NCERT Class 12 Mathematics — Probability, Exercise 13.1. All 17 questions solved.

Conditional probability is one formula:

$$\mathrm{P}(\mathrm{E}\,|\,\mathrm{F}) = \frac{\mathrm{P}(\mathrm{E} \cap \mathrm{F})}{\mathrm{P}(\mathrm{F})}, \qquad \mathrm{P}(\mathrm{F}) \ne 0$$

Everything else in the exercise is a matter of identifying $\mathrm{E}$ and $\mathrm{F}$ correctly and counting.

Key insight. $\mathrm{P}(\mathrm{E}\,|\,\mathrm{F})$ says: given that $\mathrm{F}$ has happened, how likely is $\mathrm{E}$? The conditioning event becomes the new sample space. So in a counting problem you can shortcut the formula entirely — just list the outcomes in $\mathrm{F}$, count how many of them are also in $\mathrm{E}$, and divide.

That shortcut turns questions 6, 7, 9, 10 and 11 into two lines each, and it makes question 15 obvious rather than laborious: the coin is only ever tossed when the die does not show a multiple of $3$, so “the coin shows a tail” and “a die shows a $3$” can never both happen.

Question 1

Given that $\mathrm{E}$ and $\mathrm{F}$ are events such that $\mathrm{P}(\mathrm{E}) = 0.6$, $\mathrm{P}(\mathrm{F}) = 0.3$ and $\mathrm{P}(\mathrm{E} \cap \mathrm{F}) = 0.2$, find $\mathrm{P}(\mathrm{E}|\mathrm{F})$ and $\mathrm{P}(\mathrm{F}|\mathrm{E})$.

Solution. Straight substitution, once each way:

$$\mathrm{P}(\mathrm{E}|\mathrm{F}) = \frac{0.2}{0.3} = \frac23, \qquad \mathrm{P}(\mathrm{F}|\mathrm{E}) = \frac{0.2}{0.6} = \frac13$$

The two are different: conditioning on the rarer event gives the larger answer.

Question 2

Compute $\mathrm{P}(\mathrm{A}|\mathrm{B})$, if $\mathrm{P}(\mathrm{B}) = 0.5$ and $\mathrm{P}(\mathrm{A} \cap \mathrm{B}) = 0.32$.

Solution.

$$\mathrm{P}(\mathrm{A}|\mathrm{B}) = \frac{0.32}{0.5} = 0.64 = \frac{16}{25}$$

Question 3

If $\mathrm{P}(\mathrm{A}) = 0.8$, $\mathrm{P}(\mathrm{B}) = 0.5$ and $\mathrm{P}(\mathrm{B}|\mathrm{A}) = 0.4$, find (i) $\mathrm{P}(\mathrm{A} \cap \mathrm{B})$ (ii) $\mathrm{P}(\mathrm{A}|\mathrm{B})$ (iii) $\mathrm{P}(\mathrm{A} \cup \mathrm{B})$.

Solution. Run the formula backwards to get the intersection first, then use it twice more.

(i) $\mathrm{P}(\mathrm{A} \cap \mathrm{B}) = \mathrm{P}(\mathrm{B}|\mathrm{A})\mathrm{P}(\mathrm{A}) = 0.4 \times 0.8 = 0.32$

(ii) $\mathrm{P}(\mathrm{A}|\mathrm{B}) = \dfrac{0.32}{0.5} = 0.64$

(iii) $\mathrm{P}(\mathrm{A} \cup \mathrm{B}) = 0.8 + 0.5 – 0.32 = 0.98$

Question 4

Evaluate $\mathrm{P}(\mathrm{A} \cup \mathrm{B})$, if $2\mathrm{P}(\mathrm{A}) = \mathrm{P}(\mathrm{B}) = \dfrac{5}{13}$ and $\mathrm{P}(\mathrm{A}|\mathrm{B}) = \dfrac{2}{5}$.

Solution. From $2\mathrm{P}(\mathrm{A}) = \tfrac{5}{13}$ we get $\mathrm{P}(\mathrm{A}) = \tfrac{5}{26}$, and $\mathrm{P}(\mathrm{B}) = \tfrac{5}{13}$.

$$\mathrm{P}(\mathrm{A} \cap \mathrm{B}) = \mathrm{P}(\mathrm{A}|\mathrm{B})\mathrm{P}(\mathrm{B}) = \frac25 \cdot \frac{5}{13} = \frac{2}{13}$$

$$\mathrm{P}(\mathrm{A} \cup \mathrm{B}) = \frac{5}{26} + \frac{10}{26} – \frac{4}{26}$$

Question 5

If $\mathrm{P}(\mathrm{A}) = \dfrac{6}{11}$, $\mathrm{P}(\mathrm{B}) = \dfrac{5}{11}$ and $\mathrm{P}(\mathrm{A} \cup \mathrm{B}) = \dfrac{7}{11}$, find (i) $\mathrm{P}(\mathrm{A} \cap \mathrm{B})$ (ii) $\mathrm{P}(\mathrm{A}|\mathrm{B})$ (iii) $\mathrm{P}(\mathrm{B}|\mathrm{A})$.

Solution.

(i) From the addition rule, $\mathrm{P}(\mathrm{A} \cap \mathrm{B}) = \tfrac{6}{11} + \tfrac{5}{11} – \tfrac{7}{11} = \tfrac{4}{11}$

(ii) $\mathrm{P}(\mathrm{A}|\mathrm{B}) = \dfrac{4/11}{5/11} = \dfrac45$

(iii) $\mathrm{P}(\mathrm{B}|\mathrm{A}) = \dfrac{4/11}{6/11} = \dfrac23$

Determine $\mathrm{P}(\mathrm{E}|\mathrm{F})$ in Exercises 6 to 9.

Question 6

A coin is tossed three times, where

(i) $\mathrm{E}$: head on third toss, $\mathrm{F}$: heads on first two tosses

(ii) $\mathrm{E}$: at least two heads, $\mathrm{F}$: at most two heads

(iii) $\mathrm{E}$: at most two tails, $\mathrm{F}$: at least one tail

Solution. The sample space has $8$ equally likely outcomes: $\{\mathrm{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}\}$.

(i) $\mathrm{F} = \{\mathrm{HHH, HHT}\}$, of which only $\mathrm{HHH}$ has a head on the third toss. So the answer is $\tfrac12$.

(ii) $\mathrm{F}$ (at most two heads) excludes only $\mathrm{HHH}$, so it has $7$ outcomes. Within $\mathrm{F}$, “at least two heads” means exactly two heads: $\mathrm{HHT, HTH, THH}$ — three of them.

(iii) $\mathrm{F}$ (at least one tail) excludes only $\mathrm{HHH}$, so it has $7$ outcomes. Within $\mathrm{F}$, “at most two tails” excludes $\mathrm{TTT}$, leaving $6$.

Question 7

Two coins are tossed once, where

(i) $\mathrm{E}$: tail appears on one coin, $\mathrm{F}$: one coin shows head

(ii) $\mathrm{E}$: no tail appears, $\mathrm{F}$: no head appears

Solution. Sample space $\{\mathrm{HH, HT, TH, TT}\}$.

(i) Both $\mathrm{E}$ and $\mathrm{F}$ describe the same set, $\{\mathrm{HT, TH}\}$ — “a tail on one coin” and “one coin shows a head” are the same event when there are two coins. So every outcome of $\mathrm{F}$ is in $\mathrm{E}$ and the answer is $1$.

(ii) $\mathrm{E} = \{\mathrm{HH}\}$ and $\mathrm{F} = \{\mathrm{TT}\}$ are disjoint, so $\mathrm{P}(\mathrm{E} \cap \mathrm{F}) = 0$.

Question 8

A die is thrown three times, $\mathrm{E}$: $4$ appears on the third toss, $\mathrm{F}$: $6$ and $5$ appear respectively on first two tosses.

Solution. $\mathrm{F}$ pins down the first two throws exactly, leaving the third free: $\mathrm{F}$ has $6$ outcomes out of $216$. Exactly one of those six has a $4$ on the third toss.

Working within the reduced sample space is far quicker than computing $\tfrac{1/216}{6/216}$.

Question 9

Mother, father and son line up at random for a family picture. $\mathrm{E}$: son on one end, $\mathrm{F}$: father in middle.

Solution. With three people there are $6$ arrangements. Putting the father in the middle leaves the mother and son for the two ends, so $\mathrm{F} = \{\text{MFS}, \text{SFM}\}$ — and in both the son is on an end.

Question 10

A black and a red die are rolled.

(a) Find the conditional probability of obtaining a sum greater than $9$, given that the black die resulted in a $5$.

(b) Find the conditional probability of obtaining the sum $8$, given that the red die resulted in a number less than $4$.

Solution.

(a) Fixing the black die at $5$ leaves $6$ outcomes, with sums $6$ to $11$. Those exceeding $9$ are the sums $10$ and $11$, from red $= 5$ and red $= 6$.

(b) The red die being less than $4$ gives $3 \times 6 = 18$ outcomes. For a sum of $8$: red $= 2$ with black $= 6$, and red $= 3$ with black $= 5$. (Red $= 1$ would need black $= 7$.) So $2$ out of $18$.

Question 11

A fair die is rolled. Consider events $\mathrm{E} = \{1, 3, 5\}$, $\mathrm{F} = \{2, 3\}$ and $\mathrm{G} = \{2, 3, 4, 5\}$. Find

(i) $\mathrm{P}(\mathrm{E}|\mathrm{F})$ and $\mathrm{P}(\mathrm{F}|\mathrm{E})$   (ii) $\mathrm{P}(\mathrm{E}|\mathrm{G})$ and $\mathrm{P}(\mathrm{G}|\mathrm{E})$   (iii) $\mathrm{P}((\mathrm{E} \cup \mathrm{F})|\mathrm{G})$ and $\mathrm{P}((\mathrm{E} \cap \mathrm{F})|\mathrm{G})$

Solution. Everything is counting within the given sets.

(i) $\mathrm{E} \cap \mathrm{F} = \{3\}$. So $\mathrm{P}(\mathrm{E}|\mathrm{F}) = \tfrac{1}{2}$ (one of $\mathrm{F}$’s two elements) and $\mathrm{P}(\mathrm{F}|\mathrm{E}) = \tfrac13$ (one of $\mathrm{E}$’s three).

(ii) $\mathrm{E} \cap \mathrm{G} = \{3, 5\}$. So $\mathrm{P}(\mathrm{E}|\mathrm{G}) = \tfrac24 = \tfrac12$ and $\mathrm{P}(\mathrm{G}|\mathrm{E}) = \tfrac23$.

(iii) $\mathrm{E} \cup \mathrm{F} = \{1, 2, 3, 5\}$, and its intersection with $\mathrm{G}$ is $\{2, 3, 5\}$ — three of $\mathrm{G}$’s four elements. $\mathrm{E} \cap \mathrm{F} = \{3\}$, giving one of four.

Question 12

Assume that each born child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that (i) the youngest is a girl, (ii) at least one is a girl?

Solution. Sample space $\{\mathrm{BB, BG, GB, GG}\}$, writing the elder child first.

(i) “Youngest is a girl” is $\{\mathrm{BG, GG}\}$ — two outcomes, one of which is $\mathrm{GG}$.

(ii) “At least one is a girl” is $\{\mathrm{BG, GB, GG}\}$ — three outcomes, one of which is $\mathrm{GG}$.

The two answers differ, which is the point of the question: knowing which child is a girl is more information than knowing that some child is.

Question 13

An instructor has a question bank consisting of $300$ easy True/False questions, $200$ difficult True/False questions, $500$ easy multiple choice questions and $400$ difficult multiple choice questions. If a question is selected at random from the question bank, what is the probability that it will be an easy question given that it is a multiple choice question?

Solution. Conditioning on “multiple choice” restricts attention to the $500 + 400 = 900$ multiple choice questions, of which $500$ are easy. The $500$ True/False questions play no part at all.

Question 14

Given that the two numbers appearing on throwing two dice are different, find the probability of the event ‘the sum of numbers on the dice is $4$’.

Solution. Of the $36$ outcomes, $6$ have equal numbers, so “different numbers” leaves $30$. A sum of $4$ comes from $(1,3)$, $(2,2)$ and $(3,1)$ — but $(2,2)$ is excluded by the condition, leaving $2$.

Question 15

Consider the experiment of throwing a die: if a multiple of $3$ comes up, throw the die again, and if any other number comes, toss a coin. Find the conditional probability of the event ‘the coin shows a tail’, given that ‘at least one die shows a $3$’.

Solution. Read the experiment carefully. The coin is tossed only when the first throw is not a multiple of $3$ — that is, only when the die shows $1$, $2$, $4$ or $5$.

But “at least one die shows a $3$” requires a $3$ somewhere, and a $3$ on the first throw is a multiple of $3$, so the die is thrown again and no coin is ever tossed.

The two events are therefore mutually exclusive: the intersection is empty.

In each of the Exercises 16 and 17 choose the correct answer.

Question 16

If $\mathrm{P}(\mathrm{A}) = \dfrac12$, $\mathrm{P}(\mathrm{B}) = 0$, then $\mathrm{P}(\mathrm{A}|\mathrm{B})$ is

Solution. The formula divides by $\mathrm{P}(\mathrm{B})$, and the definition of conditional probability explicitly requires $\mathrm{P}(\mathrm{B}) \ne 0$. With $\mathrm{P}(\mathrm{B}) = 0$ the quantity does not exist.

Option (A) is the trap: $\mathrm{P}(\mathrm{A} \cap \mathrm{B}) = 0$ here, but $\tfrac00$ is not $0$.

Question 17

If $\mathrm{A}$ and $\mathrm{B}$ are events such that $\mathrm{P}(\mathrm{A}|\mathrm{B}) = \mathrm{P}(\mathrm{B}|\mathrm{A})$, then

Solution. Both sides have the same numerator:

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

Provided $\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \ne 0$, cancelling gives $\mathrm{P}(\mathrm{A}) = \mathrm{P}(\mathrm{B})$.

Equal probabilities do not force the events to be equal as sets, so (B) is too strong.

Common mistakes

• Conditioning the wrong way round. $\mathrm{P}(\mathrm{E}|\mathrm{F})$ and $\mathrm{P}(\mathrm{F}|\mathrm{E})$ are different numbers whenever $\mathrm{P}(\mathrm{E}) \ne \mathrm{P}(\mathrm{F})$; question 1 shows both.
• Counting outcomes from the full sample space. Once you condition on $\mathrm{F}$, the denominator is the number of outcomes in $\mathrm{F}$, not $36$ or $8$.
• Reading “at least one is a girl” as “the elder is a girl”. Question 12 gives $\tfrac13$ and $\tfrac12$ for exactly this reason.
• Treating $\tfrac00$ as zero. Question 16 is undefined, not $0$.
• Missing that an event can be impossible. Question 15’s answer is $0$ because the experiment’s rules make the two events incompatible — not because of any computation.
• Forgetting to exclude outcomes the condition rules out. In question 14 the pair $(2,2)$ sums to $4$ but is excluded because the numbers must differ.

Practise next

• Exercise 13.2 — independence, which is exactly the case where conditioning changes nothing.
• Exercise 13.3 — Bayes’ theorem, which reverses the direction of conditioning.