NCERT Solutions Class 11 Mathematics Chapter 16

Class 11 students mostly refer to NCERT Solutions to have a better understanding of the concepts, so that they can perform well in board and competitive exams.

NCERT solutions make learning easy for students by putting out concepts in a simplified manner.  Students can access the NCERT Solutions  for all the subjects on Extramarks. The team of subject-matter experts has prepared the solutions strictly as per the syllabus and guidelines of CBSE.

NCERT Solutions for Class 11 Mathematics Chapter 16 

If you are finding it difficult to understand Chapter 16 Class 11 Mathematics, you can look at the NCERT solutions provided by Extramarks.

Chapter 16 – Probability is an important topic from the exam point of view and it is important that the students prepare it thoroughly. This chapter is highly related to real-life events. The concept of probability is used by meteorologists for predicting the weather and conducting weather analysis. 

NCERT Solutions for Class 11 Mathematics Chapter 16 – Probability

Chapter 16 Mathematics Class 11 is one of the simplest topics in the syllabus. Probability is beneficial for students not only in academics but also in practical life. Students from different fields can utilise the learnings in their day-to-day life and work smartly. Students can access the NCERT solutions class 11 Mathematics chapter 16 on Extramarks to clear their doubts and can practise the difficult topics to perform better in the Mathematics examination. 

You can access solutions for all the questions available in the textbook through the link given below – 


Access NCERT Solutions for Class-11 Mathematics Chapter 16 Probability


Class 11 Mathematics NCERT Solutions Chapter 16 will help you study the chapter anytime, anywhere irrespective of whether you have an internet connection or not. You can save a lot of time looking for accurate answers to all the questions given at the end of Chapter 16, as Class 11 NCERT Solutions by Extramarks covers them all.

Students can attempt the questions given at the end of Chapter 16, and check the NCERT Solutions for Chapter 16 Mathematics to ensure they are moving on the right track. In addition, access to NCERT solutions accelerates the pace of students in terms of scheduling and building a timeline for their exams. It will also help students in creating a base for their higher studies in the future. 


NCERT Solutions for Class 11 Mathematics Chapter 16 – All Exercises


NCERT Solutions of Class 11 Mathematics Chapter 16

The NCERT solutions for Class 11 Mathematics Chapter 16 are prepared by teachers who have been in the field of education for a long time. With years of experience, they have tried to help students clear their base in Mathematics, and learn some simple and smart tips to save  time. 

A few of the topics that are covered in the Chapter 16 are Random Experiments, Outcomes and Sample Space, Events, Occurrence of an event, Types of events and much more. 

The concept of probability solely depends on chances. It analyses the data to foresee the chances of an event happening.

 The basics of these chapters are taught to students in earlier grades as well to let them get an understanding of the concept. 


Types of Events

Impossible events – These are the events that are highly unlikely to occur for a given experiment. They are denoted by an empty set Φ. The chances of these events are almost null which is why they are called impossible events. 

Simple or elementary event – These are the events that have every outcome of a random experiment. It has all the possible results that might occur from a random experiment.

Compound events: These events have a result with more than one outcome. In this event, there will be multiple outcomes that might occur from an experiment. 

Complementary events: If there is an event A, the complement of A is the event that consists of all the sample space outcomes that do not correspond to the occurrence of A. 

Mutually Exclusive Events: The possibility of any event excludes the possibility of the other event that is called a mutually exclusive event. If there are two events, A and B, and the occurrence of event A completely eliminates the possibility of event B then, A and B are said to be mutually exclusive events. It will be denoted as P(A ∩ B) = 0. 



Let P and Q be two events. The probability formulas are listed below:

All Probability Formulas List in Mathematics
Probability Range 0 ≤ P(P) ≤ 1
Rule of Addition P(P∪Q) = P(P) + P(Q) – P(P∩Q)
Rule of Complementary Events P(P’) + P(Q) = 1
Disjoint Events P(P∩Q) = 0
Independent Events P(P∩Q) = P(P) ⋅ P(Q)
Conditional Probability P(P | Q) = P(P∩Q) / P(Q)
Bayes Formula P(P | Q) = P(Q | P) ⋅ P(P) / P(Q)


Let A and B are two events. The probability formulae are listed below:

Probability Range 0 ≤ P(P) ≤ 1

Rule of Addition P(P∪Q) = P(P) + P(Q) – P(P∩Q)

Rule of Complementary Events P(P’) + P(P) = 1

Disjoint Events P(P∩Q) = 0

Independent Events P(P∩Q) = P(P) ⋅ P(Q)

Conditional Probability P(P | Q) = P(P∩Q) / P(Q)

Bayes’ Theorem P(P | Q) = P(Q | P) ⋅ P(P) / P(Q)


NCERT Solutions For Class 11 Chapter 16 Probability Weightage Marks

The questions from Chapter 16 hold a weightage of 10 – 12 marks approximately. For students appearing for JEE, this chapter is more important as 2 -3 questions on probability are asked in JEE.


Benefits of Probability Class 11 NCERT Solutions

NCERT Solutions for Class 11  Chapter 16 Mathematics benefit students in many ways:

  • Solutions will help students with their predictive analysis. 
  • Students can save a lot of their time instead of solving those complex problems. 

Students can access solutions anytime, even when studying offline.


Related Questions

These related questions will help you gain an insight into what you can expect from the chapter. The questions require some intelligent understanding but if you have your concepts clear, you’ll find them easy to solve.

Q.1 In each of the following Exercises 1 to 7, describe the sample space for the indicated experiment.

1. A coin is tossed three times.
2. A die is thrown two times.
3. A coin is tossed four times.
4. A coin is tossed and a die is thrown.
5. A coin is tossed and then a die is rolled only in case a head is shown on the coin.
6. 2 boys and 2 girls are in Room X, and 1 boy and 3 girls in Room Y. Specify the sample space for the experiment in which a room is selected and then a person.
7. One die of red colour, one of white colour and one of blue colour are placed in a bag. One die is selected at random and rolled, its colour and the number on its uppermost face is noted. Describe the sample space.


1. Sample space for a coin tossed three times is:

2. Sample space for a die thrown two times is:
{(x, y): x, y=1, 2, 3, 4, 5, 6} In other ways:

{(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)}

3. Sample space for a coin tossed four times is:


4. Sample space for a tossed coin thrown die is:


5. Sample space when a coin is tossed and then a die is rolled only when head is shown on the coin:


6. The sample space for the experiment in which a room is selected and then a person is:


7. Dice in red colour, white colour and blue colour are shown by R, W and B respectively. Then the required sample space is


Q.2 An experiment consists of recording boy–girl composition of families with 2 children.

(i) What is the sample space if we are interested in knowing whether it is a boy or girl in the order of their births?
(ii) What is the sample space if we are interested in the number of girls in the family?


The boy and girl are represented by B and G respectively.
(i) The required sample space is:

(ii) Sample space according to number of children in family is:
{ 0, 1, 2 }

Q.3 A box contains 1 red and 3 identical white balls. Two balls are drawn at random in succession without replacement. Write the sample space for this experiment.


A red ball and white ball are represented by R and W respectively.
The required sample space is:
{ RW, WR, WW }.

Q.4 An experiment consists of tossing a coin and then throwing it second time if a head occurs. If a tail occurs on the first toss, then a die is rolled once. Find the sample space.


The head and tail are represented by H and T respectively.
The required sample space, when coin and die is thrown according to given condition, is:
{ HH, HT, T1, T2, T3, T4, T5, T6 }

Q.5 Suppose 3 bulbs are selected at random from a lot. Each bulb is tested and classified as defective (D) or non – defective(N). Write the sample space of this experiment.


The required sample space is:

Q.6 A coin is tossed. If the outcome is a head, a die is thrown. If the die shows up an even number, the die is thrown again. What is the sample space for the experiment?


Sample space according to given conditions is:
{ T, H1, H3, H5, H21, H22, H23, H24, H25, H26, H41, H42, H43, H44, H45, H46, H61, H62, H63, H64, H65, H66 }

Q.7 The numbers 1, 2, 3 and 4 are written separatly on four slips of paper. The slips are put in a box and mixed thoroughly. A person draws two slips from the box, one after the other, without replacement. Describe the sample space for the experiment.


The required sample space is:
{(1, 2), (1, 3), (1, 4), (2, 1), (2, 3), (2, 4), (3, 1), (3, 2), (3, 4), (4, 1), (4, 2), (4, 3)}

Q.8 An experiment consists of rolling a die and then tossing a coin once if the number on the die is even. If the number on the die is odd, the coin is tossed twice. Write the sample space for this experiment.


The required sample space is:
{1HH, 1HT, 1TH, 1TT, 2H, 2T, 3HH, 3HT, 3TH, 3TT, 4H, 4T, 5HH, 5HT, 5TH, 5TT, 6H, 6T }

Q.9 A coin is tossed. If it shows a tail, we draw a ball from a box which contains 2 red and 3 black balls. If it shows head, we throw a die. Find the sample space for this experiment.


The required sample space is:
{TR1, TR2, TB1, TB2, TB3, H1, H2, H3, H4, H5, H6}

Q.10 A die is thrown repeatedly until a six comes up. What is the sample space for this experiment?


Sample space when a die is thrown until a six is obtained.
{ 6, (1, 6), (2,6), (3,6), (4,6), (5,6), (1, 1, 6), (1, 2, 6), (1, 3, 6), (1, 4, 6), (1, 5, 6), (2,1,6), (2, 2, 6),…, (2,5,6), (3,1,6), (3, 2, 6), (3,3,6), (3, 4, 6), (3, 5, 6) …}

Q.11 A die is rolled. Let E be the event “die shows 4” and F be the event “die shows even number”. Are E and F mutually exclusive?


Here, E = {4}, F = {2, 4, 6}
E ∩ F = {4}
Since, E ∩ F ≠


So, E and F are not mutually exclusive events.

Q.12 A die is thrown. Describe the following events:

(i) A: a number less than 7
(ii) B: a number greater than 7
(iii) C: a multiple of 3
(iv) D: a number less than 4
(v) E: an even number greater than 4
(vi) F: a number not less than 3

Also find A ∪ B, A ∩ B, B ∪ C, E ∩ F, D ∩ E, A – C, D – E, E ∩ F′, F′


(i) A = {1, 2, 3, 4, 5, 6}
(ii) B = {} = Φ
(iii) C = {3, 6}
(iv) D = {1, 2, 3}
(v) E = {6}
(vi) F = {3, 4, 5, 6}

A ∪ B = {1, 2, 3, 4, 5, 6},
A ∩ B = {} = Φ
B ∪ C = {3, 6}
E ∩ F = {6}
D ∩ E = {} = Φ
A – C = {1, 2, 4, 5}
D – E = {1, 2, 3}
E ∩ F’ = {6} ∩ [{1, 2, 3, 4, 5, 6} – {3, 4, 5, 6}]
= {6} ∩ {1, 2}
= {} = Φ
F’ = {1, 2}

Q.13 An experiment involves rolling a pair of dice and recording the numbers that come up. Describe the following events:

A: the sum is greater than 8,
B: 2 occurs on either die
C: the sum is at least 7 and a multiple of 3.

Which pairs of these events are mutually exclusive?



A = {(3, 6), (4, 5), (4, 6),(5, 4), (5, 5),(5, 6), (6, 3),(6, 4), (6, 5),(6, 6)}
B = {(1, 2), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (4, 2), (5, 2), (6, 2)}
C = {(3, 6), (4, 5), (5, 4), (6, 3), (6, 6)}
Since, A ∩ B =


. So, A and B are mutually exclusive.
And B ∩ C =


. So, B and C are mutually exclusive.

Q.14 Three coins are tossed once. Let A denote the event ‘three heads show”, B denote the event “two heads and one tail show”, C denote the event” three tails show and D denote the event ‘a head shows on the first coin”.
Which events are (i) mutually exclusive? (ii) simple? (iii) Compound?


S={(HHH),(HHT),(HTH),(THH),(HTT),(THT),(TTH),(TTT)}A={(HHH)}B={(HHT),(HTH),(THH)}C={(TTT)}D={(HTT),(HHH),(HHT),(HTH),(HHT)}(i) Since,​ AB = ϕ, so, A and B are Mutually exclusive.     Since,​ AC = ϕ so, A and C are Mutually exclusive.     Since,​ BC = ϕ so, B and C are Mutually exclusive.     Since,​ CD = ϕ so, C and D are Mutually exclusive.(ii) A and C are simple events because they have single element. (iii) Compound events are B and D.

Q.15 Three coins are tossed. Describe

(i) Two events which are mutually exclusive.
(ii) Three events which are mutually exclusive and exhaustive.
(iii) Two events, which are not mutually exclusive.
(iv) Two events which are mutually exclusive but not exhaustive.
(v) Three events which are mutually exclusive but not exhaustive.


S={(HHH),(HHT),(HTH),(THH),(HTT),(THT),(TTH),(TTT)}(i) Two mutually exclusive events are: {(HHH)} and {(TTT)}.(ii) Three events which are mutually exclusive and  exhaustive:  A={(HHH),(HHT)}      B={(HTH),(THH)}  and     C=  {(HTT),(THT),(TTH),(TTT)}AB=ϕ,BC=ϕ,AC=ϕ and ABC=S(iii) D={(HHH),(HHT),(HTH),(THH),(HTT)}and  E={(HTH),(THH),(HTT),(THT),(TTH)}DEϕ, so, D and E are not mutually exclusive.iv F={(HHH),(HHT),(HTH),(THH),(HTT),(THT),}       G={(TTH)}Since,FG=ϕ but FG=SSo, F and G are exclusive but not exhaustive.(v) H={(HHH),(HHT)}         K={(HTH),(THH)}         L={(HTT),(THT)} HK=ϕ,HL=ϕ      KL=ϕ and HKLSSo, H, K and L are mutually exclusive but not exhaustive.

Q.16 Two dice are thrown. The events A, B and C are as follows:

A: getting an even number on the first die.
B: getting an odd number on the first die.
C: getting the sum of the numbers on the dice ≤ 5.

Describe the events
(i) A′ (ii) not B ,
(iii) A or B (iv) A and B
(v) A but not C (vi) B or C
(vii) B and C (viii) A ∩ B′∩C′


S={(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)}A={(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}B={(1,1),  (1,2),  (1,3),(1,4),(1,5),(1,6)(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)}C={(1,1),  (1,2),  (1,3),(1,4),(2,1),(2,2),(2,3),(3,1),(3,2),(4,1)} (i) A’=SA                      ={(1,1),  (1,2),  (1,3),(1,4),(1,5),(1,6)(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)}=B(ii) not B=SB                      ={(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}=Aiii A or B=AB                      ={(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)}=S(iv) A and B=AB  =ϕ(v) A but not C=AC                                ={(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}{(1,1),  (1,2),  (1,3),(1,4),(2,1),(2,2),(2,3),(3,1),(3,2),(4,1)} MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbba9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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                                ={(2,4),(2,5),(2,6),(4,2),(4,3),(4,4),(4,5),(4,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}(vi) B or C=BC                                ={(1,1),  (1,2),  (1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)}(vii) B and C=BC                                ={(1,1),(1,2),(1,3),(1,4),(3,1),(3,2)}(viii) AB=A(SB)                                =AA[From(ii)]                               =A                               ={(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}                         C=(SC)                               ={(1,5),(1,6),(2,4),(2,5),(2,6),(3,3),(3,4),(3,5),(3,6),(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)}        ABC={(2,4),(2,5),(2,6),(4,2),(4,3),(4,4),(4,5),(4,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}

Q.17 Refer to question 6 above, state true or false: (Give reason for your answer)

(i) A and B are mutually exclusive
(ii) A and B are mutually exclusive and exhaustive
(iii) A = B′
(iv) A and C are mutually exclusive
(v) A and B′ are mutually exclusive.
(vi) A′, B′, C are mutually exclusive and exhaustive.


1. True, because A ∩ B = Φ
2. True, because A ∩ B = Φ and A È B = S.
3. True, because B’= S – B = A.
4. False, because A ∩ C ≠ Φ
5. False, because A ∩ B’ = A ∩ A ≠ Φ [Here, B’ = A]
6. False, because A’ ∩ B’ = Φ, B’ ∩ C ≠ Φ


Which of the following cannot be valid assignment of probabilitiesfor outcomes of sample Space S = {ω1, ω2, ω3, ω4, ω5, ω6, ω7}Assignmentω1ω2ω3ω4ω5ω6ω7(a)


a ω1 + ω2 + ω3 + ω4 + ω5 + ω6 + ω7 = 0.1 + 0.01 + 0.05 + 0.03 + 0.01 + 0.2 + 0.6 = 1Yes, it is a valid assignment of probabilities.b ω1 + ω2 + ω3 + ω4 + ω5 + ω6 + ω7 = 17 + 17 + 17 + 17 + 17 + 17 + 17 = 77 = 1Yes, it is a valid assignment of probabilities.c ω1 + ω2 + ω3 + ω4 + ω5 + ω6 + ω7 = 0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.6 + 0.7 = 2.8No, it is not a valid assignment of probabilities. Because sum of probabilities is greater than 1.d ω1 + ω2 + ω3 + ω4 + ω5 + ω6 + ω7 = - 0.1 + 0.2 + 0.3 + 0.4 + – 0.2 +  0.1 + 0.3No, it is not a valid assignment of probabilities. Because probability can not be negative.e ω1 + ω2 + ω3 + ω4 + ω5 + ω6 + ω7 = 114 +  214 + 314 + 414 + 514 + 614 + 1514 = 1 + 2 + 3 + 4 + 5 + 6 + 1514

Q.19 A coin is tossed twice, what is the probability that at least one tail occurs?


Sample space when a coin is tossed twice,
S(E) = {HH, HT, TH, TT}
Favorable events of getting at least one tail
E = {HT, TH, TT}

P(E)=n(E)S(E) =34Thus, the probability of at least one tail is 34.

Q.20 A die is thrown; find the probability of following events:

(i) A prime number will appear,
(ii) A number greater than or equal to 3 will appear,
(iii) A number less than or equal to one will appear,
(iv) A number more than 6 will appear,
(v) A number less than 6 will appear.


Sample space of a thrown die is:
S(E) = {1,2,3,4,5,6}
(i) Prime number (E) = {2, 3, 5}

P(Prime number)=36=12

(ii) Events of getting a number greater than or equal to 3
= {3, 4, 5, 6}

P(E)=46=23Thus, the required probability is 23.

(iii) Event of getting a number less than or equal to one
= {1}

P(E)=16Thus, the required probability is 13.

(iv) Event of getting a number more than 6
= {}

P(E)=06=0Thus, the required probability is 0.

(v) Event of getting a number less than 6
= {1, 2, 3, 4, 5}

P(E)=56Thus, the required probability is 56.

Q.21 A card is selected from a pack of 52 cards.

(a) How many points are there in the sample space?
(b) Calculate the probability that the card is an ace of spades.
(c) Calculate the probability that the card is
(i) an ace (ii) black card.


(a) There are 52 points in the sample space.
(b) Number of ace of spades = 1

P(ace of spades)=152Thus, the probability of an ace of spades is 152.

(c) (i) Number of aces in a deck of 52 cards = 4

P(ace)=452=113Thus, the probability of an ace card is 113.

(ii) Number of black cards in a deck of 52 cards

P(ace)=2652=12Thus, the probability of an ace card is 12.

Q.22 A fair coin with 1 marked on one face and 6 on the other and a fair die are both tossed, find the probability that the sum of numbers that turn up is (i) 3 (ii) 12


Sample events when one coin(faced 1 and 6) and a dice: S={(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}n(S)=12(i) Events in which sum is 3: E={(1,2)}P(E)=112 Thus, the required probability is 112.(ii) Events in which sum is 12: E={(6,6)}P(E)=112Thus, the required probability is 112.

Q.23 There are four men and six women on the city council. If one council member is selected for a committee at random, how likely is it that it is a woman?


Number of people in a council = 10
Number of women in a council = 6

P(woman)=610=35Thus, the required probability is 35.

Q.24 A fair coin is tossed four times, and a person win ₹ 1 for each head and lose ₹1.50 for each tail that turns up. From the sample space calculate how many different amounts of money you can have after four tosses and the probability of having each of these amounts.


Sample space when a coin is tossed four times:

A person gain on getting each head = ₹ 1
A person loses on getting each tail = ₹ 1.50
Gain on getting HHHH = ₹ 4
Gain on getting HHHT = ₹ 1.50
Gain on getting HHTH = ₹ 1.50
Gain on getting HTHH = ₹ 1.50
Gain on getting THHH = ₹ 1.50
Loss on getting HHTT = ₹ 1.00
Loss on getting HTHT = ₹ 1.00
Loss on getting THHT = ₹ 1.00
Loss on getting THTH = ₹ 1.00
Loss on getting TTHH = ₹ 1.00
Loss on getting HTTH = ₹ 1.00
Loss on getting TTTH = ₹ 3.50
Loss on getting TTHT = ₹ 3.50
Loss on getting THTT = ₹ 3.50
Loss on getting HTTT = ₹ 3.50
Loss on getting TTTT = Rs. 6

       P(Winning4)=116P(Winning1.50)=416   =14   P(Losing1.00)=616   =38   P(Losing3.50)=416=14    P(Losing6.00)=116

Q.25 Three coins are tossed once. Find the probability of getting

(i) 3 heads
(ii) 2 heads
(iii) at least 2 heads
(iv) at most 2 heads
(v) no head (vi) 3 tails
(vii) exactly two tails
(viii) no tail
(ix) at most two tails


Sample space of tossing three coins is:

i P3 heads=18ii P2 heads=38iii Pat least 2 heads=48=12iv Pat most2 heads=78v PNo head=18vi P3 tails=18vii Pexactly 2 tails=38viii PNo tail=18ix Pat most2 tails=78


If ​ 211 is the probability of an event, what is the probability of the event ‘not A’.


Let probability of an event be P(A).Then,       P(A)=211P(notA)=1P(A)          =1211          =11211          =911Thus, the probability of the event ‘not A’ is 911.

Q.27 A letter is chosen at random from the word ‘ASSASSINATION’. Find the probability that letter is
(i) a vowel (ii) a consonant


Number of letters in the word “ASSASSINATION=13Number of vowels in the word “ASSASSINATION=6Number of consonants in the word “ASSASSINATION=136 =7(i)            P(a vowel)=613 [P(E)=n(E)S(E)](ii) P(a consonant)=713

Q.28 In a lottery, a person choses six different natural numbers at random from 1 to 20, and if these six numbers match with the six numbers already fixed by the lottery committee, he wins the prize. What is the probability of winning the prize in the game?


Total numbers from 1 to 20 = 20
Numbers to be selected = 6

Number of ways to select 6 numbers from 20 numbers =20C6 =20!6!(206)! =20×19×18×17×16×15×14!6×5×4×3×2×1×14! =38760Number of ways to fix 6 numbers for winning lottery = 1   P(Winning the prize)=138760

Q.29 Check whether the following probabilities P(A) and P(B) are consistently defined

(i) P(A) = 0.5, P(B) = 0.7, P(A ∩ B) = 0.6
(ii) P(A) = 0.5, P(B) = 0.4, P(A ∪ B) = 0.8


(i) Here,​ P(AB)>P(A) [P(AB)=0.6 and P(A)=0.5]    So, the probabilities P(A) and P(B) are not consistently defined.(ii) Here,P(AB)>P(A) and P(AB)>P(B).    So, the probabilities P(A) and P(B) are consistently defined.


Fill in the blanks in following table:P(A)P(B)P(AB)P(AB)(i)1315115.……(ii)0.35.……0.250.6(iii)0.50.35.……0.7


i PAB=PA+PBPAB                   =13+15115                   =5+3115                   =715ii PAB=PA+PBPAB               0.6=0.35+PB0.25        0.60.1=PB               PB=0.5iii  PAB=PA+PBPAB                         0.7=0.5+0.35PAB          PAB=0.850.7                                 =0.15


Given P(A) = 35  and P(B) = 15. Find P(A or B), if A and B are mutually exclusive events.


Since, A and B are mutually exclusive events. So,AB=ϕP(AB)=0P(AB)=P(A)+P(B)P(AB)     =35+150     =45Therefore, P(AorB) is 45.


If E and F are events such that P(E) = 14, P(F) = 12 and P(E and F) = 18, find (i)P(E or F), (ii)P(not E and not F).


Given: P(E)=14, P(F)=12 and P(EF)=18(i) P(EF)=P(E)+P(F)P(EF)          =14+1218          =2+418          =58Therefore, P(EorF) is 58.(ii) P(notE and not F)=P(EF)        =P(EF)[By Demorgan’s Law]        =1P(EF)        =158        =858        =38Therefore, P(notE and not F) is equal to 38.

Q.33 Events E and F are such that P(not E or not F) = 0.25, State whether E and F are mutually exclusive.


P(not E or not F) =0.25                    P(EF)=0.25                    P(EF)=0.25[By Demorgarn’s law]             1P(EF)=0.25[P(E)=1P(E)]     P(EF)=10.25     P(EF)=0.75            EF0So, E and F are not mutually exclusive.

Q.34 A and B are events such that P(A) = 0.42,
P(B) = 0.48 and P(A and B) = 0.16.
Determine (i) P(not A), (ii) P(not B) and (iii) P(A or B)


Given:P(A)=0.42, P(B)=0.48 and P(AB)=0.16(i)    P(notA)=P(A)         =1P(A)         =10.42         =0.58(ii)    P(notB)=P(B)       =1P(B)       =10.48       =0.52(iii)P(AorB)=P(AB) =P(A)+P(B)P(AB) =0.42+0.480.16 =0.74

Q.35 In Class XI of a school 40% of the students study Mathematics and 30% study Biology. 10% of the class study both Mathematics and Biology. If a student is selected at random from the class, find the probability that he will be studying Mathematics or Biology.


PMathematics​ = 40%               PM​​​ = 0.40PBiology​ = 30%         PB ​= 0.30 PMathematics and Biology=10% PMB =0.10 PMathematics or Biology=PMB     =PM+PBPMB      =0.40+0.300.10      =0.60Therefore, probability that student wiall be studying Mathematics or Biology is 0.60.

Q.36 In an entrance test that is graded on the basis of two examinations, the probability of a randomly chosen student passing the first examination is 0.8 and the probability of passing the second examination is 0.7. The probability of passing at least one of them is 0.95. What is the probability of passing both?


Let probability of passing in both examinations  be  P(A) and P(B) respectively.Then, we haveP(A)=0.8, P(B)=0.7 and P(AB)=0.95 P(AB)=P(A)+P(B)P(AB)      =0.8+0.70.95      =0.55Thus, the probability of passing in both the examination is 0.55.

Q.37 The probability that a student will pass the final examination in both English and Hindi is 0.5 and the probability of passing neither is 0.1. If the probability of passing the English examination is 0.75, what is the probability of passing the Hindi examination?


Let the probability that a student will pass the final examination in both English and Hindi =P(EH) P(EH)=0.5Let the probability that a student will not pass the final examination in neither English nor Hindi =P(EH)        P(EH)=0.1         P(EH)=0.1[By Demorgan’s law]   1  P(EH)=0.1           P(EH)=10.1           P(EH)=0.9The probability of passing the English examination        =P(E)        =0.75                                   P(EH)=P(E)+P(H)P(EH)         0.9 =0.75+P(H)0.5       P(H)=0.90.75+0.5        =0.65Therefore, the probability of passing the Hindi examination is 0.65.

Q.38 In a class of 60 students, 30 opted for NCC, 32 opted for NSS and 24 opted for both NCC and NSS. If one of these students is selected at random, find the probability that

(i) The student opted for NCC or NSS.
(ii) The student has opted neither NCC nor NSS.
(iii) The student has opted NSS but not NCC.


               Number of students in a class=60                     Number of students in NCC=30                     Number of students in NSS=32Number of students in NCC and NSS=24      P(NCC)=3060=12      P(NSS)=3260=815 P(NCCNSS)=2460=615(i)P(NCC or NSS)=P(NCCNSS)      =P(NCC)+P(NSS)P(NCCNSS)      =12+815615      =15+161230      =1930(ii)P(Neither ​NCC nor NSS)      =P{(NCC)(NSS)}      =P(NCCNSS)      =1P(NCCNSS)      =11930      =1130(iii) P(NSS but not NCC)=P{(NSS)(NCC)}   =P(NSS)P(NSSNCC)   =815615   =215

Q.39 A box contains 10 red marbles, 20 blue marbles and 30 green marbles. 5 marbles are drawn from the box, what is the probability that

(i) all will be blue?
(ii) at least one will be green?


     Number of red balls in box=10   Number of blue balls in box=20Number of green balls in box=30  Total number of balls in box=60Number of ways to select 5 balls      =60C5(i) Number of ways to select 5 blue balls      =20C5    P(5blueballs)=20C560C5(ii) P(at least one ball will be green)     =1P(no green ball)     =130C560C5

Q.40 4 cards are drawn from a well – shuffled deck of 52 cards. What is the probability of obtaining 3 diamonds and one spade?


Total cards in well-shuffled deck=52Number of drawn cards from deck=4Number of diamonds in deck=13Number of spade in deck=13P3diamonds and 1 spade=13C3×13C152C4

Q.41 A die has two faces each with number ‘1’, three faces each with number ‘2’ and one face with number ‘3’. If die is rolled once, determine

(i) P(2)
(ii) P(1 or 3)
(iii) P(not 3)


Possible events in throwing a given dice,S(E)={1,1,2,2,2,3}(i) P(2)=n(E)S(E)                  =36                   =12(ii) P(1 or 3)=P(1)+P(3)      =26+16           =36           =12(iii) P(not3)=n(E)S(E)           =56

Q.42 In a certain lottery 10,000 tickets are sold and ten equal prizes are awarded.
What is the probability of not getting a prize if you buy (a) one ticket (b) two tickets (c) 10 tickets.


          Total number of lottery tickets=10,000              Total number of equal prizes=10Number of tickets not having prize=10,00010        =9990(a)            P(1 prize ticket)=10C110000C1        =1010,000        =11000                 P(1 ticket of without prize)=1P(1 prize ticket)       =111000                 P(1 ticket of without prize)=9991000(b)         P(2 tickets of without prize)=9990C210000C2(c)      P(10 tickets of without prize)=9990C1010000C10

Q.43 Out of 100 students, two sections of 40 and 60 are formed. If you and your friend are among the 100 students, what is the probability that

(a) you both enter the same section?
(b) you both enter the different sections?


Number of students in two sections=100       Number of students in section A=40       Number of students in section B=60Total number of ways to fill both section by students         =100C40×60C60         =100!40!×60!×1         =100!40!×60!(a) When both enter in the same section:Probability(Both are in same section)=98C38×60C60+98C58×40C40(100!40!×60!)         =98!38!60!×1+98!58!40!×1(100!40!×60!)         =98!38!58!(160×59+140×39)(100×99×98!40×39×38!×60×59×58!)         =(40×39+60×5960×59×40×39)(100×9940×39×60×59)         =1560+35409900         =51009900         =5199         =1733Therefore, when both students are in the same section.Then,the required probability is 1733.(b)P(When both are not in same section)         =1Probability(Both are in same section)         =11733=1633 Then, the required probability is 1633.

Q.44 Three letters are dictated to three persons and an envelope is addressed to each of them, the letters are inserted into the envelopes at random so that each envelope contains exactly one letter. Find the probability that at least one letter is in its proper envelope.


Number of letters = 3
Number of persons = 3
Total number of ways to insert 3
letters in 3 envelops = 3!
= 6

Number of ways of inserting 3 letters in 3 envelops
so that no letter is in proper envelope = 1
Total number of ways of inserting 3 letters in 3
envelopes = 3! – one letter is in proper
envelopes and two letters are in wrong envelops – one letter in right envelopes and two letters are in
wrong envelopes – Number of ways in which all are in proper envelopes
= 6 – 3C1 x 1 – 1
= 6 – 3 – 1
= 2

P(at least one letter is in its proper envelope) =1P(No letter is in its proper envelope) =126 =113 =23Thus, the required propbability is 23.

Q.45 A and B are two events such that P(A) = 0.54, P(B) = 0.69 and P(A ∩ B) = 0.35.
Find (i) P (A∪B) (ii) P(A´∩B´) (iii) P(A ∩ B´) (iv) P(B ∩ A´)


We have, P(A)=0.54,  P(B)=0.69 and P(AB)=0.35(i)    P(AB)=P(A)+P(B)P(AB)         =0.54+0.690.35         =0.88(ii)P(AB)=P(AB)[By Demorgan’s theorem]        =1P(AB)        =10.88        =0.12(iii) P(AB)=P(A)P(AB)        =0.540.35        =0.19(iv) P(BA)=P(B)P(AB)        =0.690.35        =0.34

Q.46 From the employees of a company, 5 persons are selected to represent them in the managing committee of the company. Particulars of five persons are as follows:

Sr. No. Name Sex Age in years
1 Harish M 30
2 Rohan M 46
3 Sheetal F 46
4 Alis F 28
5 Salim M 41

A person is selected at random from this group to act as a spokesperson. What is the probability that the spokesperson will be either male or over 35 years?


     Number of male=3Number of female over 35 years =1Total number of people =5P(Male or over 35 years people)=45

Q.47 If 4-digit numbers greater than 5,000 are randomly formed from the digits 0, 1, 3, 5, and 7, what is the probability of forming a number divisible by 5 when,
(i) the digits are repeated?
(ii) the repetition of digits is not allowed?


Given digits are 0,1,3,5 and 7.(i) When digits are repeated.Total number of 4 digits number greater            or equal to 5000=2×5×5×5              =250 Therefore, number of 4-digits number greater than 5000             =2501             =249A number is divisible by 5 if its one’s digit is either 0 or 5.Therefore, 4-digits number formed by given digits and         greater or equal to 5000=2×5×5×2              =100Therefore, number of 4-digits number greater than 5000             =1001             =99P(When digits are repeated)=99249              =3383(ii)When repeatition of digits is not allowed.Total number of 4-digit numbers formed from the digits                               0,1,3,5 and 7=2×4×3×2              =48Number of 4-digit number divisible by 5:               When one’s digit is 0=2×3×2×1              =12Number of 4-digit number divisible by 5:               When one’s digit is 5=1×3×2×1              =6Total numbers divisible by 5=12+6              =18 P(When digits are not repeated)              =1848              =38Hence, probability of forming a number divisible by 5 is 38.

Q.48 The number lock of a suitcase has 4 wheels, each labelled with ten digits i.e., from 0 to 9. The lock opens with a sequence of four digits with no repeats. What is the probability of a person getting the right sequence to open the suitcase?


Number of ways to select 4 digits out of 10 digits=10C4          =10!4!6!          =10×9×8×74×3×2×1          =210Number of ways to arrange 4 selected digits=210×4!          =5040Since, there is only one sequence to open the lock. P(right sequence to open the lock)=15040

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