Probability 13.2 NCERT Solutions

1. If \displaystyle P(A)=\frac{3}{5} and \displaystyle P(B)=\frac{1}{5}, find \displaystyle P(A\cap B) if A and B are independent events.

Solution:

Think First… Since A and B are independent events, we can directly use the independence condition \displaystyle P(A\cap B)=P(A)P(B).

We are given

\displaystyle P(A)=\frac{3}{5},\quad P(B)=\frac{1}{5}

Since A and B are independent events,

\displaystyle \therefore P(A\cap B)=P(A)P(B)

Substituting the given values, we get

\displaystyle P(A\cap B)=\frac{3}{5}\times\frac{1}{5}=\frac{3}{25}

Hence,

\displaystyle \boxed{P(A\cap B)=\frac{3}{25}}

2. Two cards are drawn at random and without replacement from a pack of 52 playing cards. Find the probability that both the cards are black.

Solution:

Think First… Since the cards are drawn without replacement, the probability for the second card depends on what happened in the first draw. So, we use the multiplication theorem with conditional probability.

Let E and F denote the events that the first and second cards drawn are black, respectively.

There are 26 black cards in a pack of 52 cards. Therefore,

\displaystyle P(E)=\frac{26}{52}=\frac{1}{2}

If the first card is black, one black card has been removed. Thus, 25 black cards remain out of 51 cards.

Therefore, probability of F given that E has aleady occured, is given by

\displaystyle P(F\mid E)=\frac{25}{51}

By the multiplication theorem of probability,

\displaystyle P(E\cap F)=P(E)P(F\mid E)

Substituting the values,

\displaystyle P(E\cap F)=\frac{1}{2}\times\frac{25}{51}=\frac{25}{102}

Therefore, the required probability is

\displaystyle \boxed{\frac{25}{102}}

3. A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If all the three oranges are good, the box is approved for sale, otherwise, it is rejected. Find the probability that a box containing 15 oranges out of which 12 are good and 3 are bad ones will be approved for sale.

Solution:

Think First… The box is approved only when all three oranges are good. Since the oranges are drawn without replacement, the probabilities change after each draw. Hence, we use the multiplication theorem.

Let E, F and G denote the events that the first, second and third oranges drawn are good, respectively.

There are 12 good oranges out of 15 oranges. Therefore,

\displaystyle P(E)=\frac{12}{15}

If the first orange is good, 11 good oranges remain out of 14 oranges. Thus, probability of F when E has aleady occured, is given by

\displaystyle P(F\mid E)=\frac{11}{14}

If the first two oranges are good, 10 good oranges remain out of 13 oranges. Therefore, probability of G when E and F have aleady occured, is given by

\displaystyle P(G\mid EF)=P(G\mid E\cap F)=\frac{10}{13}

By the multiplication theorem for three events,

\displaystyle P(E\cap F\cap G)=P(E)P(F\mid E)P(G\mid E\cap F)

Substituting the values, we get

\displaystyle P(E\cap F\cap G)=\frac{12}{15}\times\frac{11}{14}\times\frac{10}{13}=\frac{44}{91}

Therefore, the probability that the box will be approved for sale is

\displaystyle \boxed{\frac{44}{91}}

4. A fair coin and an unbiased die are tossed. Let A be the event ‘head appears on the coin’ and B be the event ‘3 on the die’. Check whether A and B are independent events or not.

Solution:

Think First… To check whether two events are independent, compare \displaystyle P(A\cap B) with \displaystyle P(A)P(B). If they are equal, the events are independent.

Since the coin has two possible outcomes and the die has six possible outcomes, the sample space is

\displaystyle S=\{(H,1),(H,2),(H,3),(H,4),(H,5),(H,6),\\(T,1),(T,2),(T,3),(T,4),(T,5),(T,6)\}

Thus, there are 12 equally likely outcomes.

Event A represents getting a head on the coin. Therefore,

\displaystyle A=\{(H,1),(H,2),(H,3),(H,4),(H,5),(H,6)\}

Hence,

\displaystyle P(A)=\frac{6}{12}=\frac{1}{2}

Event B represents getting 3 on the die. Therefore,

\displaystyle B=\{(H,3),(T,3)\}

Hence,

\displaystyle P(B)=\frac{2}{12}=\frac{1}{6}

Now, \displaystyle A\cap B represents the outcomes in which the coin shows a head and the die shows 3. The only such outcome is \displaystyle(H,3).

Therefore,

\displaystyle A\cap B=\{(H,3)\}

Thus,

\displaystyle P(A\cap B)=\frac{1}{12}

Also,

\displaystyle P(A)P(B)=\frac{1}{2}\times\frac{1}{6}=\frac{1}{12}

Therefore,

\displaystyle P(A\cap B)=P(A)P(B)

Therefore, A and B are independent events.

5. A die marked 1, 2, 3 in red and 4, 5, 6 in green is tossed. Let A be the event, ‘the number is even,’ and B be the event, ‘the number is red.’ Are A and B independent?

Solution:

Think First… To check whether A and B are independent, compare \displaystyle P(A\cap B) with \displaystyle P(A)P(B). If they are equal, A and B are independent.

The sample space is

\displaystyle S=\{1,2,3,4,5,6\}

Event A represents an even number. Therefore,

\displaystyle A=\{2,4,6\}

Hence, \displaystyle P(A)=\frac{3}{6}=\frac{1}{2}

Event B represents a red number. Therefore,

\displaystyle B=\{1,2,3\}

Hence, \displaystyle P(B)=\frac{3}{6}=\frac{1}{2}

Now, \displaystyle A\cap B=\{2\}

Therefore, \displaystyle P(A\cap B)=\frac{1}{6}

Also, \displaystyle P(A)P(B)=\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}

Clearly, \displaystyle P(A\cap B)\neq P(A)P(B)

Therefore, A and B are not independent events.

6. Let E and F be events with \displaystyle P(E)=\frac{3}{5},\; P(F)=\frac{3}{10} and \displaystyle P(E\cap F)=\frac{1}{5}. Are E and F independent?

Solution:

Think First… To check whether two events are independent, compare \displaystyle P(E\cap F) with \displaystyle P(E)P(F). If they are equal, E and F are independent.

We are given

\displaystyle P(E)=\frac{3}{5},\; P(F)=\frac{3}{10},\; P(E\cap F)=\frac{1}{5}

For independent events, we must have

\displaystyle P(E\cap F)=P(E)P(F)

Now, \displaystyle P(E)P(F)=\frac{3}{5}\times\frac{3}{10}=\frac{9}{50}

But, \displaystyle P(E\cap F)=\frac{1}{5}=\frac{10}{50}

Therefore, \displaystyle P(E\cap F)\neq P(E)P(F)

Hence, E and F are not independent.

7. Given that the events A and B are such that \displaystyle P(A)=\frac{1}{2},\; P(A\cup B)=\frac{3}{5} and \displaystyle P(B)=p. Find p if they are (i) mutually exclusive (ii) independent.

Solution:

Think First… We are given the probability of the union. For part (i), use the condition for mutually exclusive events. For part (ii), use the multiplication rule for independent events.

We are given

\displaystyle P(A)=\frac{1}{2},\; P(A\cup B)=\frac{3}{5},\; P(B)=p

(i) When A and B are mutually exclusive

For mutually exclusive events,

\displaystyle A\cap B=\varnothing

Therefore,

\displaystyle P(A\cup B)=P(A)+P(B)

Substituting the given values,

\displaystyle \frac{3}{5}=\frac{1}{2}+p

\displaystyle \Rightarrow p=\frac{3}{5}-\frac{1}{2}=\frac{1}{10}

Therefore,

\displaystyle \boxed{p=\frac{1}{10}}

(ii) When A and B are independent

For any two events,

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

Since A and B are independent,

\displaystyle P(A\cap B)=P(A)P(B)

Therefore,

\displaystyle P(A\cup B)=P(A)+P(B)-P(A)P(B)

Substituting the given values,

We get, \displaystyle \frac{3}{5}=\frac{1}{2}+p-\frac{1}{2}p

or \displaystyle \frac{3}{5}=\frac{1}{2}+\frac{p}{2}

or \displaystyle \frac{p}{2}=\frac{3}{5}-\frac{1}{2}=\frac{1}{10}

Therefore,

\displaystyle \boxed{p=\frac{1}{5}}

8. Let A and B be independent events with \displaystyle P(A)=0.3 and \displaystyle P(B)=0.4. Find (i) \displaystyle P(A\cap B) (ii) \displaystyle P(A\cup B) (iii) \displaystyle P(A\mid B) (iv) \displaystyle P(B\mid A)

Solution:

Think First… Since A and B are independent, first use \displaystyle P(A\cap B)=P(A)P(B). Then use the addition theorem for the union. For conditional probabilities, independence gives \displaystyle P(A\mid B)=P(A) and \displaystyle P(B\mid A)=P(B).

We are given

\displaystyle P(A)=0.3,\quad P(B)=0.4

(i) Finding \displaystyle P(A\cap B)

Since A and B are independent,

Using, \displaystyle P(A\cap B)=P(A)P(B)

We get,

\displaystyle P(A\cap B)=0.3\times0.4=0.12

Hence,

\displaystyle \boxed{P(A\cap B)=0.12}

(ii) Finding \displaystyle P(A\cup B)

Using the addition theorem,

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

Substituting the values,

\displaystyle P(A\cup B)=0.3+0.4-0.12=0.58

Hence,

\displaystyle \boxed{P(A\cup B)=0.58}

(iii) Finding \displaystyle P(A\mid B)

For independent events, the occurrence of B does not affect the probability of A. Therefore,

\displaystyle P(A\mid B)=P(A)=0.3

Hence,

\displaystyle \boxed{P(A\mid B)=0.3}

We can also verify this using the formula for conditional probability.

\displaystyle P(A\mid B)=\frac{P(A\cap B)}{P(B)}=\frac{0.12}{0.4}=0.3

Thus, the result is consistent with the fact that A and B are independent events.

(iv) Finding \displaystyle P(B\mid A)

Similarly, the occurrence of A does not affect the probability of B. Therefore,

\displaystyle P(B\mid A)=P(B)=0.4

Hence,

\displaystyle \boxed{P(B\mid A)=0.4}

Just like (iii) above, we can verify this also using the formula for conditional probability.

\displaystyle P(B\mid A)=\frac{P(A\cap B)}{P(A)}=\frac{0.12}{0.3}=0.4

Thus, the result is consistent with the fact that A and B are independent events.

9. If A and B are two events such that \displaystyle P(A)=\frac{1}{4},\; P(B)=\frac{1}{2} and \displaystyle P(A\cap B)=\frac{1}{8}, find the probability that neither A nor B occurs.

Solution:

Think First… “Neither A nor B” means that \displaystyle A' and \displaystyle B' both occur. This is the complement of \displaystyle A\cup B. So, first find \displaystyle P(A\cup B) using the addition theorem.

We are given

\displaystyle P(A)=\frac{1}{4},\; P(B)=\frac{1}{2},\; P(A\cap B)=\frac{1}{8}

Using the addition theorem of probability,

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

Substituting the given values, we get

\displaystyle P(A\cup B)=\frac{1}{4}+\frac{1}{2}-\frac{1}{8}

\displaystyle =\frac{2}{8}+\frac{4}{8}-\frac{1}{8}=\frac{5}{8}

The event “neither A nor B” is the complement of \displaystyle A\cup B. Therefore,

\displaystyle P(\text{neither A nor B})=1-P(\text{either A or B})=1-P(A\cup B)

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

Hence,

\displaystyle \boxed{P(\text{neither A nor B})=\frac{3}{8}}

10. Events A and B are such that \displaystyle P(A)=\frac{1}{2},\; P(B)=\frac{7}{12} and \displaystyle P(\text{not A or not B})=\frac{1}{4}. State whether A and B are independent.

Solution:

Think First… “Not A or not B” means \displaystyle A'\cup B'. By De Morgan’s law, its complement is \displaystyle A\cap B. So, first find \displaystyle P(A\cap B) and then compare it with \displaystyle P(A)P(B).

We are given

\displaystyle P(A)=\frac{1}{2},\; P(B)=\frac{7}{12},\; P(A'\cup B')=\frac{1}{4}

By De-Morgan’s Law, we know

\displaystyle (A'\cup B')'=A\cap B

Therefore, using the complement rule,

\displaystyle P((A'\cup B')')=1-P(A'\cup B')

Hence,

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

Therefore,

\displaystyle P(A\cap B)=1-\frac{1}{4}=\frac{3}{4}

Now, if A and B were independent, then

\displaystyle P(A\cap B)=P(A)P(B)

But

\displaystyle P(A)P(B)=\frac{1}{2}\times\frac{7}{12}=\frac{7}{24}

Clearly,

\displaystyle P(A\cap B)=\frac{3}{4}\neq\frac{7}{24}=P(A)P(B)

Therefore, A and B are not independent.

11. Given two independent events A and B such that \displaystyle P(A)=0.3,\; P(B)=0.6. Find (i) \displaystyle P(A\text{ and }B) (ii) \displaystyle P(A\text{ and not }B) (iii) \displaystyle P(A\text{ or }B) (iv) \displaystyle P(\text{neither }A\text{ nor }B)

Solution:

Think First… Since A and B are independent, use the multiplication rule \displaystyle P(A\cap B)=P(A)P(B). For events involving complements, remember that \displaystyle P(B')=1-P(B). The same independence rule can then be applied to the complementary events.

We are given

\displaystyle P(A)=0.3,\qquad P(B)=0.6

(i) Finding \displaystyle P(A\text{ and }B)\text{ or }P(A\cap B)

Since A and B are independent,

\displaystyle P(A\cap B)=P(A)P(B)

Therefore,

\displaystyle P(A\cap B)=0.3\times0.6=0.18

Hence,

\displaystyle \boxed{P(A\text{ and }B)=0.18}

(ii) Finding \displaystyle P(A\text{ and not }B)\text{ or }P(A\cap B')

First,

\displaystyle P(B')=1-P(B)=1-0.6=0.4

Since A and B are independent, A and B’ are also independent. Therefore,

\displaystyle P(A\cap B')=P(A)P(B')

Thus,

\displaystyle P(A\cap B')=0.3\times0.4=0.12

Hence,

\displaystyle \boxed{P(A\text{ and not }B)=0.12}

(iii) Finding \displaystyle P(A\text{ or }B)\text{ or }P(A\cup B)

Using the addition theorem,

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

Substituting the values,

\displaystyle P(A\cup B)=0.3+0.6-0.18=0.72

Hence,

\displaystyle \boxed{P(A\text{ or }B)=0.72}

(iv) Finding \displaystyle P(\text{neither }A\text{ nor }B)\text{ or }P(A'\cap B')

“Neither A nor B” means that both A and B do not occur. Therefore,

\displaystyle P(\text{neither }A\text{ nor }B)=P(A'\cap B')

Since A and B are independent, A’ and B’ are also independent. Hence,

\displaystyle P(A'\cap B')=P(A')P(B')

Now,

\displaystyle P(A')=1-0.3=0.7,\quad P(B')=1-0.6=0.4

Therefore,

\displaystyle P(A'\cap B')=0.7\times0.4=0.28

Hence,

\displaystyle \boxed{P(\text{neither }A\text{ nor }B)=0.28}

12. A die is tossed thrice. Find the probability of getting an odd number at least once.

Solution:

Think First… “At least once” is usually easier to find using the complement. The complement of getting an odd number at least once is getting an even number on all three tosses.

Let E be the event that an odd number is obtained at least once.

Then, the complement event \displaystyle E' is the event that an even number is obtained on all three tosses.

For one toss of a die,

\displaystyle P(\text{even number})=\frac{3}{6}=\frac{1}{2}

Since the three tosses are independent,

\displaystyle P(E')=\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}=\frac{1}{8}

Therefore,

\displaystyle P(E)=1-P(E')

\displaystyle =1-\frac{1}{8}=\frac{7}{8}

Therefore, the required probability is

\displaystyle \boxed{\frac{7}{8}}

13. Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that (i) \displaystyle \text{both balls are red} (ii) \displaystyle \text{the first ball is black and the second is red} (iii) \displaystyle \text{one of them is black and the other is red}

Solution:

Think First… Since the first ball is replaced before the second draw, the composition of the box remains unchanged. Therefore, the two draws are independent.

The box contains 10 black and 8 red balls, so there are 18 balls in total.

Therefore,

\displaystyle P(\text{black})=\frac{10}{18}=\frac{5}{9}

and

\displaystyle P(\text{red})=\frac{8}{18}=\frac{4}{9}

(i) Both balls are red

Since the draws are independent, because the balls are replaced (with replacement case)

\displaystyle P(\text{both red})=P(\text{red})P(\text{red})

Therefore,

\displaystyle P(\text{both red})=\frac{4}{9}\times\frac{4}{9}=\frac{16}{81}

Hence, the required probability is

\displaystyle \boxed{\frac{16}{81}}

(ii) First ball is black and second is red

Using the multiplication rule for independent events,

\displaystyle P(\text{first black and second red})=P(\text{black})P(\text{red})

Therefore,

\displaystyle =\frac{5}{9}\times\frac{4}{9}=\frac{20}{81}

Hence, the required probability is

\displaystyle \boxed{\frac{20}{81}}

(iii) One ball is black and the other is red

This can happen in two mutually exclusive ways:

\displaystyle \text{(a) First black and second red}

or

\displaystyle \text{(b) First red and second black}

Therefore, \displaystyle P(\text{one black and one red})

\displaystyle =P(BR)+P(RB)

\displaystyle =\frac{5}{9}\times\frac{4}{9}+\frac{4}{9}\times\frac{5}{9}

i.e. \displaystyle P(BR\cup RB)=\frac{20}{81}+\frac{20}{81}=\frac{40}{81}

Hence, the required probability is

\displaystyle \boxed{\frac{40}{81}}

14. Probability of solving a specific problem independently by A and B are \displaystyle \frac{1}{2} and \displaystyle \frac{1}{3} respectively. If both try to solve the problem independently, find the probability that (i) \displaystyle \text{the problem is solved} (ii) \displaystyle \text{exactly one of them solves the problem}

Solution:

Think First… Since A and B try independently, the events of their solving the problem are independent. For “at least one”, it is easier to find the complement: neither solves the problem.

Let A denote the event that A solves the problem and B denote the event that B solves the problem.

We are given

\displaystyle P(A)=\frac{1}{2},\quad P(B)=\frac{1}{3}

Therefore, the probabilities that they do not solve the problem are

\displaystyle P(A')=1-\frac{1}{2}=\frac{1}{2}

and

\displaystyle P(B')=1-\frac{1}{3}=\frac{2}{3}

(i) Probability that the problem is solved

The problem is solved if at least one of A and B solves it.

Therefore,

\displaystyle P(\text{problem is solved})=P(A\cup B)

It is easier to use the complement:

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

Since A and B are independent, A’ and B’ are also independent. Hence,

\displaystyle P(A'\cap B')=P(A')P(B')

Therefore,

\displaystyle P(A\cup B)=1-\frac{1}{2}\times\frac{2}{3}

\displaystyle =1-\frac{1}{3}=\frac{2}{3}

Therefore,

\displaystyle \boxed{P(\text{problem is solved})=\frac{2}{3}}

(ii) Probability that exactly one of them solves the problem

Exactly one of them solves the problem in two mutually exclusive ways:

\displaystyle \text{(a) A solves and B does not solve}

or

\displaystyle \text{(b) A does not solve and B solves}

Therefore,

\displaystyle P(\text{exactly one solves})=P(A\cap B')+P(A'\cap B)

Since the events are independent,

\displaystyle =P(A)P(B')+P(A')P(B)

Substituting the values,

\displaystyle =\frac{1}{2}\times\frac{2}{3}+\frac{1}{2}\times\frac{1}{3}

\displaystyle =\frac{1}{3}+\frac{1}{6}=\frac{1}{2}

Therefore,

\displaystyle \boxed{P(\text{exactly one solves})=\frac{1}{2}}

15. One card is drawn at random from a well shuffled deck of 52 cards. In which of the following cases are the events E and F independent?

(i) E: ‘the card drawn is a spade’

F: ‘the card drawn is an ace’

(ii) E: ‘the card drawn is black’

F: ‘the card drawn is a king’

(iii) E: ‘the card drawn is a king or queen’

F: ‘the card drawn is a queen or jack’

Solution:

Think First… To check whether E and F are independent, compare \displaystyle P(E\cap F) with \displaystyle P(E)P(F). If they are equal, the events are independent.

(i) E: card is a spade, F: card is an ace

There are 13 spades and 4 aces in a deck of 52 cards. Therefore,

\displaystyle P(E)=\frac{13}{52}=\frac{1}{4},\quad P(F)=\frac{4}{52}=\frac{1}{13}

The only card which is both a spade and an ace is the ace of spades. Hence,

\displaystyle P(E\cap F)=\frac{1}{52}

Also,

\displaystyle P(E)P(F)=\frac{1}{4}\times\frac{1}{13}=\frac{1}{52}

Thus,

\displaystyle P(E\cap F)=P(E)P(F)

Therefore, E and F are independent.

(ii) E: card is black, F: card is a king

There are 26 black cards and 4 kings in a deck of 52 cards. Therefore,

\displaystyle P(E)=\frac{26}{52}=\frac{1}{2},\quad P(F)=\frac{4}{52}=\frac{1}{13}

There are 2 black kings. Hence,

\displaystyle P(E\cap F)=\frac{2}{52}=\frac{1}{26}

Also,

\displaystyle P(E)P(F)=\frac{1}{2}\times\frac{1}{13}=\frac{1}{26}

Thus,

\displaystyle P(E\cap F)=P(E)P(F)

Therefore, E and F are independent.

(iii) E: card is a king or queen, F: card is a queen or jack

There are 4 kings and 4 queens. Therefore,

\displaystyle P(E)=\frac{8}{52}=\frac{2}{13}

Similarly, there are 4 queens and 4 jacks. Hence,

\displaystyle P(F)=\frac{8}{52}=\frac{2}{13}

The common cards in E and F are the 4 queens. Therefore,

\displaystyle P(E\cap F)=\frac{4}{52}=\frac{1}{13}

But

\displaystyle P(E)P(F)=\frac{2}{13}\times\frac{2}{13}=\frac{4}{169}

Clearly,

\displaystyle P(E\cap F)\neq P(E)P(F)

Therefore, E and F are not independent.

Hence, the events are independent in cases (i) and (ii).

16. In a girls’ hostel, 60% of the students read Hindi newspaper, 40% read English newspaper and 20% read both Hindi and English newspapers. A student is selected at random.

(a) Find the probability that she reads neither Hindi nor English newspapers.

(b) If she reads Hindi newspaper, find the probability that she reads English newspaper.

(c) If she reads English newspaper, find the probability that she reads Hindi newspaper.

Solution:

Think First… Let H and E represent the events that the student reads Hindi and English newspapers, respectively. For part (a), “neither” means the complement of \displaystyle H\cup E. Parts (b) and (c) require conditional probability.

Let H be the event that the student reads Hindi newspaper and E be the event that the student reads English newspaper.

We are given

\displaystyle P(H)=0.6,\; P(E)=0.4,\; P(H\cap E)=0.2

(a) Probability that she reads neither Hindi nor English newspapers

First, using the addition theorem,

\displaystyle P(H\cup E)=P(H)+P(E)-P(H\cap E)

Substituting the given values,

\displaystyle P(H\cup E)=0.6+0.4-0.2=0.8

Therefore, the probability that she reads neither Hindi nor English newspapers is

\displaystyle P(\text{neither H nor E})=1-P(H\cup E)

\displaystyle =1-0.8=0.2=\frac{1}{5}

Therefore,

\displaystyle \boxed{P(\text{neither H nor E})=0.2}

(b) Probability that she reads English newspaper, given that she reads Hindi newspaper

We need to find \displaystyle P(E\mid H).

Using the formula of conditional probability,

\displaystyle P(E\mid H)=\frac{P(E\cap H)}{P(H)}

Substituting the given values,

\displaystyle P(E\mid H)=\frac{0.2}{0.6}=\frac{1}{3}

Therefore,

\displaystyle \boxed{P(E\mid H)=\frac{1}{3}}

(c) Probability that she reads Hindi newspaper, given that she reads English newspaper

We need to find \displaystyle P(H\mid E).

Using the formula of conditional probability,

\displaystyle P(H\mid E)=\frac{P(H\cap E)}{P(E)}

Substituting the given values,

\displaystyle P(H\mid E)=\frac{0.2}{0.4}=\frac{1}{2}

Therefore,

\displaystyle \boxed{P(H\mid E)=\frac{1}{2}}

17. The probability of obtaining an even prime number on each die, when a pair of dice is rolled is

• (A) \displaystyle 0
• (B) \displaystyle \frac{1}{3}
• (C) \displaystyle \frac{1}{12}
• (D) \displaystyle \frac{1}{36}

Solution

Think First… There is only one even prime number, namely \displaystyle 2. So, we need the probability of getting \displaystyle 2 on both dice.

The only even prime number is \displaystyle 2.

For one die,

\displaystyle P(\text{even prime number})=P(2)=\frac{1}{6}

Since the two dice are independent, the required probability is

\displaystyle P(2\text{ on first die and }2\text{ on second die})

\displaystyle =\frac{1}{6}\times\frac{1}{6}=\frac{1}{36}

Hence, the correct answer is (D).

18. Two events A and B will be independent, if

• (A) A and B are mutually exclusive
• (B) \displaystyle P(A'B')=[1-P(A)][1-P(B)]
• (C) \displaystyle P(A)=P(B)
• (D) \displaystyle P(A)+P(B)=1

Solution

Think First… For independent events, the occurrence of one event does not affect the probability of the other. Independence can also be expressed using the complementary events.

For two independent events A and B, we have

\displaystyle P(A\cap B)=P(A)P(B)

If A and B are independent, their complementary events A’ and B’ are also independent. Therefore,

\displaystyle P(A'\cap B')=P(A')P(B')

Using \displaystyle P(A')=1-P(A) and \displaystyle P(B')=1-P(B), we get

\displaystyle P(A'B')=[1-P(A)][1-P(B)]

Thus, option (B) is the correct condition.

Hence, the correct answer is (B).