NCERT Solutions for Exercise 16.3 Class 11 Maths Chapter 16 - Probability

# NCERT Solutions for Exercise 16.3 Class 11 Maths Chapter 16 - Probability

Edited By Ravindra Pindel | Updated on Jul 12, 2022 04:46 PM IST

In the previous classes, you have learned about the classical theory of probability which is defined as the ratio of the number of favorable outcomes to the total number of equally likely outcomes. Also, in the previous chapter, you have learned about the statistical approach to probability. Both the above-mentioned theories are not applicable in the random experiment when the number of total outcomes is infinities. In NCERT solutions for Class 11 Maths chapter 16 exercise 16.3 you will learn about the axiomatic approach of probability.

In the previous exercises of this NCERT chapter, you have already learned about the application of outcomes, random experiments, events, sample space, etc which will be needed to understand this approach of probability. The subtopics like the probability of an event, probabilities of equally likely outcomes, probability of compliment of an event are also covered in NCERT book exercise 16.3 Class 11 Maths. Also, you can go through NCERT Solutions, if are needed the NCERT solutions for Class 6 to Class 12 for Science and Maths.

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## Question:1(a) Which of the following can not 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 $\omega _1$ $\omega _2$ $\omega _3$ $\omega _4$ $\omega _5$ $\omega _6$ $\omega _7$ (a) $0.1$ $0.01$ $0.05$ $0.03$ $0.01$ $0.2$ $0.6$

(a) Condition (i): Each of the number p( $\dpi{100} \omega_i$ ) is positive and less than one.

Condition (ii): Sum of probabilities = 0.1 + 0.01 + 0.05 + 0.03 + 0.01 + 0.2 + 0.6 = 1

Therefore, the assignment is valid

 Assignment $\omega _1$ $\omega _2$ $\omega _3$ $\omega _4$ $\omega _5$ $\omega _6$ $\omega _7$ (b) $\frac{1}{7}$ $\frac{1}{7}$ $\frac{1}{7}$ $\frac{1}{7}$ $\frac{1}{7}$ $\frac{1}{7}$ $\frac{1}{7}$

(b) Condition (i): Each of the number p( $\dpi{100} \omega_i$ ) is positive and less than one.

Condition (ii): Sum of probabilities = $\dpi{100} \frac{1}{7}+\frac{1}{7}+\frac{1}{7}+\frac{1}{7}+\frac{1}{7}+\frac{1}{7}+\frac{1}{7} = 1$

Therefore, the assignment is valid

 Assignment $\omega _1$ $\omega _2$ $\omega _3$ $\omega _4$ $\omega _5$ $\omega _6$ $\omega _7$ (d) $-0.1$ $0.2$ $0.3$ $0.4$ $-0.2$ $0.1$ $0.3$

(d) Two of the probabilities p( $\omega_1$ ) and p( $\omega_5$ ) are negative, hence condition(i) is not satisfied.

Therefore, the assignment is not valid.

 Assignment $\omega _1$ $\omega _2$ $\omega _3$ $\omega _4$ $\omega _5$ $\omega _6$ $\omega _7$ (e) $\frac{1}{14}$ $\frac{2}{14}$ $\frac{3}{14}$ $\frac{4}{14}$ $\frac{5}{14}$ $\frac{6}{14}$ $\frac{15}{14}$

(e) Each of the number p( $\dpi{100} \omega_i$) is positive but p( $\dpi{100} \omega_7$) is not less than one. Hence the condition is not satisfied.

Therefore, the assignment is not valid.

Sample space when a coin is tossed twice, S = {HH, HT, TH, TT}

[Note: A coin tossed twice is same as two coins tossed at once]

$\therefore$ Number of possible outcomes n(S) = 4

Let E be the event of getting at least one tail = {HT, TH, TT}

$\therefore$ n(E) = 3

$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$ $= \frac{3}{4}$

= 0.75

A prime number will appear

Sample space when a die is thrown, S = {1,2,3,4,5,6}

$\therefore$ Number of possible outcomes n(S) = 6

Let E be the event of getting a prime number = {2,3,5}

$\therefore$ n(E) = 3

$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$ $= \frac{3}{6}$

= 0.5

Question:3.(ii) A die is thrown, find the probability of following events:

A number greater than or equal to $3$ will appear

Sample space when a die is thrown, S = {1,2,3,4,5,6}

$\therefore$ Number of possible outcomes n(S) = 6

Let E be the event of getting a number greater than or equal to 3 = {3,4,5,6}

$\therefore$ n(E) = 4

$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$ $= \frac{4}{6} = \frac{2}{3}$

= 0.67

Question:3.(iii) A die is thrown, find the probability of following events:

A number less than or equal to one will appear

Sample space when a die is thrown, S = {1,2,3,4,5,6}

$\therefore$ Number of possible outcomes n(S) = 6

Let E be the event of getting a number less than or equal to one = {1}

$\therefore$ n(E) = 1

$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$ $= \frac{1}{6}$

= 0.167

Question:3.(iv) A die is thrown, find the probability of following events:

A number more than $\small 6$ will appear

Sample space when a die is thrown, S = {1,2,3,4,5,6}

$\therefore$ Number of possible outcomes n(S) = 6

Let E be the event of getting a number more than 6 will appear = $\phi$

$\therefore$ n(E) = 0

$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$ $= \frac{0}{6}$

= 0

A number less than $\small 6$ will appear.

Sample space when a die is thrown, S = {1,2,3,4,5,6}

$\therefore$ Number of possible outcomes n(S) = 6

Let E be the event of getting a number less than 6 will appear = {1,2,3,4,5}

$\therefore$ n(E) = 5

$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$ $= \frac{5}{6}$

= 0.83

(a) Number of points(events) in the sample space = Number of cards in the pack = 52

Calculate the probability that the card is an ace of spades.

Number of possible outcomes, n(S) = 52

Let E be the event that the card is an ace of spades

$\therefore$ n(E) = 1

$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$ $= \frac{1}{52}$

The required probability that the card is an ace of spades is $\dpi{80} \frac{1}{52}$.

Question:4(c)(i) A card is selected from a pack of 52 cards.

Calculate the probability that the card is an ace

Number of possible outcomes, n(S) = 52

Let E be the event that the card is an ace. There are 4 aces.

$\therefore$ n(E) = 4

$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$ $= \frac{4}{52} = \frac{1}{13}$

The required probability that the card is an ace is $\dpi{80} \frac{1}{13}$.

Calculate the probability that the card is black card.

Number of possible outcomes, n(S) = 52

Let E be the event that the card is a black card. There are 26 black cards. (Diamonds and Clubs)

$\therefore$ n(E) = 26

$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$ $= \frac{26}{52} = \frac{1}{2}$

The required probability that the card is an ace is $\dpi{80} \frac{1}{2}$.

The coin and die are tossed together.

The coin can have only 1 or 6 as possible outcomes and the die can have {1,2,3,4,5,6} as poosible outcomes

Sample space, S = {(x,y): x $\dpi{80} \in$ {1,6} and y $\dpi{80} \in$ {1,2,3,4,5,6}}

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

Number of possible outcomes, n(S) = 12

(i) Let E be the event having sum of numbers as 3 = {(1, 2)}

$\therefore$ n(E) = 1

$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$ $= \frac{1}{12}$

The required probability of having 3 as sum of numbers is $\dpi{80} \frac{1}{12}$.

The coin and die are tossed together.

The coin can have only 1 or 6 as possible outcomes and the die can have {1,2,3,4,5,6} as poosible outcomes

Sample space, S = {(x,y): x $\dpi{80} \in$ {1,6} and y $\dpi{80} \in$ {1,2,3,4,5,6}}

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

Number of possible outcomes, n(S) = 12

(ii) Let E be the event having sum of numbers as 12 = {(6, 6)}

$\therefore$ n(E) = 1

$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$ $= \frac{1}{12}$

The required probability of having 12 as sum of numbers is $\dpi{80} \frac{1}{12}$.

There are four men and six women on the city council

$\therefore$ n(S) = n(men) + n(women) = 4 + 6 = 10

Let E be the event of selecting a woman

$\therefore$ n(E) = 6

$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$ $= \frac{6}{10} = \frac{3}{5}$

Therefore, the required probability of selecting a woman is 0.6

Here the sample space is,

S = {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, THHT, HTTH, THTH, TTHH, TTTH, TTHT, THTT, HTTT, TTTT}

According to question,

1.) 4 heads = 1 + 1 + 1 + 1 = Rs. 4

2.) 3 heads and 1 tail = 1 + 1 + 1 - 1.50 = Rs. 1.50

3.) 2 heads and 2 tails = 1 + 1 - 1.50 - 1.50 = - Rs. 1 : he will lose Re. 1

4.) 1 head and 3 tails = 1 – 1.50 – 1.50 – 1.50 = - Rs. 3.50 : he will lose Rs. 3.50

5.) 4 tails = – 1.50 – 1.50 – 1.50 – 1.50 = - Rs. 6 = he will lose Rs. 6

Now, sample space of amounts corresponding to S:

S' = {4, 1.50, 1.50, 1.50, 1.50, - 1, - 1, - 1, - 1, - 1, - 1, - 3.50, - 3.50, - 3.50, - 3.50, - 6}

$\therefore$ n(S') = 12

$\therefore$ Required Probabilities are:

$\dpi{100} P(Winning\ Rs.\ 4) = \frac{n(Winning\ Rs.\ 4)}{n(S')}$ $= \frac{1}{16}$

$\dpi{100} P(Winning\ Rs.\ 1.50) = \frac{n(Winning\ Rs.\ 1.50)}{n(S')}$ $= \frac{4}{16} = \frac{1}{4}$

$\dpi{100} P(Losing\ Re.\ 1) = \frac{n(Losing\ Re.\ 1)}{n(S')}$ $= \frac{6}{16} = \frac{3}{8}$

$\dpi{100} P(Losing\ Rs.\ 3.50) = \frac{n(Losing\ Rs.\ 3.50)}{n(S')}$ $= \frac{4}{16} = \frac{1}{4}$

$\dpi{100} P(Losing\ Rs.\ 6) = \frac{n(Losing\ Rs.\ 6)}{n(S')}$ $= \frac{1}{16}$

$\small 3$ heads

Sample space when three coins are tossed: [Same as a coin tossed thrice!]

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

Number of possible outcomes, n(S) = 8 [Note: 2x2x2 = 8]

Let E be the event of getting 3 heads = {HHH}

$\therefore$ n(E) = 1

$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$ $= \frac{1}{8}$

The required probability of getting 3 heads is $\dpi{80} \frac{1}{8}$.

Question:8.(ii) Three coins are tossed once. Find the probability of getting

$\small 2$ heads

Sample space when three coins are tossed: [Same as a coin tossed thrice!]

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

Number of possible outcomes, n(S) = 8 [Note: 2x2x2 = 8]

Let E be the event of getting 2 heads = {HHT, HTH, THH}

$\therefore$ n(E) = 3

$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$ $= \frac{3}{8}$

The required probability of getting 2 heads is $\dpi{80} \frac{3}{8}$.

Question:8.(iii) Three coins are tossed once. Find the probability of getting

atleast $2$ heads

Sample space when three coins are tossed: [Same as a coin tossed thrice!]

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

Number of possible outcomes, n(S) = 8 [Note: 2x2x2 = 8]

Let E be the event of getting atleast 2 heads = Event of getting 2 or more heads = {HHH, HHT, HTH, THH}

$\therefore$ n(E) = 4

$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$ $= \frac{4}{8} = \frac{1}{2}$

The required probability of getting atleast 2 heads is $\dpi{80} \frac{1}{2}$.

Question:8.(iv) Three coins are tossed once. Find the probability of getting

atmost $\small 2$ heads

Sample space when three coins are tossed: [Same as a coin tossed thrice!]

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

Number of possible outcomes, n(S) = 8 [Note: 2x2x2 = 8]

Let E be the event of getting atmost 2 heads = Event of getting 2 or less heads = {HHT, HTH, THH, TTH, HTT, THT}

$\therefore$ n(E) = 6

$= \frac{6}{8} = \frac{3}{4}$$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$

The required probability of getting almost 2 heads is $\dpi{80} \frac{3}{4}$.

Sample space when three coins are tossed: [Same as a coin tossed thrice!]

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

Number of possible outcomes, n(S) = 8 [Note: 2x2x2 = 8]

Let E be the event of getting no head = Event of getting only tails = {TTT}

$\therefore$ n(E) = 1

$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$ $= \frac{1}{8}$

The required probability of getting no head is $\dpi{80} \frac{1}{8}$.

Question:8.(vi) Three coins are tossed once. Find the probability of getting

$\small 3$ tails

Sample space when three coins are tossed: [Same as a coin tossed thrice!]

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

Number of possible outcomes, n(S) = 8 [Note: 2x2x2 = 8]

Let E be the event of getting 3 tails = {TTT}

$\therefore$ n(E) = 1

$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$ $= \frac{1}{8}$

The required probability of getting 3 tails is $\dpi{80} \frac{1}{8}$.

Question:8(vii) Three coins are tossed once. Find the probability of getting

exactly two tails

Sample space when three coins are tossed: [Same as a coin tossed thrice!]

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

Number of possible outcomes, n(S) = 8 [Note: 2x2x2 = 8]

Let E be the event of getting exactly 2 tails = {TTH, HTT, THT}

$\therefore$ n(E) = 3

$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$ $= \frac{3}{8}$

The required probability of getting exactly 2 tails is $\dpi{80} \frac{3}{8}$.

Question:8.(viii) Three coins are tossed once. Find the probability of getting

no tail

Sample space when three coins are tossed: [Same as a coin tossed thrice!]

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

Number of possible outcomes, n(S) = 8 [Note: 2x2x2 = 8]

Let E be the event of getting no tail = Event of getting only heads = {HHH}

$\therefore$ n(E) = 1

$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$ $= \frac{1}{8}$

The required probability of getting no tail is $\dpi{80} \frac{1}{8}$.

Question:8.(ix) Three coins are tossed once. Find the probability of getting

atmost two tails

Sample space when three coins are tossed: [Same as a coin tossed thrice!]

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

Number of possible outcomes, n(S) = 8 [Note: 2x2x2 = 8]

Let E be the event of getting atmost 2 tails = Event of getting 2 or less tails = {HHT, HTH, THH, TTH, HTT, THT}

$\therefore$ n(E) = 6

$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$ $= \frac{6}{8} = \frac{3}{4}$

The required probability of getting atmost 2 tails is $\dpi{80} \frac{3}{4}$.

Given,

P(E) = $\small \frac{2}{11}$

We know,

P(not E) = P(E') = 1 - P(E)

= $\dpi{80} 1 - \small \frac{2}{11}$

= $\dpi{80} \frac{9}{11}$

Given, ‘ASSASSINATION’

No. of A's = 3; No. of S's = 4; No. of I's = 2; No. of N's = 2; No. of T = 1; No. of O = 1

No. of letters = 13

No. of vowels = {3 A's,2 I's,O} = 6

One letter is selected:

n(S) = $\dpi{100} ^{13}\textrm{C}_{1}$ = 13

Let E be the event of getting a vowel.

n(E) = $\dpi{100} ^{6}\textrm{C}_{1}$ = 6

$\dpi{100} \therefore$ $\dpi{100} P(E) = \frac{6}{13}$

Given, ‘ASSASSINATION’

No. of A's = 3; No. of S's = 4; No. of I's = 2; No. of N's = 2; No. of T = 1; No. of O = 1

No. of letters = 13

No. of consonants = {4 S's,2 N's,T} = 7

One letter is selected:

n(S) = $\dpi{100} ^{13}\textrm{C}_{1}$ = 13

Let E be the event of getting a consonant.

n(E) = $\dpi{100} ^{7}\textrm{C}_{1}$ = 7

$\dpi{100} \therefore$ $\dpi{100} P(E) = \frac{7}{13}$

Total numbers of numbers in the draw = 20

Numbers to be selected = 6

$\dpi{100} \therefore$ n(S) = $\dpi{100} ^{20}\textrm{C}_{6}$

Let E be the event that six numbers match with the six numbers fixed by the lottery committee.

n(E) = 1 (Since only one prize to be won.)

$\dpi{100} \therefore$ Probability of winning =

$\dpi{100} P(E) = \frac{n(E)}{n(S)}$$\dpi{100} = \frac{1}{^{20}\textrm{C}_{6}} = \frac{6!14!}{20!}$

$\dpi{100} = \frac{6.5.4.3.2.1.14!}{20.19.18.17.16.15.14!}$

$\dpi{100} = \frac{1}{38760}$

$P(A)=0.5,P(B)=0.7.P(A\cap B)=0.6$

(i) Given, $P(A)=0.5,P(B)=0.7.P(A\cap B)=0.6$

Now P(A $\cap$ B) > P(A)

(Since A $\cap$ B is a subset of A, P(A $\cap$ B) cannot be more than P(A))

Therefore, the given probabilities are not consistently defined.

$P(A)=0.5,P(B)=0.4,P(A\cup B)=0.8$

(ii) Given, $P(A)=0.5,P(B)=0.4,P(A\cup B)=0.8$

We know,

P(A $\cup$ B) = P(A)+ P(B) - P(A $\cap$ B)

$\implies$ 0.8 = 0.5 + 0.4 - P(A $\cap$ B)

$\implies$ P(A $\cap$ B) = 0.9 - 0.8 = 0.1

Therefore, P(A $\cap$ B) < P(A) and P(A $\cap$ B) < P(B) , which satisfies the condition.

Hence, the probabilities are consistently defined

Question:13 Fill in the blanks in following table:

 $P(A)$ $P(B)$ $P(A\cap B)$ $P(A\cup B)$ (i) $\frac{1}{3}$ $\frac{1}{5}$ $\frac{1}{15}$ $...$ (ii) $0.35$ $...$ $0.25$ $0.6$ (iii) $0.5$ $0.35$ $...$ $0.7$

We know,

$\dpi{100} P(A \cup B) = P(A)+ P(B) - P(A \cap B)$

(i) $\dpi{100} P(A \cup B)$ = $\dpi{100} \frac{1}{3}+\frac{1}{5}-\frac{1}{15}$ = $\dpi{100} \frac{5+3-1}{15} = \frac{7}{15}$

(ii) $\dpi{100} 0.6 = 0.35 + P(B) - 0.25$

$\implies$ $\dpi{100} P(B) = 0.6 - 0.1 = 0.5$

(iii) $\dpi{100} 0.7 = 0.5 + 0.35 - P(A \cap B)$

$\implies$ $\dpi{100} P(A \cap B) = 0.85 - 0.7 = 0.15$

 $P(A)$ $P(B)$ $P(A\cap B)$ $P(A\cup B)$ (i) $\frac{1}{3}$ $\frac{1}{5}$ $\frac{1}{15}$ $\dpi{100} \boldsymbol{\frac{7}{15}}$ (ii) $0.35$ 0.5 $0.25$ $0.6$ (iii) $0.5$ $0.35$ 0.15 $0.7$

Given, $\dpi{80} P(A)=\frac{3}{5}$ and $\dpi{80} P(B)=\frac{1}{5}$

To find : $\dpi{100} P(A or B) = P(A \cup B)$

We know,

$P(A \cup B) = P(A)+ P(B) - P(A \cap B) = P(A)+ P(B)$ [Since A and B are mutually exclusive events.]

$\implies$ $P(A \cup B) = \frac{3}{5}+\frac{1}{5} = \frac{4}{5}$

Therefore, $\dpi{100} P(A \cup B) = \frac{4}{5}$

Given, $P(E)=\frac{1}{4}$, $P(F)=\frac{1}{2}$ and $P(E\hspace{1mm}and\hspace{1mm}F)=\frac{1}{8}$

To find : $P(E or F) = P(E \cup F)$

We know,

$P(A \cup B) = P(A)+ P(B) - P(A \cap B)$

$\implies$ $P(E \cup F) =$ $\frac{1}{4}+\frac{1}{2} -\frac{1}{8} = \frac{2+4-1}{8}$

$= \frac{5}{8}$
Therefore, $P(E \cup F) =$ $\frac{5}{8}$

Given, $P(E)=\frac{1}{4}$, $P(F)=\frac{1}{2}$ and $P(E\hspace{1mm}and\hspace{1mm}F)=\frac{1}{8}$

To find :$P(not\ E\ and\ not\ F) = P(E' \cap F')$

We know,

$P(A' \cap B') = P(A \cup B)' = 1 - P(A \cup B)$

And $P(A\cup B) = P(A)+ P(B) - P(A \cap B)$

$\implies$ $P(E \cup F) =$$\frac{1}{4}+\frac{1}{2} -\frac{1}{8} = \frac{2+4-1}{8}$

$= \frac{5}{8}$

$\implies$ $P(E' \cap F') = 1 - P(E \cup F)$

$= 1- \frac{5}{8}= \frac{3}{8}$

Therefore, $P(E' \cap F') =$ $\frac{3}{8}$

Given, $P(not\ E\ or\ not\ F) = 0.25$

For A and B to be mutually exclusive, $P(A \cap B) = 0$

Now, $P(not\ E\ or\ not\ F) = P(E' \cup F') = 0.25$

We know,

$P(A' \cup B') = P(A \cap B)' = 1 - P(A \cap B)$
$\\ \implies 0.25 = 1 - P(E \cap F) \\ \implies P(E \cap F) = 1 - 0.25 = 0.75 \neq 0$

Hence, E and F are not mutually exclusive.

Given, P(A) = 0.42, P(B) = 0.48 and P(A and B) = 0.16

(i) $P(not\ A) = P(A') = 1 - P(A)$

$\implies$ $P(not\ A) = 1 - 0.42 = 0.58$

Therefore, P(not A) = 0.58

Given, P(A) = 0.42, P(B) = 0.48 and P(A and B) = 0.16

(ii) $P(not\ B) = P(B') = 1 - P(B)$

$\implies$ $P(not\ B) = 1 - 0.48 = 0.52$

Therefore, P(not B) = 0.52

Given, P(A) = 0.42, P(B) = 0.48 and P(A and B) = 0.16

(iii) We know,

$P(A \cup B) = P(A)+ P(B) - P(A \cap B)$

$\implies$ $P(A \cup B) = 0.42 + 0.48 - 0.16 = 0.9 - 0.16$

= 0.74

Let M denote the event that the student is studying Mathematics and B denote the event that the student is studying Biology

And total students in the class be 100.

Given, n(M) = 40 $\implies$ P(M) = $\dpi{100} \frac{40}{100} = \frac{2}{5}$

n(B) = 30$\implies$ P(M) = $\dpi{100} \frac{30}{100} = \frac{3}{10}$

n(M $\cap$ B) = 10$\implies$ P(M) = $\dpi{100} \frac{10}{100} = \frac{1}{10}$

We know,

P(A $\cup$ B) = P(A)+ P(B) - P(A $\cap$ B)

$\implies$ P(M $\cup$ B) = 0.4 + 0.3 - 0.1 = 0.6

Hence, the probability that he will be studying Mathematics or Biology is 0.6

Let A be the event that the student passes the first examination and B be the event that the students passes the second examination.

P(A $\cup$ B) is probability of passing at least one of the examination.

Therefore,

P(A $\cup$ B) = 0.95 , P(A)=0.8, P(B)=0.7

We know,

P(A $\cup$ B) = P(A)+ P(B) - P(A $\cap$ B)

$\implies$ P(A $\cap$ B) = 0.8 + 0.7 - 0.95 = 1.5 -0.95 = 0.55

Hence,the probability that the student will pass both the examinations is 0.55

Let A be the event that the student passes English examination and B be the event that the students pass Hindi examination.

Given,

P(A)=0.75, P(A $\cap$ B) = 0.5, P(A' $\cap$ B') =0.1

We know,

P(A' $\cap$ B') = 1 - P(A $\cup$ B)

$\implies$ P(A $\cup$ B) = 1 - 0.1 = 0.9

Also,

P(A $\cup$ B) = P(A)+ P(B) - P(A $\cap$ B)

$\implies$ P(B) = 0.9 - 0.75 + 0.5 = 0.65

Hence,the probability of passing the Hindi examination is 0.65

The student opted for NCC or NSS.

Let A be the event that student opted for NCC and B be the event that the student opted for NSS.

Given,

n(S) = 60, n(A) = 30, n(B) =32, n(A $\cap$ B) = 24

Therefore, P(A) = $\inline \dpi{100} \frac{30}{60} = \frac{1}{2}$

P(B) = $\inline \frac{32}{60} = \frac{8}{15}$

P(A $\cap$ B) = $\inline \frac{24}{60} = \frac{2}{5}$

(i) We know,

P(A $\cup$ B) = P(A)+ P(B) - P(A $\cap$ B)

$\frac{1}{2} + \frac{8}{15} - \frac{2}{5} = \frac{15+16-12}{30}$

$= \frac{19}{30}$

Hence,the probability that the student opted for NCC or NSS is $\frac{19}{30}$

The student has opted neither NCC nor NSS.

Let A be the event that student opted for NCC and B be the event that the student opted for NSS.

Given,

n(S) = 60, n(A) = 30, n(B) =32, n(A $\cap$ B) = 24

Therefore, P(A) = $\inline \dpi{100} \frac{30}{60} = \frac{1}{2}$

P(B) = $\inline \frac{32}{60} = \frac{8}{15}$

P(A $\cap$ B) = $\inline \frac{24}{60} = \frac{2}{5}$

(ii) Now,

Probability that the student has opted neither NCC nor NSS = P(A' $\cap$ B' )

We know,

P(A' $\cap$ B' ) = 1 - P(A $\cup$ B) [De morgan's law]

And, P(A $\cup$ B) = P(A)+ P(B) - P(A $\cap$ B)

$= \frac{19}{30}$

$\therefore$ P(A' $\cap$ B' )

$= 1- \frac{19}{30} = \frac{11}{30}$

Hence,the probability that the student has opted neither NCC nor NSS. is $\frac{11}{30}$

Let A be the event that student opted for NCC and B be the event that the student opted for NSS.

Given,

n(S) = 60, n(A) = 30, n(B) =32, n(A $\cap$ B) = 24

Therefore, P(A) = $\inline \dpi{100} \frac{30}{60} = \frac{1}{2}$

P(B) = $\inline \frac{32}{60} = \frac{8}{15}$

P(A $\cap$ B) = $\inline \frac{24}{60} = \frac{2}{5}$

(iii) Now,

Probability that the student has opted NSS but not NCC = P(B $\cap$ A' ) = P(B-A)

We know,

P(B-A) = P(B) - P(A $\cap$ B)

$= \frac{8}{15}- \frac{2}{5} = \frac{8-6}{15}$

$= \frac{2}{15}$

Hence,the probability that the student has opted NSS but not NCC is $\frac{2}{15}$

## More About NCERT Solutions for Class 11 Maths Chapter 16 Exercise 16.3:-

Class 11 Maths chapter 16 exercise 16.3 solutions consist of questions related to finding the probability of events, probability of compliment of an event, probability of mutually exclusive events, probability of exhaustive events, etc. There are 21 long-answer types questions given in Class 11 Maths chapter 16 exercise 16.3 which you must try to solve by yourself. You can solve the solved examples given before the Class 11 Maths chapter 16 exercise 16.3. Solving these examples will help you to solve Class 11 Maths chapter 16 exercise 16.3 problems very easily.

Also Read| Probability Class 11 Notes

## Benefits of NCERT Solutions for Class 11 Maths Chapter 16 Exercise 16.3:-

• The previous two exercises of this chapter were the building blocks for this exercise which is needed to solve Class 11 Maths chapter 16 exercise 16.3 problems.
• Class 11 Maths chapter 16 exercise 16.3 is very important for the exam as well as applicability point of view in the upcoming classes.
• In Class 11 Maths chapter 16 exercise 16.3 solutions, you will also learn about the venn diagram which is an important concept to solve probability problems.
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## Subject Wise NCERT Exampler Solutions

Happy learning!!!

1. If the probability of event A is 0.6 then find the probability of event "not A" ?

Given probability of event A p(A) = 0.6

p(A') = p( not A) = 1- p(A) = 0.4

2. What is venn diagram ?

Venn diagram is a diagram that represents the mathematical set as the circles. The intersections of these circles are used to represent the common elements of the sets.

3. What is the use of venn diagram ?

Venn diagram is used to show the logical relation between sets. In the probability theory it is most useful tool to find the common outcomes of the events.

4. If the event A and event B are mutually exclusive events then find the probability of event A and event B ?

As the event A and event B are mutually exclusive so the probability of  A and B i .e. p(A and B) = 0.

5. If the event A and event B are two mutually exclusive events and p(A) = 0.4 and p(B) = 0.3 then what is p(A or B) ?

P(A) = 0.4

p(B) = 0.3

p(A or B) = 0.4 + 0.3 - p(A and B)

p(A or B) = 0.7 - 0 = 0.7

6. Does event A and event not A are mutually exclusive events ?

Yes, event A and event not A are mutually exclusive events.

7. A die is thrown, find the probability that A prime number will appear ?

Prime numbers on a die = { 2, 3, 5}

The probability of appearing a prime number = 1/2

8. A die is thrown, find the probability that a number more than 6 will appear ?

The probability of appearing the number more than 6 is zero when a die is thrown.

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