NCERT Solutions for Class 11 Maths Chapter 14 Exercise 14.2 - Probability

Question 1:(a) Which of the following can not be valid assignment of probabilities for outcomes of sample Space S $=\{{\omega }_1,{\omega }_2,{\omega }_3,{\omega }_4,{\omega }_5,{\omega }_6,{\omega }_7\}$

Answer:

(a) Condition (i): Each of the number p( ${\omega }_i$ ) 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

Question 1:(b) Which of the following cannot be the valid assignment of probabilities for outcomes of sample Space $S=\{{\omega }_1,{\omega }_2,{\omega }_3,{\omega }_4,{\omega }_5,{\omega }_6,{\omega }_7\}$

Answer:

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

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

Therefore, the assignment is valid

Question 1:(c) Which of the following can not be valid assignment of probabilities for outcomes of sample Space $S=\{{\omega }_1,{\omega }_2,{\omega }_3,{\omega }_4,{\omega }_5,{\omega }_6,{\omega }_7\}$

Answer:

(c) Since sum of probabilities = 0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.6 + 0.7 = 2.8 > 1

Hence, Condition (ii) is not satisfied.

Therefore, the assignment is not valid

Question 1:(d) Which of the following can not be valid assignment of probabilities for outcomes of sample Space $S=\{{\omega }_1,{\omega }_2,{\omega }_3,{\omega }_4,{\omega }_5,{\omega }_6,{\omega }_7\}$

Answer:

(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.

Question 1:(e) Which of the following can not be valid assignment of probabilities for outcomes of sample Space $S=\{{\omega }_1,{\omega }_2,{\omega }_3,{\omega }_4,{\omega }_5,{\omega }_6,{\omega }_7\}$

Answer:

(e) Each of the number p( ${\omega }_i$) is positive but p( $\pi 100{\omega }_7$) is not less than one. Hence, the condition is not satisfied.

Therefore, the assignment is not valid.

Question 2: A coin is tossed twice, what is the probability that atleast one tail occurs?

Answer:

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

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

= 0.75

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

A prime number will appear

Answer:

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

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

= 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

Answer:

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

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

= 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

Answer:

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

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

= 0.167

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

A number more than $6$ will appear

Answer:

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 = $\varphi$

$\therefore$ n(E) = 0

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

= 0

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

A number less than $6$ will appear.

Answer:

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

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

= 0.83

Question 4:(a). A card is selected from a pack of $52$ cards.
How many points are there in the sample space?

Answer:

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

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

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

Answer:

Number of possible outcomes, n(S) = 52

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

$\therefore$ n(E) = 1

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

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

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

Calculate the probability that the card is an ace

Answer:

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

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

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

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

Calculate the probability that the card is black card.

Answer:

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

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

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

Question 5:(i) 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 $3$

Answer:

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 $\in$ {1,6} and y $\in$ {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

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

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

Question 5:(ii) 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 $12$

Answer:

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 $\in$ {1,6} and y $\in$ {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

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

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

Question 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?

Answer:

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

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

Therefore, the required probability of selecting a woman is 0.6

Question 7: A fair coin is tossed four times, and a person win Re $1$ 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.

Answer:

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:

$P(WinningRs.4)=\frac{n(WinningRs.4)}{n(S^{\prime })}$ $=\frac{1}{16}$

$P(WinningRs.1.50)=\frac{n(WinningRs.1.50)}{n(S^{\prime })}$ $=\frac{4}{16}=\frac{1}{4}$

$P(LosingRe.1)=\frac{n(LosingRe.1)}{n(S^{\prime })}$ $=\frac{6}{16}=\frac{3}{8}$

$P(LosingRs.3.50)=\frac{n(LosingRs.3.50)}{n(S^{\prime })}$ $=\frac{4}{16}=\frac{1}{4}$

$P(LosingRs.6)=\frac{n(LosingRs.6)}{n(S^{\prime })}$ $=\frac{1}{16}$

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

$3$ heads

Answer:

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

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

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

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

$2$ heads

Answer:

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

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

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

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

atleast $2$ heads

Answer:

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

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

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

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

atmost $2$ heads

Answer:

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$ $P(E)=\frac{n(E)}{n(S)}$

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

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

no head

Answer:

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

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

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

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

$3$ tails

Answer:

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

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

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

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

exactly two tails

Answer:

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

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

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

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

no tail

Answer:

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

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

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

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

atmost two tails

Answer:

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

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

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

Question 9: If $\frac{2}{11}$ is the probability of an event, what is the probability of the event ‘not A’.

Answer:

Given,

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

We know,

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

= $1-\frac{2}{11}$

= $\frac{9}{11}$

Question 10:(i) A letter is chosen at random from the word ‘ASSASSINATION’. Find the probability that letter is a vowel

Answer:

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) = ${}^{13}{\text{C}}_1$ = 13

Let E be the event of getting a vowel.

n(E) = ${}^6{\text{C}}_1$ = 6

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

Question 10:(ii) A letter is chosen at random from the word ‘ASSASSINATION’. Find the probability that letter is a consonant

Answer:

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) = ${}^{13}{\text{C}}_1$ = 13

Let E be the event of getting a consonant.

n(E) = ${}^7{\text{C}}_1$ = 7

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

Question 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.]

Answer:

Total numbers of numbers in the draw = 20

Numbers to be selected = 6

$\therefore$ n(S) = $\pi {100}^{20}{\text{C}}_6$

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.)

$\therefore$ Probability of winning =

$P(E)=\frac{n(E)}{n(S)}$$=\frac{1}{{}^{20}{\text{C}}_6}=\frac{6!14!}{20!}$

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

$=\frac{1}{38760}$

Question 12:(i) Check whether the following probabilities $P(A)$ and $P(B)$ are consistently defined:

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

Answer:

(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.

Question 12:(ii) Check whether the following probabilities $P(A)$ and $P(B)$ are consistently defined

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

Answer:

(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)

$⟹$ 0.8 = 0.5 + 0.4 - P(A $\cap$ B)

$⟹$ 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:

Answer:

We know,

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

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

(ii) $0.6=0.35+P(B)-0.25$

$⟹$ $P(B)=0.6-0.1=0.5$

(iii) $0.7=0.5+0.35-P(A\cap B)$

$⟹$ $P(A\cap B)=0.85-0.7=0.15$

Question 14: Given $P(A)=\frac{3}{5}$ and $P(B)=\frac{1}{5}$. Find $P(AorB)$, if $A$ and $B$ are mutually exclusive events.

Answer:

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

To find : $P(AorB)=P(A\cup B)$

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.]

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

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

Question 15:(i) If E and F are events such that $P(E)=\frac{1}{4}$, $P(F)=\frac{1}{2}$ and $P(EandF)=\frac{1}{8}$, find (i) P(E or F)

Answer:

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

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

We know,

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

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

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

Question 15:(ii) If E and F are events such that $P(E)=\frac{1}{4}$ , $P(F)=\frac{1}{2}$ and $P(EandF)=\frac{1}{8}$ find P(not E and not F).

Answer:

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

To find :$P(notEandnotF)=P(E^{\prime }\cap F^{\prime })$

We know,

$P(A^{\prime }\cap B^{\prime })=P(A\cup B{)}^{\prime }=1-P(A\cup B)$

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

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

$=\frac{5}{8}$

$⟹$ $P(E^{\prime }\cap F^{\prime })=1-P(E\cup F)$

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

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

Question 16: Events E and F are such that P(not E or not F) $=0.25$, State whether E and F are mutually exclusive.

Answer:

Given, $P(notEornotF)=0.25$

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

Now, $P(notEornotF)=P(E^{\prime }\cup F^{\prime })=0.25$

We know,

$P(A^{\prime }\cup B^{\prime })=P(A\cap B{)}^{\prime }=1-P(A\cap B)$
$$\begin{gathered} ⟹0.25=1-P(E\cap F) \\ ⟹P(E\cap F) \end{gathered}$$

$=1-0.25=0.75\ne 0$

Hence, E and F are not mutually exclusive.

Question 17:(i) 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)

Answer:

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

(i) $P(notA)=P(A^{\prime })=1-P(A)$

$⟹$ $P(notA)=1-0.42=0.58$

Therefore, P(not A) = 0.58

Question 17:(ii) A and B are events such that P(A) $=0.42$, P(B) $=0.48$ and P(A and B) $=0.16$. Determine P(not B)

Answer:

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

(ii) $P(notB)=P(B^{\prime })=1-P(B)$

$⟹$ $P(notB)=1-0.48=0.52$

Therefore, P(not B) = 0.52

Question 17:(iii) A and B are events such that P(A) $=0.42$ , P(B) $=0.48$ and P(A and B) $=0.16$. Determine P(A or B)

Answer:

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

(iii) We know,

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

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

= 0.74

Question 18: In Class XI of a school $40\mathrm{\%}$ of the students study Mathematics and $30\mathrm{\%}$ study Biology. $10\mathrm{\%}$ 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.

Answer:

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 $⟹$ P(M) = $\frac{40}{100}=\frac{2}{5}$

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

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

We know,

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

$⟹$ 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

Question 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 atleast one of them is $0.95$ . What is the probability of passing both?

Answer:

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)

$⟹$ 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

Question 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?

Answer:

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)

$⟹$ P(A $\cup$ B) = 1 - 0.1 = 0.9

Also,

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

$⟹$ P(B) = 0.9 - 0.75 + 0.5 = 0.65

Hence,the probability of passing the Hindi examination is 0.65

Question 21:(i) 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

The student opted for NCC or NSS.

Answer:

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) = $\frac{30}{60}=\frac{1}{2}$

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

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

(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 $\frac{19}{30}$

Question 21:(ii) 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

The student has opted neither NCC nor NSS.

Answer:

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) = $\frac{30}{60}=\frac{1}{2}$

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

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

(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 $\frac{11}{30}$

Question 21:(iii) 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 The student has opted NSS but not NCC.

Answer:

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) = $\frac{30}{60}=\frac{1}{2}$

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

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

(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 $\frac{2}{15}$