(B) But two events are mutually exclusive events if the occurrence of one vent precludes the occurrence of the other event.


98. If A and B are two independent events then P(A and B)=P(A)P(B).

Ans: True

Two events A and B are said to be independent if the occurrence of one does not affect the probability of the probability of the occurrence of the other. Thus if A and B are two independent events, the P(AB)=P(A).P(B) 


99. Another name for the mean of a probability distribution is expected value.

Ans: True

E(X)=XP(X)=μ

Mean gives the average of values and if it is related with probability or random variable it is often called expected value.


100. If A and B are independent events, then P(AB)=1P(A)P(B)

Ans: True

P(AB)=1P(AB)=1P(A)P(B)


101. If A and B are independent, then P(exactly one of A,B occurs)=P(A)P(B)+P(B)P(A) 

Ans: False

We have, P(exactly one of A,B occurs)

=P[(AB)(AB)]=P(AB)+P(AB)

=P(A)P(AB)+P(B)P(AB)

=P(A)+P(B)2P(AB)

=P(AB)P(AB)

Also, P (exactly one of A,B occurs)

=[1P(AB)][1P(AB)]

=P(AB)P(AB)

=P(A)+P(B)2P(AB)


102. If A and B are two events such that P(A)>0 and P(A)+P(B)>1, then P(BA)1P(B)P(A)

Ans: False

P(BA)=P(AB)P(A)

=P(A)+P(B)P(AB)P(A)>1P(AB)P(A)

So, P(BA)>1P(AB)P(A)


103. If A,B and C are three independent events such that P(A)=P(B)=P(C)=p, then P( At least two of A, B, C occur )=3p22p3

Ans: True

P (At least two of A, B, C occur)

=p×p×(1p)+(1p)pp+p(1p)p+ppp

=p2[1p+1p+1p+p]

=p2(33p)+p3

=3p23p3+p3=3p22p3


Fill in the Blanks in Each of the Following Questions:

104. If A and B are two events such that

P(A/B)=p,P(A)=p,P(B)=13 And P(AB)=59, then p=......

Ans:  Given, P(A)=p,P(B)=13 and P(AB)=59 

P(AB)=P(AB)P(B)=pP(AB)=p3

and P(AB)=P(A)+P(B)P(AB)

59=p+13p35913=2p3

539=2p3p=29×32=13


105. If A and B are such that

P(AB)=23 and P(AB)=59, then P(A)+P(B)=.......

Ans:  Given, P(AB)=23 and P(AB)=59

P(AB)=1P(AB)

23=1P(AB)

P(AB)=123=13

P(A)+P(B)=1P(A)+1P(B)

=2[P(A)+P(B)]

=2[P(AB)+P(AB)]

=2(59+13)=2(5+39)

=1889=109


106. If X follows binomial distribution with parameters n=5,p and P(X=2)=9,P(X=3), then p=....

Ans: because P(X=2)=9P(X=3) (where, n=5 and q=1p )

5C2p2(1p)3=95C3P3(1p)2

5!2!3!p2(1p)3=95!3!2!p3(1p)25!2!3!p2(1p)3=95!3!2!p3(1p)2

p2(1p)3p3(1p)2=9

p2(1p)3p3(1p)2=9

(1p)p=99p+p=1(1p)p=99p+p=1

p=110p=110


107. Let X be a random variable taking values x1,x2,,xn with probabilities P1,P2,,Pn, respectively. Then var (X)=.......

Ans: Because, Var(X)=(X2)[E(X)]2

=i=1nX2P(X)[i=1nXP(X)]2

=Pixi2(Pixi)2


108. Let A and B be two events. If P(A}B)=P(A), then A is.......... of B.

Ans: P(AB)=P(AB)P(B)

P(A)=P(AB)P(B)

P(A)P(B)=P(AB)

So, A is independent of B


About NCERT Exemplar

The NCERT Exemplar for Class 12 Maths Chapter 12 - Probability consists of problems and solutions which is focused on enabling students to get an in-depth understanding of basic concepts of Mathematics. It is recommended that students diligently practise NCERT Class 12 Maths Exemplar Solutions for Chapter 12 - Probability in order to fully understand and grasp complex mathematical problems, and be thoroughly prepared for the Exam. These Exemplar solutions are prepared by our subject experts for Class 12 students.

WhatsApp Banner

FAQs on NCERT Exemplar for Class 12 Maths Chapter-13 (Book Solutions)

1. How to download the NCERT Exemplar for Class 12 Maths Chapter 13 - Probability from the Vedantu Website?

To download the NCERT Book for Class 12 Maths Chapter 12 - Probability in PDF, students can go to the Vedantu website and type the keyword on the search bar. Alternatively, the students can click here to access it. After opening the link, click on the “Download PDF” option. You can also find an array of NCERT Exam preparation materials on the website. Students are encouraged to access multiple study materials and learning tools in order to have the required level of practice for successfully clearing the Examination.

2. What are the Advantages of Solving NCERT Exemplar Questions and Solutions?

There are various benefits that students will get by solving the NCERT Exemplar problem:

  • It Offers a Large Number of Questions of Various Difficulty Levels: This will help students to develop problem-solving abilities and tackle any type of question in the Exams.

  • Develops In-Depth Understanding of Difficult Concepts: The advanced level questions present in the NCERT Exemplar help students to develop a deeper understanding of the respective subjects.

  • Helps to Prepare for Board as well as various Competitive entrance Exams 

  • Many questions asked by the NCERT Board is taken directly from the NCERT Exemplar

3. What are the key differences between NCERT books and NCERT Exemplar problems?

The NCERT textbooks enable students to lay the basic foundation for the subjects. They explain topics in simple language, and consist of easy questions.

While the Exemplar provides complex questions that are of a higher level than that of the NCERT. The questions which are available in NCERT textbook are of basic level meant for understanding the concepts in simple terms. NCERT Exemplar must be referred to by students in order to master the concepts on an advanced level. In addition, the NCERT Exemplar helps students with preparation for competitive Examinations.

4. How does NCERT Exemplar help students for competitive Exams?

The NCERT Exemplar book consists of questions which judge a student’s in-depth understanding of key concepts. While NCERT books help students with their preparation for the Board Examination, the NCERT Exemplar books are considered essential for various competitive Exams such as JEE (Mains and Advanced), AIIMS Exam, NEET etc. As the NCERT Exemplar book contains questions that are of a higher difficulty level, it is considered as an important book for students preparation for the competitive Exams. 

5. Is CBSE the same as NCERT?

No, they’re not the same. CBSE refers to the Central Board for Secondary Education. It is one of the most prestigious and prominent education boards in India. It is a government-affiliated education board. The Central Board for Secondary Education aims to holistically educate children and equip them with the skills required in the real world. The Board is responsible for the publication of the National Council of Educational Research and Training (NCERT) books. In addition, the Board is also responsible for creating the syllabus for students affiliated with the CBSE board.