NCERT Class 12 Mathematics — Probability, Exercise 13.2. All 18 questions solved.

Two ideas run through the whole exercise.

The multiplication theorem. For any two events,

$$\mathrm{P}(\mathrm{A} \cap \mathrm{B}) = \mathrm{P}(\mathrm{A})\,\mathrm{P}(\mathrm{B}|\mathrm{A})$$

which is just the conditional probability formula rearranged. It is what handles drawing without replacement, where the second probability depends on the first.

Independence. $\mathrm{A}$ and $\mathrm{B}$ are independent when

$$\mathrm{P}(\mathrm{A} \cap \mathrm{B}) = \mathrm{P}(\mathrm{A})\,\mathrm{P}(\mathrm{B})$$

and then the multiplication theorem simplifies to a plain product.

Key insight. Independence is a numerical condition, not an intuitive one. Questions 4, 5, 6, 10 and 15 all ask you to decide it, and the only correct method is to compute all three of $\mathrm{P}(\mathrm{A})$, $\mathrm{P}(\mathrm{B})$ and $\mathrm{P}(\mathrm{A} \cap \mathrm{B})$ and check whether the product law holds. Question 15(iii) looks as independent as (i) and (ii) and is not.

Independence is also not the same as mutual exclusivity — in fact two events of non-zero probability cannot be both. Question 7 asks for the same quantity under each assumption, and the two answers differ.

Question 1

If $\mathrm{P}(\mathrm{A}) = \dfrac35$ and $\mathrm{P}(\mathrm{B}) = \dfrac15$, find $\mathrm{P}(\mathrm{A} \cap \mathrm{B})$ if $\mathrm{A}$ and $\mathrm{B}$ are independent events.

Solution. Independence means the intersection is the product:

$$\mathrm{P}(\mathrm{A} \cap \mathrm{B}) = \frac35 \times \frac15$$

Question 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. “Without replacement” is the signal for the multiplication theorem. There are $26$ black cards:

$$\mathrm{P} = \frac{26}{52} \times \frac{25}{51}$$

The second fraction has both a smaller numerator and a smaller denominator, because one black card has already gone.

Question 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. Three good oranges in succession, without replacement:

$$\frac{12}{15} \times \frac{11}{14} \times \frac{10}{13} = \frac{1320}{2730}$$

Question 4

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

Solution. The sample space has $12$ equally likely outcomes.

$$\mathrm{P}(\mathrm{A}) = \frac{6}{12} = \frac12, \qquad \mathrm{P}(\mathrm{B}) = \frac{2}{12} = \frac16, \qquad \mathrm{P}(\mathrm{A} \cap \mathrm{B}) = \frac{1}{12}$$

Since $\tfrac12 \times \tfrac16 = \tfrac{1}{12}$, the product law holds.

Question 5

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

Solution. $\mathrm{A} = \{2, 4, 6\}$ so $\mathrm{P}(\mathrm{A}) = \tfrac12$; $\mathrm{B} = \{1, 2, 3\}$ so $\mathrm{P}(\mathrm{B}) = \tfrac12$; and $\mathrm{A} \cap \mathrm{B} = \{2\}$ so $\mathrm{P}(\mathrm{A} \cap \mathrm{B}) = \tfrac16$.

$$\frac12 \times \frac12 = \frac14 \ne \frac16$$

Question 6

Let $\mathrm{E}$ and $\mathrm{F}$ be events with $\mathrm{P}(\mathrm{E}) = \dfrac35$, $\mathrm{P}(\mathrm{F}) = \dfrac{3}{10}$ and $\mathrm{P}(\mathrm{E} \cap \mathrm{F}) = \dfrac15$. Are $\mathrm{E}$ and $\mathrm{F}$ independent?

Solution.

$$\mathrm{P}(\mathrm{E})\,\mathrm{P}(\mathrm{F}) = \frac35 \times \frac{3}{10} = \frac{9}{50} = 0.18$$

but $\mathrm{P}(\mathrm{E} \cap \mathrm{F}) = 0.2$.

Question 7

Given that the events $\mathrm{A}$ and $\mathrm{B}$ are such that $\mathrm{P}(\mathrm{A}) = \dfrac12$, $\mathrm{P}(\mathrm{A} \cup \mathrm{B}) = \dfrac35$ and $\mathrm{P}(\mathrm{B}) = p$. Find $p$ if they are (i) mutually exclusive (ii) independent.

Solution.

(i) Mutually exclusive means $\mathrm{P}(\mathrm{A} \cap \mathrm{B}) = 0$, so the addition rule is simply additive:

$$\frac35 = \frac12 + p \quad\Longrightarrow\quad p = \frac{1}{10}$$

(ii) Independent means $\mathrm{P}(\mathrm{A} \cap \mathrm{B}) = \tfrac12 p$:

$$\frac35 = \frac12 + p – \frac{p}{2} = \frac12 + \frac{p}{2} \quad\Longrightarrow\quad p = \frac15$$

The same data give different answers under the two assumptions — proof that they are different conditions.

Question 8

Let $\mathrm{A}$ and $\mathrm{B}$ be independent events with $\mathrm{P}(\mathrm{A}) = 0.3$ and $\mathrm{P}(\mathrm{B}) = 0.4$. Find (i) $\mathrm{P}(\mathrm{A} \cap \mathrm{B})$ (ii) $\mathrm{P}(\mathrm{A} \cup \mathrm{B})$ (iii) $\mathrm{P}(\mathrm{A}|\mathrm{B})$ (iv) $\mathrm{P}(\mathrm{B}|\mathrm{A})$.

Solution.

(i) $0.3 \times 0.4 = 0.12$

(ii) $0.3 + 0.4 – 0.12 = 0.58$

(iii) $\dfrac{0.12}{0.4} = 0.3 = \mathrm{P}(\mathrm{A})$

(iv) $\dfrac{0.12}{0.3} = 0.4 = \mathrm{P}(\mathrm{B})$

Parts (iii) and (iv) illustrate what independence means: knowing that one event occurred leaves the other’s probability unchanged.

Question 9

If $\mathrm{A}$ and $\mathrm{B}$ are two events such that $\mathrm{P}(\mathrm{A}) = \dfrac14$, $\mathrm{P}(\mathrm{B}) = \dfrac12$ and $\mathrm{P}(\mathrm{A} \cap \mathrm{B}) = \dfrac18$, find $\mathrm{P}$(not $\mathrm{A}$ and not $\mathrm{B}$).

Solution. By De Morgan’s law, “not $\mathrm{A}$ and not $\mathrm{B}$” is the complement of $\mathrm{A} \cup \mathrm{B}$:

$$\mathrm{P}(\mathrm{A} \cup \mathrm{B}) = \frac14 + \frac12 – \frac18 = \frac58$$

$$\mathrm{P}(\mathrm{A}’ \cap \mathrm{B}’) = 1 – \frac58$$

Question 10

Events $\mathrm{A}$ and $\mathrm{B}$ are such that $\mathrm{P}(\mathrm{A}) = \dfrac12$, $\mathrm{P}(\mathrm{B}) = \dfrac{7}{12}$ and $\mathrm{P}$(not $\mathrm{A}$ or not $\mathrm{B}$) $= \dfrac14$. State whether $\mathrm{A}$ and $\mathrm{B}$ are independent.

Solution. “Not $\mathrm{A}$ or not $\mathrm{B}$” is the complement of $\mathrm{A} \cap \mathrm{B}$, so

$$\mathrm{P}(\mathrm{A} \cap \mathrm{B}) = 1 – \frac14 = \frac34$$

But $\mathrm{P}(\mathrm{A})\,\mathrm{P}(\mathrm{B}) = \tfrac12 \times \tfrac{7}{12} = \tfrac{7}{24}$, which is nowhere near $\tfrac34$.

(In fact $\tfrac34$ exceeds $\mathrm{P}(\mathrm{B}) = \tfrac{7}{12}$, which is already impossible for an intersection — but the question only asks about independence.)

Question 11

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

Solution. If $\mathrm{A}$ and $\mathrm{B}$ are independent, so are $\mathrm{A}$ and $\mathrm{B}’$, and $\mathrm{A}’$ and $\mathrm{B}’$.

(i) $0.3 \times 0.6 = 0.18$

(ii) $0.3 \times 0.4 = 0.12$

(iii) $0.3 + 0.6 – 0.18 = 0.72$

(iv) $0.7 \times 0.4 = 0.28$, which is also $1 – 0.72$.

Question 12

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

Solution. “At least once” is almost always easier through the complement. The probability of no odd number in a single throw is $\tfrac12$, and the three throws are independent:

$$\mathrm{P}(\text{no odd number}) = \left(\frac12\right)^3 = \frac18$$

Question 13

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

Solution. “With replacement” makes the two draws independent, each with $\mathrm{P}(\text{red}) = \tfrac{8}{18} = \tfrac49$ and $\mathrm{P}(\text{black}) = \tfrac{10}{18} = \tfrac59$.

(i) $\left(\tfrac49\right)^2 = \tfrac{16}{81}$

(ii) $\tfrac59 \times \tfrac49 = \tfrac{20}{81}$

(iii) Either order counts, so double part (ii): $\tfrac{40}{81}$.

The word “first” in (ii) fixes the order; its absence in (iii) is what doubles the answer.

Question 14

Probability of solving a specific problem independently by $\mathrm{A}$ and $\mathrm{B}$ are $\dfrac12$ and $\dfrac13$ respectively. If both try to solve the problem independently, find the probability that (i) the problem is solved (ii) exactly one of them solves the problem.

Solution.

(i) “Solved” means at least one succeeds, so take the complement of both failing:

$$1 – \left(\frac12\right)\left(\frac23\right) = 1 – \frac13 = \frac23$$

(ii) Exactly one means $\mathrm{A}$ succeeds and $\mathrm{B}$ fails, or the other way round:

$$\frac12 \cdot \frac23 + \frac12 \cdot \frac13 = \frac13 + \frac16 = \frac12$$

Question 15

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

(i) $\mathrm{E}$: ‘the card drawn is a spade’, $\mathrm{F}$: ‘the card drawn is an ace’

(ii) $\mathrm{E}$: ‘the card drawn is black’, $\mathrm{F}$: ‘the card drawn is a king’

(iii) $\mathrm{E}$: ‘the card drawn is a king or queen’, $\mathrm{F}$: ‘the card drawn is a queen or jack’

Solution. In each case compute all three probabilities and compare.

(i) $\mathrm{P}(\mathrm{E}) = \tfrac{13}{52} = \tfrac14$, $\mathrm{P}(\mathrm{F}) = \tfrac{4}{52} = \tfrac{1}{13}$, and the intersection is the ace of spades: $\tfrac{1}{52}$. Since $\tfrac14 \times \tfrac{1}{13} = \tfrac{1}{52}$ — independent.

(ii) $\mathrm{P}(\mathrm{E}) = \tfrac12$, $\mathrm{P}(\mathrm{F}) = \tfrac{1}{13}$, intersection is the two black kings: $\tfrac{2}{52} = \tfrac{1}{26}$. Since $\tfrac12 \times \tfrac{1}{13} = \tfrac{1}{26}$ — independent.

(iii) $\mathrm{P}(\mathrm{E}) = \tfrac{8}{52} = \tfrac{2}{13}$, $\mathrm{P}(\mathrm{F}) = \tfrac{2}{13}$, and the intersection is the four queens: $\tfrac{4}{52} = \tfrac{1}{13}$. But $\tfrac{2}{13} \times \tfrac{2}{13} = \tfrac{4}{169}$, while $\tfrac{1}{13} = \tfrac{13}{169}$ — not independent.

Question 16

In a 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. Write $\mathrm{H}$ and $\mathrm{E}$ for the two reading events, with $\mathrm{P}(\mathrm{H}) = 0.6$, $\mathrm{P}(\mathrm{E}) = 0.4$ and $\mathrm{P}(\mathrm{H} \cap \mathrm{E}) = 0.2$.

(a) $\mathrm{P}(\mathrm{H} \cup \mathrm{E}) = 0.6 + 0.4 – 0.2 = 0.8$, so neither is $1 – 0.8 = 0.2 = \tfrac15$.

(b) $\mathrm{P}(\mathrm{E}|\mathrm{H}) = \dfrac{0.2}{0.6} = \dfrac13$

(c) $\mathrm{P}(\mathrm{H}|\mathrm{E}) = \dfrac{0.2}{0.4} = \dfrac12$

Note that $\mathrm{P}(\mathrm{E}|\mathrm{H}) = \tfrac13 \ne 0.4 = \mathrm{P}(\mathrm{E})$, so these two events are not independent.

Choose the correct answer in Exercises 17 and 18.

Question 17

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

Solution. There is exactly one even prime number: $2$. Every other even number has $2$ as a factor and so is not prime.

So the required event is “$2$ on both dice”, a single outcome out of $36$.

Option (A) is chosen by students who think no even number can be prime; the whole question turns on the exception.

Question 18

Two events $\mathrm{A}$ and $\mathrm{B}$ will be independent, if

Solution. Option (B) says $\mathrm{A}’$ and $\mathrm{B}’$ are independent, and that is equivalent to $\mathrm{A}$ and $\mathrm{B}$ being independent — a standard result proved in the chapter.

Option (A) is the opposite: mutually exclusive events with non-zero probabilities are never independent, because $\mathrm{P}(\mathrm{A} \cap \mathrm{B}) = 0$ while the product is not. Options (C) and (D) are conditions on the individual probabilities and say nothing about the intersection.

Common mistakes

• Assuming independence because the events “feel” unrelated. Question 15(iii) is the counterexample: two ordinary card events that fail the product test.
• Confusing independent with mutually exclusive. They are opposites in practice, and question 7 gives different answers under each.
• Multiplying probabilities when drawing without replacement. Questions 2 and 3 need the second probability to change; questions 13 and 14 do not, because replacement or independence is stated.
• Forgetting to double for order. In question 13(iii) “one black and one red” admits two orders; part (ii) fixes the order and does not.
• Mishandling “not $\mathrm{A}$ or not $\mathrm{B}$”. By De Morgan’s law it is the complement of $\mathrm{A} \cap \mathrm{B}$, not of $\mathrm{A} \cup \mathrm{B}$. Questions 9 and 10 use each version once.
• Attacking “at least once” directly. The complement is almost always shorter, as in questions 12 and 14.

Practise next

• Exercise 13.3 — Bayes’ theorem, which uses the multiplication theorem in reverse.
• Exercise 13.1 — the conditional probability that independence is defined against.