NCERT Solutions for Class 11 Maths Chapter 14 Probability Ex 14.2

Exercise 14.2

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

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 \[ = 1 + 1 + 1--1.50\]

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,

\[n\left( S \right){ = ^{20}}{C_6}\]

\[n\left( S \right)=\frac{\left| \underline{20} \right.}{\left| \underline{6\left| \underline{20-6} \right.} \right.}\]

\[n\left( S \right) = 38760\]

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,

\[P(A \cup B) = P\left( A \right) + P\left( B \right)--P(A \cap B)\]

\[0.8 = 0.5 + 0.4--P(A \cap B)\]

\[0.8 - 0.9 = --P(A \cap B)\]

\[P(A \cap B) = 0.1\]

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:

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, 

\[P\left( {A' \cap B'} \right) = 1--P(A \cup B)\]

\[P(A \cup B) = 1--P\left( {A' \cap B'} \right)\]

\[P(A \cup B) = 1 - 0.1\]

\[P(A \cup B) = 0.9\]

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: \[P\left( {not{\text{ }}A{\text{ }}and{\text{ }}not{\text{ }}B} \right) = \;P\left( {A' \cap B'} \right)\]

\[P\left( {A' \cap B'} \right) = 1--P(A \cup B)\]

\[P\left( {A' \cap B'} \right) = 1--\dfrac{{19}}{{30}}\]

\[P\left( {A' \cap B'} \right) = \dfrac{{11}}{{30}}\]

(iii) The student has opted NSS but not NCC.

Ans: P(student opted NSS but not NCC)

\[n\left( {B--A} \right) = n\left( B \right)--n\left( {A \cap B} \right)\]

\[n\left( {B--A} \right) = 32--24\]

\[n\left( {B--A} \right) = 8\]

The probability that the selected student has opted for NSS and not NCC is

\[P\left( {B--A} \right) = \dfrac{{n\left( {B--A} \right)}}{{n\left( S \right)}}\]

\[P\left( {B--A} \right) = \dfrac{8}{{60}}\]

\[P\left( {B--A} \right) = \dfrac{2}{{15}}\]

NCERT Solutions for Class 11 Maths Chapter 14 Probability Exercise 14.2

Opting for the NCERT solutions for Ex 14.2 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 14.2 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 in-house 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 well-versed 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 14 Exercise 14.2 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 14 Exercise 14.2, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house 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. 

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NCERT Solution Class 11 Maths of Chapter 14 All Exercises

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