NCERT Solutions for Class 11 Maths Chapter 16 Probability
Free PDF download of NCERT Solutions for Class 11 Math Chapter 16 Exercise 16.3 (Ex 16.3) and all chapter exercises at one place prepared by expert teachers as per NCERT (CBSE) books guidelines. Class 11 Math Chapter 16 Probability Exercise 16.3 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all exercise solutions in your emails.
Class:  
Subject:  
Chapter Name:  
Exercise:  Exercise  16.3 
ContentType:  Text, Videos, Images and PDF Format 
Academic Year:  202425 
Medium:  English and Hindi 
Available Materials: 

Other Materials 

What is Exercise 16.3 of Class 11 Maths Chapter 16 Probability About?
Class 11 Maths Chapter 16 Probability contains three exercises and one miscellaneous exercise. Here, the NCERT Solutions for Class 11 Chapter 16 Exercise 16.3 are given, which has 21 questions  MCQs, word problems and short type questions.
The exercise begins with basic questions to establish the main concept of the chapter, that is, the sum of probabilities of all individual elements in a given sample space is always equal to one. The level of difficulty gradually increases with questions asking the probabilities of different events, which may be mutually exclusive or mutually exclusive but exhaustive or event B, such that P(B) = 1  P(A).
In Chapter 16 Probability Ex 16.3, students will learn the following important topics of Probability.
Axiomatic approach to Probability
Probability of an event
Probabilities of equally likely outcomes
Probability of the event ‘A or B’
Probability of event ‘not A’
The three primary concepts under the axiomatic theory of probability that students must remember to solve questions in this exercise are:
The probability of any event is either greater or equal to zero.
The probability of sample space is equal to 1.
For two mutually exclusive events A and B, P (A OR B) = P(A) + P(B).
Questions now combine the knowledge of sample space and events with axioms of probability. This provides us with a list of questions that help us practice and also deepen our understanding of this axiomatic approach to probability.
Access NCERT Solutions for Class 11 Chapter 16  Probability
Exercise 16.3
1. Which of the following cannot be valid assignment of probabilities for outcomes of sample space \[S = \left( {{\omega _1},{\omega _2},{\omega _3},{\omega _4},{\omega _5},{\omega _6},{\omega _7}} \right)\]?
Assignment
Assignment  ω1  ω2  ω3  ω4  ω5  ω6  ω7 
(a)  0.1  0.01  0.05  0.03  0.01  0.2  0.6 
(b)  1/7  1/7  1/7  1/7  1/7  1/7  1/7 
(c)  0.1  0.2  0.3  0.4  0.5  0.6  0.7 
(d)  0.1  0.2  0.3  0.4  0.2  0.1  0.3 
(e)  1/14  2/14  3/14  4/14  5/14  6/14  15/14 
Ans:
(a) Probability should be less than 1 and positive.
\[P\left( E \right) = 0.01 + 0.05 + 0.03 + 0.01 + 0.2 + 0.6\]
\[P\left( E \right) = 1\]
Therefore, the given assignment is valid.
b) Probability should be less than 1 and positive.
\[P\left( E \right) = \dfrac{1}{7} + \dfrac{1}{7} + \dfrac{1}{7} + \dfrac{1}{7} + \dfrac{1}{7} + \dfrac{1}{7} + \dfrac{1}{7}\]
\[P\left( E \right) = 1\]
Therefore, the given assignment is valid.
c) Probability should be less than 1 and positive.
\[P\left( E \right) = 0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.6 + 0.7\]
\[P\left( E \right) = 2.8 > 1\]
Here probability is greater than 1.
Therefore, the given assignment is not valid.
d) Probability can never be negative.
Therefore, the assignment is not valid.
e) Probability should be less than 1 and positive.
\[P\left( E \right) = \dfrac{1}{{14}} + \dfrac{2}{{14}} + \dfrac{3}{{14}} + \dfrac{4}{{14}} + \dfrac{5}{{14}} + \dfrac{6}{{14}} + \dfrac{{15}}{{14}}\]
\[P\left( E \right) = \dfrac{{28}}{{14}} > 1\]
Therefore, the assignment is not valid.
2. A coin is tossed twice, what is the probability that at least one tail occurs?
Ans: Here coin is tossed twice, then sample space is \[S = \left( {TT,HH,TH,HT} \right)\]
Number of possible outcomes \[n\left( S \right) = 4\]
Let A be the event of getting at least one tail
\[\therefore n\left( A \right) = 3\]
\[P\left( E \right){\text{ }} = {\text{ }}\dfrac{{Number{\text{ }}of{\text{ }}outcomes{\text{ }}favorable{\text{ }}to{\text{ }}event}}{{Total{\text{ }}number{\text{ }}of{\text{ }}possible{\text{ }}outcomes}}\]
\[P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}\]
\[P\left( A \right) = \dfrac{3}{4}\]
3. A die is thrown, find the probability of following events:
(i) A prime number will appear,
Ans: Here die is thrown, therefore the sample space of outcomes will be \[S = \left\{ {1,2,3,4,5,6} \right\}\]
Let A be the event of getting a prime number,
\[A = \left\{ {2,3,5} \right\}\]
Then, \[n\left( A \right) = 3\]
\[P\left( E \right){\text{ }} = {\text{ }}\dfrac{{Number{\text{ }}of{\text{ }}outcomes{\text{ }}favorable{\text{ }}to{\text{ }}event}}{{Total{\text{ }}number{\text{ }}of{\text{ }}possible{\text{ }}outcomes}}\]
\[P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}\]
\[P\left( A \right) = \dfrac{3}{6}\]
\[P\left( A \right) = \dfrac{1}{2}\]
(ii) A number greater than or equal to 3 will appear,
Ans: Let B be the event of getting a number greater than or equal to 3,
\[B = \left\{ {3,4,5,6} \right\}\]
Then, \[n\left( B \right) = 4\]
\[P\left( E \right){\text{ }} = {\text{ }}\dfrac{{Number{\text{ }}of{\text{ }}outcomes{\text{ }}favorable{\text{ }}to{\text{ }}event}}{{Total{\text{ }}number{\text{ }}of{\text{ }}possible{\text{ }}outcomes}}\]
\[P\left( B \right) = \dfrac{4}{6}\]
\[P\left( B \right) = \dfrac{2}{3}\]
(iii) A number less than or equal to one will appear,
Ans: Let A be the event of getting a number less than or equal to 1,
\[A = \left\{ 1 \right\}\]
Then, \[n\left( A \right) = 1\]
\[P\left( E \right){\text{ }} = {\text{ }}\dfrac{{Number{\text{ }}of{\text{ }}outcomes{\text{ }}favorable{\text{ }}to{\text{ }}event}}{{Total{\text{ }}number{\text{ }}of{\text{ }}possible{\text{ }}outcomes}}\]
\[P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}\]
\[P\left( A \right) = \dfrac{1}{6}\]
(iv) A number more than 6 will appear,
Ans: Let A be the event of getting a number more than 6, then
\[A = \left\{ 0 \right\}\]
Then, \[n\left( A \right) = 0\]
\[P\left( E \right){\text{ }} = {\text{ }}\dfrac{{Number{\text{ }}of{\text{ }}outcomes{\text{ }}favorable{\text{ }}to{\text{ }}event}}{{Total{\text{ }}number{\text{ }}of{\text{ }}possible{\text{ }}outcomes}}\]
\[P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}\]
\[P\left( A \right) = \dfrac{0}{6}\]
\[P\left( A \right) = 0\]
(v) A number less than 6 will appear.
Ans: Let E be the event of getting a number less than 6, then
\[A = \left( {1,2,3,4,5} \right)\]
Then, \[n\left( A \right) = 5\]
\[P\left( E \right){\text{ }} = {\text{ }}\dfrac{{Number{\text{ }}of{\text{ }}outcomes{\text{ }}favorable{\text{ }}to{\text{ }}event}}{{Total{\text{ }}number{\text{ }}of{\text{ }}possible{\text{ }}outcomes}}\]
\[P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}\]
\[P\left( A \right) = \dfrac{5}{6}\]
4. A card is selected from a pack of 52 cards.
(a) How many points are there in the sample space?
Ans: Number of points in the sample space = 52
\[\therefore {\text{ }}n\left( S \right) = 52\]
(b) Calculate the probability that the card is an ace of spades.
Ans: Let A be the event of drawing an ace of spades.
\[A = \left\{ 1 \right\}\]
Then, \[n\left( A \right) = 1\]
\[P\left( E \right){\text{ }} = {\text{ }}\dfrac{{Number{\text{ }}of{\text{ }}outcomes{\text{ }}favorable{\text{ }}to{\text{ }}event}}{{Total{\text{ }}number{\text{ }}of{\text{ }}possible{\text{ }}outcomes}}\]
\[P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}\]
\[P\left( A \right) = \dfrac{1}{{52}}\]
(c) Calculate the probability that the card is
(i) an ace
Ans: Let A be the event of drawing an ace. There are four aces.
Then, \[n\left( A \right) = 4\]
\[P\left( E \right){\text{ }} = {\text{ }}\dfrac{{Number{\text{ }}of{\text{ }}outcomes{\text{ }}favorable{\text{ }}to{\text{ }}event}}{{Total{\text{ }}number{\text{ }}of{\text{ }}possible{\text{ }}outcomes}}\]
\[P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}\]
\[P\left( A \right) = \dfrac{4}{{52}}\]
\[P\left( A \right) = \dfrac{1}{{13}}\]
(ii) black card
Ans: Let C be the event of drawing a black card. There are 26 black cards.
Then, \[n\left( A \right) = 26\]
\[P\left( E \right){\text{ }} = {\text{ }}\dfrac{{Number{\text{ }}of{\text{ }}outcomes{\text{ }}favorable{\text{ }}to{\text{ }}event}}{{Total{\text{ }}number{\text{ }}of{\text{ }}possible{\text{ }}outcomes}}\]
\[P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}\]
\[P\left( A \right) = \dfrac{{26}}{{52}}\]
\[P\left( A \right) = \dfrac{1}{2}\]
5. 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
Ans: As die and a coin marked 1 is tossed therefore, the sample space of possible outcomes will be,
\[S = \left\{ {\left( {1,1} \right),\left( {1,2} \right),\left( {1,3} \right),\left( {1,4} \right),\left( {1,5} \right),\left( {1,6} \right),\left( {6,1} \right),\left( {6,2} \right),\left( {6,3} \right),\left( {6,4} \right),\left( {6,5} \right),\left( {6,6} \right)} \right\}\]
Therefore, \[n\left( S \right) = 12\]
Let A be the event having sum of numbers as 3.
\[A = \left\{ {\left( {1,2} \right)} \right\}\]
\[n\left( A \right) = 1\]
\[P\left( E \right){\text{ }} = {\text{ }}\dfrac{{Number{\text{ }}of{\text{ }}outcomes{\text{ }}favorable{\text{ }}to{\text{ }}event}}{{Total{\text{ }}number{\text{ }}of{\text{ }}possible{\text{ }}outcomes}}\]
\[P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}\]
\[P\left( A \right) = \dfrac{1}{{12}}\]
(ii) 12
Ans: Let A be the event having sum of number as 12.
\[A = \left\{ {\left( {6,6} \right)} \right\}\]
\[n\left( A \right) = 1\]
\[P\left( E \right){\text{ }} = {\text{ }}\dfrac{{Number{\text{ }}of{\text{ }}outcomes{\text{ }}favorable{\text{ }}to{\text{ }}event}}{{Total{\text{ }}number{\text{ }}of{\text{ }}possible{\text{ }}outcomes}}\]
\[P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}\]
\[P\left( A \right) = \dfrac{1}{{12}}\]
6. 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?
Ans: \[{\text{Total members in the council}} = 4 + 6\]
\[{\text{Total members in the council}} = 10\]
Therefore,
\[n\left( S \right) = 10\]
Number of women are 6
Let A be the event of selecting a woman
Therefore, \[n\left( A \right) = 6\]
\[P\left( E \right){\text{ }} = {\text{ }}\dfrac{{Number{\text{ }}of{\text{ }}outcomes{\text{ }}favorable{\text{ }}to{\text{ }}event}}{{Total{\text{ }}number{\text{ }}of{\text{ }}possible{\text{ }}outcomes}}\]
\[P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}\]
\[P\left( A \right) = \dfrac{6}{{10}}\]
\[P\left( A \right) = \dfrac{3}{5}\]
7. A fair coin is tossed four times, and a person win Rs 1 for each head and lose Rs 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.
Ans: Here, coin is tossed four times so the possible sample space contains,
S={HHHH,HHHT,HHTH,HTHH,THHH,HHTT,HTHT,THHT,HTTH,THTH,TTHH,TTTH,TTHT,THTT,HTTT,TTTT}
According to question, a person will win or lose money depending up on the face of the coin so,
(i) For 4 heads \[ = 1 + 1 + 1 + 1\] = ₹ 4
Therefore, he wins ₹ 4
(ii) For 3 heads and 1 tail \[ = 1 + 1 + 11.50\]
Therefore, he wins ₹ 1.50
(iii) For 2 heads and 2 tails = 1 + 1 – 1.50 – 1.50
= 2 – 3
= – ₹ 1
Therefore, he will be losing ₹ 1
(iv) For 1 head and 3 tails = 1 – 1.50 – 1.50 – 1.50
= 1 – 4.50
= – ₹ 3.50
Therefore, he will be losing Rs. 3.50
(v) For 4 tails = – 1.50 – 1.50 – 1.50 – 1.50
= – ₹ 6
Thererfore, he will be losing Rs. 6
So, the new sample space will be
\[S = {\text{ }}\left\{ {4,1.50,1.50,1.50,1.50,1,1,1,1,1,1,3.50,3.50,3.50,3.50,6} \right\}\]
Then, \[n\left( S \right) = 16\]
P (winning ₹ 4) \[ = \dfrac{1}{{16}}\]
P (winning ₹ 1.50) \[ = \dfrac{4}{{16}}\]
P (winning ₹ 1.50) \[ = \dfrac{1}{4}\]
P (loosing ₹ 1) \[ = \dfrac{6}{{16}}\]
P (loosing ₹ 1) \[ = \dfrac{3}{8}\]
P (loosing ₹ 3.50) \[ = \dfrac{4}{{16}}\]
P (loosing ₹ 3.50) \[ = \dfrac{1}{4}\]
P (loosing ₹ 6) \[ = \dfrac{1}{{16}}\]
8. Three coins are tossed once. Find the probability of getting
(i) 3 heads
Ans: Three coin is tossed so the possible sample space contains,
\[S = \left\{ {HHH,HHT,HTH,THH,TTH,HTT,TTT,THT} \right\}\]
\[n\left( S \right) = 8\]
Let A be the event of getting 3 heads
\[n\left( A \right) = 1\]
\[P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}\]
\[P\left( A \right) = \dfrac{1}{8}\]
(ii) 2 heads
Ans: Three coin is tossed so the possible sample space contains,
\[S = \left\{ {HHH,HHT,HTH,THH,TTH,HTT,TTT,THT} \right\}\]
\[n\left( S \right) = 8\]
Let A be the event of getting 2 heads
\[n\left( A \right) = 3\]
\[P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}\]
\[P\left( A \right) = \dfrac{3}{8}\]
(iii) at least 2 heads
Ans: Three coin is tossed so the possible sample space contains,
\[S = \left\{ {HHH,HHT,HTH,THH,TTH,HTT,TTT,THT} \right\}\]
\[n\left( S \right) = 8\]
Let A be the event of getting at least 2 head
\[n\left( A \right) = 4\]
\[P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}\]
\[P\left( A \right) = \dfrac{4}{8}\]
\[P\left( A \right) = \dfrac{1}{2}\]
(iv) at most 2 heads
Ans: Three coin is tossed so the possible sample space contains,
\[S = \left\{ {HHH,HHT,HTH,THH,TTH,HTT,TTT,THT} \right\}\]
\[n\left( S \right) = 8\]
Let A be the event of getting at most 2 heads
\[n\left( A \right) = 7\]
\[P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}\]
\[P\left( A \right) = \dfrac{7}{8}\]
(v) no head
Ans: Three coin is tossed so the possible sample space contains,
\[S = \left\{ {HHH,HHT,HTH,THH,TTH,HTT,TTT,THT} \right\}\]
\[n\left( S \right) = 8\]
Let A be the event of getting no heads
\[n\left( A \right) = 1\]
\[P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}\]
\[P\left( A \right) = \dfrac{1}{8}\]
(vi) 3 tails
Ans: Three coin is tossed so the possible sample space contains,
\[S = \left\{ {HHH,HHT,HTH,THH,TTH,HTT,TTT,THT} \right\}\]
\[n\left( S \right) = 8\]
Let A be the event of getting 3 tails
\[n\left( A \right) = 1\]
\[P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}\]
\[P\left( A \right) = \dfrac{1}{8}\]
(vii) Exactly two tails
Ans: Three coin is tossed so the possible sample space contains,
\[S = \left\{ {HHH,HHT,HTH,THH,TTH,HTT,TTT,THT} \right\}\]
\[n\left( S \right) = 8\]
Let A be the event of getting exactly 2 tails
\[n\left( A \right) = 3\]
\[P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}\]
\[P\left( A \right) = \dfrac{3}{8}\]
(viii) no tail
Ans: Three coin is tossed so the possible sample space contains,
\[S = \left\{ {HHH,HHT,HTH,THH,TTH,HTT,TTT,THT} \right\}\]
\[n\left( S \right) = 8\]
Let A be the event of getting no tails
\[n\left( A \right) = 1\]
\[P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}\]
\[P\left( A \right) = \dfrac{1}{8}\]
(ix) at most two tails
Ans: Three coin is tossed so the possible sample space contains,
\[S = \left\{ {HHH,HHT,HTH,THH,TTH,HTT,TTT,THT} \right\}\]
\[n\left( S \right) = 8\]
Let A be the event of getting at most 2 tails
\[n\left( A \right) = 7\]
\[P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}\]
\[P\left( A \right) = \dfrac{7}{8}\]
9. If 2/11 is the probability of an event, what is the probability of the event ‘not A’.
Ans: According to question, it is given that, 2/11 is the probability of an event A,
i.e. \[P\left( A \right) = \dfrac{2}{{11}}\]
Therefore,
\[P\left( {{\text{not }}A} \right) = 1  \dfrac{2}{{11}}\]
\[P\left( {{\text{not }}A} \right) = \dfrac{9}{{11}}\]
10. A letter is chosen at random from the word ‘ASSASSINATION’. Find the probability that letter is
(i) a vowel
Ans: Total letters in the given word is 13
\[n\left( S \right) = 13\]
Number of vowels in the given word is 6
Let A be the event of selecting a vowel
\[n\left( A \right) = 6\]
\[P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}\]
\[P\left( A \right) = \dfrac{6}{{13}}\]
(ii) a consonant
Ans: Total letters in the given word is 13
\[n\left( S \right) = 13\]
Number of consonants in the given word = 7
Let A be the event of selecting the consonant
\[n\left( A \right) = 7\]
\[P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}\]
\[P\left( A \right) = \dfrac{7}{{13}}\]
11. 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? (Hint: order of the numbers is not important.)
Ans: According to question.
Total numbers of numbers in the draw is 20
Numbers to be selected is 6
Therefore,
\[n\left( S \right){ = ^{20}}{C_6}\]
\[n\left( S \right)=\frac{\left \underline{20} \right.}{\left \underline{6\left \underline{206} \right.} \right.}\]
\[n\left( S \right) = 38760\]
Let be the event that six numbers match with the six numbers already fixed by the lottery committee.
\[n\left( A \right){ = ^6}{C_6}\]
\[n\left( A \right) = 1\]
Therefore, the probability of winning the prize,
\[P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}\]
\[P\left( A \right) = \dfrac{1}{{38760}}\]
12. Check whether the following probabilities P(A) and P(B) are consistently defined
(i) \[P\left( A \right) = 0.5,{\text{ }}P\left( B \right) = 0.7,{\text{ }}P\left( {A \cap B} \right) = 0.6\]
Ans:\[P\left( {A \cap B} \right) > P\left( A \right)\]
Therefore, the given probabilities are not consistently defined.
(ii) \[P\left( A \right) = 0.5,{\text{ }}P\left( B \right) = 0.4,{\text{ }}P\left( {A \cup B} \right) = 0.8\]
Ans: As we know that,
\[P(A \cup B) = P\left( A \right) + P\left( B \right)P(A \cap B)\]
\[0.8 = 0.5 + 0.4P(A \cap B)\]
\[0.8  0.9 = P(A \cap B)\]
\[P(A \cap B) = 0.1\]
Therefore, \[P\left( {A \cap B} \right) < P\left( A \right)\] and \[P\left( {A \cap B} \right) < P\left( B \right)\]
So, the given probabilities are consistently defined.
13. Fill in the blanks in following table:
P(A)  P(B)  P(A ⋂ B)  P(A ⋃ B)  
(i)  1/3  1/5  1/15  …. 
(ii)  0.35  …..  0.25  0.6 
(iii)  0.5  0.35  ….  0.7 
Ans:
(i) We know that,
\[P(A \cup B) = P\left( A \right) + P\left( B \right)P(A \cap B)\]
\[P(A \cup B) = \dfrac{1}{3} + \dfrac{1}{5} + \dfrac{1}{{15}}\]
\[P(A \cup B) = \dfrac{7}{{15}}\]
Ans:
(ii) We know that,
\[P(A \cup B) = P\left( A \right) + P\left( B \right)P(A \cap B)\]
\[0.6 = 0.35 + P\left( B \right)  0.25\]
\[P\left( B \right) = 0.6  0.35 + 0.25\]
\[P\left( B \right) = 0.5\]
Ans:
(iii) We know that,
\[P(A \cup B) = P\left( A \right) + P\left( B \right)P(A \cap B)\]
\[0.7 = 0.5 + 0.35  P(A \cap B)\]
\[P(A \cap B) = 0.5 + 0.35  0.7\]
\[P(A \cap B) = 0.15\]
14. Given P(A) = 5/3 and P(B) = 1/5 . Find P(A or B), if A and B are mutually exclusive events.
Ans: \[P\left( A \right) = \dfrac{3}{5}\] and \[P\left( B \right) = \dfrac{1}{5}\]
If A and B are mutually exclusive
\[P(A \cup B) = P\left( A \right) + P\left( B \right)\]
\[P(A \cup B) = \dfrac{3}{5} + \dfrac{1}{5}\]
\[P(A \cup B) = \dfrac{4}{5}\]
15. If E and F are events such that P(E) = ¼ , P(F) = ½ and P(E and F) = 1/8 , find
(i) P(E or F)
Ans: We have, \[P\left( E \right) = \dfrac{1}{4}\], \[P\left( F \right) = \dfrac{1}{2}\], \[P\left( {E \cap F} \right) = \dfrac{1}{8}\]
We know that,
\[P(A \cup B) = P\left( A \right) + P\left( B \right)P(A \cap B)\]
\[P(A \cup B) = \dfrac{1}{4} + \dfrac{1}{2}  \dfrac{1}{8}\]
\[P(A \cup B) = \dfrac{5}{8}\]
(ii) P(not E and not F)
Ans: \[P\left( {{\text{not E and not F}}} \right) = P\left( {\overline E \cap \overline F } \right)\]
\[P\left( {{\text{not E and not F}}} \right) = P\left( {\overline {E \cup F} } \right)\]
\[P\left( {{\text{not E and not F}}} \right) = 1  P\left( {E \cup F} \right)\]
\[P\left( {{\text{not E and not F}}} \right) = 1  \dfrac{5}{8}\]
\[P\left( {{\text{not E and not F}}} \right) = \dfrac{3}{8}\]
16. Events E and F are such that P(not E or not F) = 0.25, State whether E and F are mutually exclusive.
Ans: We have,
\[P\left( {{\text{not E or not F}}} \right) = 0.25\]
\[P\left( {\overline {\text{E}} \cup \overline {\text{F}} } \right) = 0.25\]
\[P\left( {\overline {E \cap F} } \right) = 0.25\]
\[1  P\left( {E \cap F} \right) = 0.25\]
\[P\left( {E \cap F} \right) = 1  0.25\]
\[P\left( {E \cap F} \right) = 0.75\]
\[P\left( {E \cap F} \right) \ne 0\]
Therefore, E and F are not mutually exclusive event.
17. 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)
Ans: We know that,
\[P\left( {{\text{not A}}} \right) = 1  P\left( A \right)\]
\[P\left( {{\text{not A}}} \right) = 1  0.42\]
\[P\left( {{\text{not A}}} \right) = 0.58\]
(ii) P(not B)
Ans: We know that,
\[P\left( {{\text{not B}}} \right) = 1  P\left( B \right)\]
\[P\left( {{\text{not B}}} \right) = 1  0.48\]
\[P\left( {{\text{not B}}} \right) = 0.52\]
(iii) P(A or B)
Ans: We know that,
\[P(A \cup B) = P\left( A \right) + P\left( B \right)P(A \cap B)\]
\[P(A \cup B) = 0.42 + 0.48  0.16\]
\[P(A \cup B) = 0.74\]
18. 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.
Ans: Let A be the event that the student is studying mathematics and B be the event that the student is studying biology,
Therefore,
\[P\left( A \right) = \dfrac{{40}}{{100}}\]
\[P\left( A \right) = \dfrac{2}{5}\]
And,
\[P\left( B \right) = \dfrac{{30}}{{100}}\]
\[P\left( B \right) = \dfrac{3}{{10}}\]
So, student studying both subjects,
\[P\left( {A \cap B} \right) = \dfrac{{10}}{{100}}\]
\[P\left( {A \cap B} \right) = \dfrac{1}{{10}}\]
Therefore, probability of studying mathematics or biology is given as,
\[P(A \cup B) = P\left( A \right) + P\left( B \right)P(A \cap B)\]
\[P(A \cup B) = \dfrac{2}{5} + \dfrac{3}{{10}}  \dfrac{1}{{10}}\]
\[P(A \cup B) = \dfrac{6}{{10}}\]
\[P(A \cup B) = \dfrac{3}{5}\]
Therefore, \[\dfrac{3}{5}\] is the probability that student will studying mathematics or biology.
19. 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?
Ans: Let probability of a randomly chosen student passing the first examination is 0.8 be \[P\left( A \right)\].
And also probability of passing the second examination is 0.7 be \[P\left( B \right)\]
Therefore,
\[P(A \cup B)\;\]is probability of passing at least one of the examination
So, We know that,
\[P(A \cup B) = P\left( A \right) + P\left( B \right)P(A \cap B)\]
\[0.95 = 0.8 + 0.7  P(A \cap B)\]
\[P(A \cap B) = 0.8 + 0.7  0.95\]
\[P(A \cap B) = 0.55\]
therefor, 0.55 is the probability that student will pass both the examinations.
20. 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?
Ans: Let probability of passing the English examination is 0.75 be \[P\left( A \right)\].
And also assume the probability of passing the Hindi examination is be \[P\left( B \right)\].
We hve given, \[P\left( A \right) = 0.75,{\text{ }}P\left( {A \cap B} \right) = 0.5,{\text{ }}P\left( {A' \cap B'} \right) = 0.1\]
We know that,
\[P\left( {A' \cap B'} \right) = 1P(A \cup B)\]
\[P(A \cup B) = 1P\left( {A' \cap B'} \right)\]
\[P(A \cup B) = 1  0.1\]
\[P(A \cup B) = 0.9\]
We know that,
\[P(A \cup B) = P\left( A \right) + P\left( B \right)P(A \cap B)\]
\[0.9 = 0.75 + P\left( B \right)  0.5\]
\[P\left( B \right) = 0.9 + 0.5  0.75\]
\[P\left( B \right) = 0.65\]
Therefore, the probability of passing the hindi examination is 0.65.
21. 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.
Ans: The total number of students in class = 60
Therefore, sample space consist of \[n\left( S \right) = 60\]
Let the students opted for NCC be ‘A’
And the students opted for NSS be ‘B’
\[n\left( A \right) = 30,{\text{ }}n\left( B \right) = 32{\text{ }},{\text{ }}n\left( {A \cap B} \right) = 24\]
Therefore,
\[P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}\]
\[P\left( A \right) = \dfrac{{30}}{{60}}\]
\[P\left( A \right) = \dfrac{1}{2}\]
\[P\left( B \right) = \dfrac{{32}}{{60}}\]
\[P\left( B \right) = \dfrac{8}{{15}}\]
And,
\[P\left( {A \cap B} \right) = \dfrac{{n\left( {A \cap B} \right)}}{{n\left( S \right)}}\]
\[P\left( {A \cap B} \right) = \dfrac{{24}}{{60}}\]
\[P\left( {A \cap B} \right) = \dfrac{2}{5}\]
We know that,
\[P(A \cup B) = P\left( A \right) + P\left( B \right)P(A \cap B)\]
\[P(A \cup B) = \dfrac{1}{2} + \dfrac{8}{{15}}  \dfrac{2}{5}\]
\[P(A \cup B) = \dfrac{{19}}{{30}}\]
(ii) The student has opted neither NCC nor NSS.
Ans: \[P\left( {not{\text{ }}A{\text{ }}and{\text{ }}not{\text{ }}B} \right) = \;P\left( {A' \cap B'} \right)\]
\[P\left( {A' \cap B'} \right) = 1P(A \cup B)\]
\[P\left( {A' \cap B'} \right) = 1\dfrac{{19}}{{30}}\]
\[P\left( {A' \cap B'} \right) = \dfrac{{11}}{{30}}\]
(iii) The student has opted NSS but not NCC.
Ans: P(student opted NSS but not NCC)
\[n\left( {BA} \right) = n\left( B \right)n\left( {A \cap B} \right)\]
\[n\left( {BA} \right) = 3224\]
\[n\left( {BA} \right) = 8\]
The probability that the selected student has opted for NSS and not NCC is
\[P\left( {BA} \right) = \dfrac{{n\left( {BA} \right)}}{{n\left( S \right)}}\]
\[P\left( {BA} \right) = \dfrac{8}{{60}}\]
\[P\left( {BA} \right) = \dfrac{2}{{15}}\]
NCERT Solutions for Class 11 Maths Chapters
NCERT Solutions for Class 11 Maths Chapter 16 Probability Exercise 16.3
Opting for the NCERT solutions for Ex 16.3 Class 11 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 16.3 Class 11 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.
Vedantu inhouse subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 11 students who are thorough with all the concepts from the Maths textbook and quite wellversed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 11 Maths Chapter 16 Exercise 16.3 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.
Besides these NCERT solutions for Class 11 Maths Chapter 16 Exercise 16.3, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our inhouse subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.
Do not delay any more. Download the NCERT solutions for Class 11 Maths Chapter 16 Exercise 16.3 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well.
NCERT Solution Class 11 Maths of Chapter 16 All Exercises
Chapter 16  Probability Exercises in PDF Format  
16 Questions & Solutions  
7 Questions & Solutions  
21 Questions & Solutions  
Miscellaneous Exercise  10 Questions & Solutions 
FAQs on NCERT Solutions for Class 11 Maths Chapter 16: Probability  Exercise 16.3
1. How many questions are there in NCERT Solutions for Class 11 Maths Chapter 16 Probability (Ex 16.3) Exercise 16.3?
Exercise 16.3 from Chapter 16 of the 11thgrade math textbook, Probability, has a total of 21 questions. NCERT questions can be viewed as a helpful tool because they are filled with tactics that will enable pupils to get better outcomes. For the purpose of assisting students in their exam preparation, Vedantu offers NCERT Solutions for all the chapters of Class 11 math at no charge. You can download these solutions in PDF format from Vedantu's official website and mobile application.
2. Where can I get NCERT Solutions for Class 11 Maths Chapter 16 Probability (Ex 16.3) Exercise 16.3?
NCERT solutions are available for all topics on Vedantu, India's top online learning platform, to assist students in getting ready for their math exams. These solutions were created by top topic specialists in accordance with the most recent CBSE curriculum. The solutions are presented stepbystep and are meticulously arranged and correct. You may access the NCERT Solutions for Class 11 Maths Chapter 16 Probability (Ex 16.3) Exercise 16.3 for free by going to Vedantu's official website (Vedantu.com) or by downloading the Vedantu mobile application.
3. What concepts are being discussed in Chapter 16 Probability of Class 11 Maths?
NCERT Chapter 16 Probability of Class 11 Math discusses the concepts of sample space, sample points, events, and their various types—such as impossible events, sure events, and complementary events—set operations on events, mutually exclusive events, and more—are explored. The probability is then determined using axiomatic principles using the events and sample space as a foundation. Because there are numerous questions in this chapter in both the board final exams and other competitive exams like the JEE Advance and JEE Mains, students should thoroughly understand each concept covered in this chapter.
4. What are the benefits of choosing Vedantu’s NCERT Solutions for Class 11 Maths Chapter 16 Probability (Ex 16.3) Exercise 16.3?
Vedantu's NCERT Solutions for Class 11 Maths can definitely help you. Finding trustworthy study material might be challenging because there are so many websites and online resources available. Due to a lack of fundamental knowledge on each topic, students may have trouble solving math exercise questions. For these solutions, which are explained from basic to advanced and are simple to understand, top subject experts with years of experience are ready at Vedantu. Additionally, students can download these PDFs whenever they want and without paying any fees.
5. How can I revise Chapter 11 Probability of NCERT Class 11 Maths?
For students taking any board test or competitive exam, a constant review is crucial during the preparation phase. After finishing each chapter, move on to the corresponding exercises to practise solving the corresponding problems. Up until your final exam, frequently practise those problems. You can use the Class 11 Math revision notes created by the subjectmatter specialists at Vedantu to help you remember the equations and distinctive characteristics of each conic section.