NCERT Class 11 Mathematics — Probability, Exercise 14.2. All 21 questions solved.

The axiomatic definition asks only two things of an assignment of probabilities:

$$\mathrm P(\omega_i) \ge 0 \text{ for every outcome}, \qquad \sum_i \mathrm P(\omega_i) = 1$$

When all outcomes are equally likely,

$$\mathrm P(\mathrm E) = \frac{\text{number of outcomes in } \mathrm E}{\text{total number of outcomes}}$$

Two results do the rest:

$$\mathrm P(\mathrm A’) = 1 – \mathrm P(\mathrm A)$$

$$\mathrm P(\mathrm A \cup \mathrm B) = \mathrm P(\mathrm A) + \mathrm P(\mathrm B) – \mathrm P(\mathrm A \cap \mathrm B)$$

the second reducing to $\mathrm P(\mathrm A) + \mathrm P(\mathrm B)$ when the events are mutually exclusive.

Key insight. $\mathrm P(\mathrm A \cap \mathrm B)$ can never exceed either $\mathrm P(\mathrm A)$ or $\mathrm P(\mathrm B)$ — the intersection is a subset of both. That single inequality is what makes question 12(i) inconsistent, and it is the quickest check to run on any set of given probabilities before doing anything with them.

Question 1

Which of the following cannot be a valid assignment of probabilities for the outcomes of the sample space $\mathrm S = \{\omega_1, \omega_2, \ldots, \omega_7\}$?

	$\omega_1$	$\omega_2$	$\omega_3$	$\omega_4$	$\omega_5$	$\omega_6$	$\omega_7$
(a)	$0.1$	$0.01$	$0.05$	$0.03$	$0.01$	$0.2$	$0.6$
(b)	$\frac17$	$\frac17$	$\frac17$	$\frac17$	$\frac17$	$\frac17$	$\frac17$
(c)	$0.1$	$0.2$	$0.3$	$0.4$	$0.5$	$0.6$	$0.7$
(d)	$-0.1$	$0.2$	$0.3$	$0.4$	$-0.2$	$0.1$	$0.3$
(e)	$\frac{1}{14}$	$\frac{2}{14}$	$\frac{3}{14}$	$\frac{4}{14}$	$\frac{5}{14}$	$\frac{6}{14}$	$\frac{15}{14}$

Solution. Test both axioms on each row.

(a) Valid. All entries are non-negative, and

$$0.1 + 0.01 + 0.05 + 0.03 + 0.01 + 0.2 + 0.6 = 1$$

(b) Valid. Seven equal entries of $\tfrac17$ sum to $1$, and none is negative.

(c) Not valid. All entries are non-negative, but

$$0.1 + 0.2 + \cdots + 0.7 = 2.8 \ne 1$$

(d) Not valid. Two entries, $-0.1$ and $-0.2$, are negative. A probability cannot be less than zero, so the first axiom fails outright.

(e) Not valid. The last entry $\tfrac{15}{14} > 1$, and the total is $\tfrac{36}{14} \ne 1$. Either failure is enough.

Question 2

A coin is tossed twice. What is the probability that at least one tail occurs?

Solution. $\mathrm S = \{\mathrm{HH, HT, TH, TT}\}$, four equally likely outcomes.

“At least one tail” excludes only $\mathrm{HH}$, so the event has three outcomes:

$$\mathrm P = \frac34$$

(Equivalently, $1 – \mathrm P(\text{no tail}) = 1 – \tfrac14 = \tfrac34$.)

Question 3

A die is thrown. Find the probability of the following events:

(i) a prime number will appear   (ii) a number greater than or equal to $3$ will appear (iii) a number less than or equal to one will appear   (iv) a number more than $6$ will appear (v) a number less than $6$ will appear

Solution. $\mathrm S = \{1,2,3,4,5,6\}$.

(i) Primes: $\{2, 3, 5\}$, so $\mathrm P = \tfrac36 = \tfrac12$.

(ii) $\{3,4,5,6\}$, so $\mathrm P = \tfrac46 = \tfrac23$.

(iii) “$\le 1$” is just $\{1\}$, so $\mathrm P = \tfrac16$.

(iv) Impossible, so $\mathrm P = 0$.

(v) $\{1,2,3,4,5\}$, so $\mathrm P = \tfrac56$.

Question 4

A card is selected from a pack of $52$ cards.

(a) How many points are there in the sample space? (b) Calculate the probability that the card is an ace of spades. (c) Calculate the probability that the card is (i) an ace, (ii) a black card.

Solution.

(a) One card from $52$, so the sample space has $52$ points.

(b) There is exactly one ace of spades:

$$\mathrm P = \frac{1}{52}$$

(c)(i) Four aces, one per suit:

$$\mathrm P = \frac{4}{52} = \frac{1}{13}$$

(c)(ii) Two black suits, clubs and spades, of $13$ cards each:

$$\mathrm P = \frac{26}{52} = \frac12$$

Question 5

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 (i) $3$, (ii) $12$.

Solution. The coin contributes $1$ or $6$; the die contributes $1$ to $6$. So there are $2 \times 6 = 12$ equally likely outcomes.

(i) Sum $3$: the coin must give $1$ and the die $2$ — one outcome, since $6$ on the coin already exceeds $3$:

$$\mathrm P = \frac{1}{12}$$

(ii) Sum $12$: the coin must give $6$ and the die $6$ — again one outcome:

$$\mathrm P = \frac{1}{12}$$

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?

Solution. Ten members in all, six of them women:

$$\mathrm P = \frac{6}{10} = \frac35$$

Question 7

A fair coin is tossed four times, and a person wins Re $1$ for each head and loses 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.

Solution. With $h$ heads and $4 – h$ tails, the net amount is

$$h(1) – (4-h)(1.5) = 2.5h – 6$$

so the amount depends only on the number of heads, and there are five possible values.

Heads $h$	Amount	Number of outcomes	Probability
$4$	$+4.00$	$^4\mathrm C_4 = 1$	$\dfrac{1}{16}$
$3$	$+1.50$	$^4\mathrm C_3 = 4$	$\dfrac{4}{16} = \dfrac14$
$2$	$-1.00$	$^4\mathrm C_2 = 6$	$\dfrac{6}{16} = \dfrac38$
$1$	$-3.50$	$^4\mathrm C_1 = 4$	$\dfrac{4}{16} = \dfrac14$
$0$	$-6.00$	$^4\mathrm C_0 = 1$	$\dfrac{1}{16}$

The probabilities sum to $1$ ✓

Question 8

Three coins are tossed once. Find the probability of getting

(i) $3$ heads   (ii) $2$ heads   (iii) at least $2$ heads   (iv) at most $2$ heads (v) no head   (vi) $3$ tails   (vii) exactly two tails   (viii) no tail   (ix) at most two tails

Solution. Eight equally likely outcomes; the number with exactly $k$ heads is $^3\mathrm C_k$, giving counts $1, 3, 3, 1$ for $k = 0, 1, 2, 3$.

(i) Exactly $3$ heads: $\tfrac18$

(ii) Exactly $2$ heads: $\tfrac38$

(iii) At least $2$ heads means $2$ or $3$: $\tfrac{3+1}{8} = \tfrac12$

(iv) At most $2$ heads is the complement of “$3$ heads”: $1 – \tfrac18 = \tfrac78$

(v) No head is the same as $3$ tails: $\tfrac18$

(vi) $3$ tails: $\tfrac18$

(vii) Exactly two tails is the same as exactly one head: $\tfrac38$

(viii) No tail is the same as $3$ heads: $\tfrac18$

(ix) At most two tails is the complement of “$3$ tails”: $1 – \tfrac18 = \tfrac78$

Question 9

If $\dfrac{2}{11}$ is the probability of an event $\mathrm A$, what is the probability of the event “not $\mathrm A$”?

Solution.

$$\mathrm P(\mathrm A’) = 1 – \mathrm P(\mathrm A) = 1 – \frac{2}{11} = \frac{9}{11}$$

Question 10

A letter is chosen at random from the word ‘ASSASSINATION’. Find the probability that the letter is (i) a vowel, (ii) a consonant.

Solution. Count the letters: ASSASSINATION has

$$\mathrm A(3),\ \mathrm S(4),\ \mathrm I(2),\ \mathrm N(2),\ \mathrm T(1),\ \mathrm O(1)$$

a total of $13$ letters. The vowels are the A’s, I’s and O:

$$3 + 2 + 1 = 6 \text{ vowels}, \qquad 13 – 6 = 7 \text{ consonants}$$

Note the sample space is the thirteen letter positions, not the six distinct letters — a repeated letter is more likely to be picked.

Question 11

In a lottery, a person chooses six different natural numbers at random from $1$ to $20$, and if these six numbers match the six numbers already fixed by the lottery committee, they win the prize. What is the probability of winning the prize in the game?

Solution. Order does not matter, so the number of possible selections is

$$^{20}\mathrm C_6 = \frac{20 \times 19 \times 18 \times 17 \times 16 \times 15}{720} = 38760$$

Exactly one of these wins:

$$\mathrm P = \frac{1}{38760}$$

Question 12

Check whether the following probabilities $\mathrm P(\mathrm A)$ and $\mathrm P(\mathrm B)$ are consistently defined:

(i) $\mathrm P(\mathrm A) = 0.5$, $\mathrm P(\mathrm B) = 0.7$, $\mathrm P(\mathrm A \cap \mathrm B) = 0.6$ (ii) $\mathrm P(\mathrm A) = 0.5$, $\mathrm P(\mathrm B) = 0.4$, $\mathrm P(\mathrm A \cup \mathrm B) = 0.8$

Solution.

(i) Not consistent. Since $\mathrm A \cap \mathrm B \subset \mathrm A$, we must have $\mathrm P(\mathrm A \cap \mathrm B) \le \mathrm P(\mathrm A)$. But $0.6 > 0.5$, so no such events exist.

(ii) Consistent. From the addition rule,

$$\mathrm P(\mathrm A \cap \mathrm B) = \mathrm P(\mathrm A) + \mathrm P(\mathrm B) – \mathrm P(\mathrm A \cup \mathrm B) = 0.5 + 0.4 – 0.8 = 0.1$$

This is non-negative and at most both $0.5$ and $0.4$, so everything is consistent.

Question 13

Fill in the blanks in the following table:

	$\mathrm P(\mathrm A)$	$\mathrm P(\mathrm B)$	$\mathrm P(\mathrm A \cap \mathrm B)$	$\mathrm P(\mathrm A \cup \mathrm B)$
(i)	$\frac13$	$\frac15$	$\frac{1}{15}$	$\ldots$
(ii)	$0.35$	$\ldots$	$0.25$	$0.6$
(iii)	$0.5$	$0.35$	$\ldots$	$0.7$

Solution. Every row is the addition rule with a different unknown.

(i)

$$\mathrm P(\mathrm A \cup \mathrm B) = \frac13 + \frac15 – \frac{1}{15} = \frac{5 + 3 – 1}{15} = \frac{7}{15}$$

(ii) Rearranged for $\mathrm P(\mathrm B)$:

$$\mathrm P(\mathrm B) = \mathrm P(\mathrm A \cup \mathrm B) + \mathrm P(\mathrm A \cap \mathrm B) – \mathrm P(\mathrm A) = 0.6 + 0.25 – 0.35 = 0.5$$

(iii) Rearranged for the intersection:

$$\mathrm P(\mathrm A \cap \mathrm B) = 0.5 + 0.35 – 0.7 = 0.15$$

Question 14

Given $\mathrm P(\mathrm A) = \dfrac35$ and $\mathrm P(\mathrm B) = \dfrac15$, find $\mathrm P(\mathrm A$ or $\mathrm B)$ if $\mathrm A$ and $\mathrm B$ are mutually exclusive events.

Solution. Mutually exclusive means $\mathrm P(\mathrm A \cap \mathrm B) = 0$, so the addition rule loses its last term:

$$\mathrm P(\mathrm A \cup \mathrm B) = \frac35 + \frac15 = \frac45$$

Question 15

If $\mathrm E$ and $\mathrm F$ are events such that $\mathrm P(\mathrm E) = \dfrac14$, $\mathrm P(\mathrm F) = \dfrac12$ and $\mathrm P(\mathrm E$ and $\mathrm F) = \dfrac18$, find

(i) $\mathrm P(\mathrm E$ or $\mathrm F)$    (ii) $\mathrm P($not $\mathrm E$ and not $\mathrm F)$

Solution.

(i)

$$\mathrm P(\mathrm E \cup \mathrm F) = \frac14 + \frac12 – \frac18 = \frac{2 + 4 – 1}{8} = \frac58$$

(ii) “Not $\mathrm E$ and not $\mathrm F$” is $\mathrm E’ \cap \mathrm F’$, which by De Morgan’s law is $(\mathrm E \cup \mathrm F)’$:

$$\mathrm P\left(\mathrm E’ \cap \mathrm F’\right) = 1 – \mathrm P(\mathrm E \cup \mathrm F) = 1 – \frac58 = \frac38$$

Recognising the De Morgan step is what turns part (ii) into one line.

Question 16

Events $\mathrm E$ and $\mathrm F$ are such that $\mathrm P($not $\mathrm E$ or not $\mathrm F) = 0.25$. State whether $\mathrm E$ and $\mathrm F$ are mutually exclusive.

Solution. “Not $\mathrm E$ or not $\mathrm F$” is $\mathrm E’ \cup \mathrm F’ = (\mathrm E \cap \mathrm F)’$ by De Morgan. So

$$\mathrm P\left((\mathrm E \cap \mathrm F)’\right) = 0.25 \quad\Longrightarrow\quad \mathrm P(\mathrm E \cap \mathrm F) = 1 – 0.25 = 0.75$$

Since $\mathrm P(\mathrm E \cap \mathrm F) = 0.75 \ne 0$, the events can occur together and are therefore not mutually exclusive.

Question 17

$\mathrm A$ and $\mathrm B$ are events such that $\mathrm P(\mathrm A) = 0.42$, $\mathrm P(\mathrm B) = 0.48$ and $\mathrm P(\mathrm A$ and $\mathrm B) = 0.16$. Determine

(i) $\mathrm P($not $\mathrm A)$   (ii) $\mathrm P($not $\mathrm B)$   (iii) $\mathrm P(\mathrm A$ or $\mathrm B)$

Solution.

$$\mathrm P(\mathrm A’) = 1 – 0.42 = 0.58$$ $$\mathrm P(\mathrm B’) = 1 – 0.48 = 0.52$$ $$\mathrm P(\mathrm A \cup \mathrm B) = 0.42 + 0.48 – 0.16 = 0.74$$

Question 18

In Class XI of a school, $40\%$ of the students study Mathematics and $30\%$ study Biology. $10\%$ of the class study both Mathematics and Biology. If a student is selected at random from the class, find the probability that they will be studying Mathematics or Biology.

Solution. Let $\mathrm M$ and $\mathrm B$ be the two events:

$$\mathrm P(\mathrm M) = 0.4, \qquad \mathrm P(\mathrm B) = 0.3, \qquad \mathrm P(\mathrm M \cap \mathrm B) = 0.1$$

$$\mathrm P(\mathrm M \cup \mathrm B) = 0.4 + 0.3 – 0.1 = 0.6$$

Subtracting the overlap is essential — the $10\%$ studying both are counted in each of the first two percentages.

Question 19

In an entrance test 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 is $0.7$. The probability of passing at least one of them is $0.95$. What is the probability of passing both?

Solution. “At least one” is the union, so rearrange the addition rule for the intersection:

$$\mathrm P(\mathrm A \cap \mathrm B) = \mathrm P(\mathrm A) + \mathrm P(\mathrm B) – \mathrm P(\mathrm A \cup \mathrm B) = 0.8 + 0.7 – 0.95 = 0.55$$

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?

Solution. “Passing neither” is $(\mathrm E \cup \mathrm H)’$, so

$$\mathrm P(\mathrm E \cup \mathrm H) = 1 – 0.1 = 0.9$$

Now the addition rule, solved for $\mathrm P(\mathrm H)$:

$$\mathrm P(\mathrm H) = \mathrm P(\mathrm E \cup \mathrm H) + \mathrm P(\mathrm E \cap \mathrm H) – \mathrm P(\mathrm E) = 0.9 + 0.5 – 0.75 = 0.65$$

Question 21

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

(i) the student opted for NCC or NSS; (ii) the student has opted neither NCC nor NSS; (iii) the student has opted NSS but not NCC.

Solution. With $\mathrm A$ = “opted NCC” and $\mathrm B$ = “opted NSS”:

$$\mathrm P(\mathrm A) = \frac{30}{60}, \qquad \mathrm P(\mathrm B) = \frac{32}{60}, \qquad \mathrm P(\mathrm A \cap \mathrm B) = \frac{24}{60}$$

(i)

$$\mathrm P(\mathrm A \cup \mathrm B) = \frac{30 + 32 – 24}{60} = \frac{38}{60} = \frac{19}{30}$$

(ii) “Neither” is the complement of “either”:

$$1 – \frac{19}{30} = \frac{11}{30}$$

(iii) “NSS but not NCC” is $\mathrm B – \mathrm A$, which contains the students in $\mathrm B$ minus those also in $\mathrm A$:

$$\mathrm P(\mathrm B) – \mathrm P(\mathrm A \cap \mathrm B) = \frac{32 – 24}{60} = \frac{8}{60} = \frac{2}{15}$$

Common mistakes

• Question 1, checking only the sum. Row (d) sums to $1.0$ but contains negative entries, so it still fails. Both axioms must be tested.
• Question 5, taking the coin as contributing $1$ to $6$. It shows only $1$ or $6$, so the sample space has $12$ points, not $36$.
• Question 7, listing sixteen different amounts. The amount depends only on how many heads appeared, so there are five distinct values with multiplicities $1, 4, 6, 4, 1$.
• Question 10, taking the sample space as the six distinct letters. The letter is chosen from the word, so all thirteen positions are equally likely — S is four times as likely as O.
• Question 11, using $^{20}\mathrm P_6$. The hint says order does not matter, so it is a combination.
• Questions 15(ii) and 16, expanding “not E and not F” directly. De Morgan’s laws turn both into complements of a single event, after which $\mathrm P(\mathrm X’) = 1 – \mathrm P(\mathrm X)$ finishes the job.
• Questions 18 and 21, adding the two figures without subtracting the overlap. $0.4 + 0.3 = 0.7$ double-counts the students doing both subjects.
• Question 21(iii), computing $\mathrm P(\mathrm B) – \mathrm P(\mathrm A)$. “NSS but not NCC” removes only the students in both, giving $\mathrm P(\mathrm B) – \mathrm P(\mathrm A \cap \mathrm B)$.

Practise next

• Miscellaneous Exercise on Chapter 14 — probability with combinations, where counting the sample space becomes the harder half of the problem.
• Exercise 6.4 — worth revisiting alongside question 11, since every “choose $r$ from $n$” probability rests on $^n\mathrm C_r$.