NCERT Class 11 Mathematics — Probability, Exercise 14.1. All 7 questions solved.

An event is a subset of the sample space $\mathrm S$, so every set operation from Chapter 1 carries over with a probabilistic reading:

An event with exactly one outcome is simple; one with more than one is compound.

Key insight. “Mutually exclusive” and “exhaustive” are separate conditions and a set of events can satisfy either without the other. Mutually exclusive means the events cannot overlap; exhaustive means between them they cover everything. Question 5 asks for pairs and triples of each combination, which only makes sense once the two ideas are kept apart.

Question 1

A die is rolled. Let $\mathrm E$ be the event “die shows $4$” and $\mathrm F$ be the event “die shows an even number”. Are $\mathrm E$ and $\mathrm F$ mutually exclusive?

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

$$\mathrm E = \{4\}, \qquad \mathrm F = \{2, 4, 6\}$$

$$\mathrm E \cap \mathrm F = \{4\} \ne \varphi$$

Since the intersection is non-empty, the events are not mutually exclusive. In fact $\mathrm E \subset \mathrm F$: whenever the die shows $4$ it necessarily shows an even number.

Question 2

A die is thrown. Describe the following events:

(i) $\mathrm A$: a number less than $7$   (ii) $\mathrm B$: a number greater than $7$ (iii) $\mathrm C$: a multiple of $3$   (iv) $\mathrm D$: a number less than $4$ (v) $\mathrm E$: an even number greater than $4$   (vi) $\mathrm F$: a number not less than $3$

Also find $\mathrm A \cup \mathrm B$, $\mathrm A \cap \mathrm B$, $\mathrm B \cup \mathrm C$, $\mathrm E \cap \mathrm F$, $\mathrm D \cap \mathrm E$, $\mathrm A – \mathrm C$, $\mathrm D – \mathrm E$, $\mathrm E \cap \mathrm F’$, $\mathrm F’$.

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

(i) Every outcome is less than $7$, so $\mathrm A = \{1,2,3,4,5,6\} = \mathrm S$ — a sure event.

(ii) No outcome exceeds $7$, so $\mathrm B = \varphi$ — an impossible event.

(iii) $\mathrm C = \{3, 6\}$

(iv) $\mathrm D = \{1, 2, 3\}$

(v) Even numbers greater than $4$: only $6$, so $\mathrm E = \{6\}$.

(vi) “Not less than $3$” means $\ge 3$, so $\mathrm F = \{3, 4, 5, 6\}$.

Now the combinations:

$$\mathrm A \cup \mathrm B = \mathrm S \cup \varphi = \{1,2,3,4,5,6\}$$ $$\mathrm A \cap \mathrm B = \varphi$$ $$\mathrm B \cup \mathrm C = \varphi \cup \{3,6\} = \{3, 6\}$$ $$\mathrm E \cap \mathrm F = \{6\} \cap \{3,4,5,6\} = \{6\}$$ $$\mathrm D \cap \mathrm E = \{1,2,3\} \cap \{6\} = \varphi$$ $$\mathrm A – \mathrm C = \{1,2,3,4,5,6\} – \{3,6\} = \{1, 2, 4, 5\}$$ $$\mathrm D – \mathrm E = \{1,2,3\} – \{6\} = \{1, 2, 3\}$$ $$\mathrm F’ = \mathrm S – \mathrm F = \{1, 2\}$$ $$\mathrm E \cap \mathrm F’ = \{6\} \cap \{1,2\} = \varphi$$

Question 3

An experiment involves rolling a pair of dice and recording the numbers that come up. Describe the following events:

$\mathrm A$: the sum is greater than $8$;   $\mathrm B$: $2$ occurs on either die; $\mathrm C$: the sum is at least $7$ and a multiple of $3$.

Which pairs of these events are mutually exclusive?

Solution. The sample space has $36$ ordered pairs.

$\mathrm A$ — sum greater than $8$, so the sum is $9$, $10$, $11$ or $12$:

$$\mathrm A = \{(3,6), (4,5), (5,4), (6,3), (4,6), (5,5), (6,4), (5,6), (6,5), (6,6)\}$$

$\mathrm B$ — a $2$ on either die. Take every pair with $2$ in the first place or the second, counting $(2,2)$ once:

$$\mathrm B = \{(1,2), (2,2), (3,2), (4,2), (5,2), (6,2), (2,1), (2,3), (2,4), (2,5), (2,6)\}$$

$\mathrm C$ — sum at least $7$ and a multiple of $3$. Multiples of $3$ that are at least $7$ are $9$ and $12$:

$$\mathrm C = \{(3,6), (6,3), (4,5), (5,4), (6,6)\}$$

Mutual exclusivity.

• $\mathrm A \cap \mathrm B$: every element of $\mathrm A$ has both entries at least $3$, so none contains a $2$. Hence $\mathrm A \cap \mathrm B = \varphi$ — mutually exclusive.
• $\mathrm B \cap \mathrm C$: the same argument applies to $\mathrm C$, all of whose entries are at least $3$. So $\mathrm B \cap \mathrm C = \varphi$ — mutually exclusive.
• $\mathrm A \cap \mathrm C = \mathrm C \ne \varphi$ (every element of $\mathrm C$ has sum $9$ or $12$, both greater than $8$), so these are not mutually exclusive. Indeed $\mathrm C \subset \mathrm A$.

Question 4

Three coins are tossed once. Let $\mathrm A$ denote the event “three heads show”, $\mathrm B$ “two heads and one tail show”, $\mathrm C$ “three tails show” and $\mathrm D$ “a head shows on the first coin”. Which events are

(i) mutually exclusive?   (ii) simple?   (iii) compound?

Solution. The sample space is

$$\mathrm S = \{\mathrm{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}\}$$

$$\mathrm A = \{\mathrm{HHH}\}, \quad \mathrm B = \{\mathrm{HHT, HTH, THH}\}, \quad \mathrm C = \{\mathrm{TTT}\}$$

$$\mathrm D = \{\mathrm{HHH, HHT, HTH, HTT}\}$$

(i) Mutually exclusive pairs. Check each of the six pairs:

• $\mathrm A \cap \mathrm B = \varphi$ ✓
• $\mathrm A \cap \mathrm C = \varphi$ ✓
• $\mathrm B \cap \mathrm C = \varphi$ ✓
• $\mathrm C \cap \mathrm D = \varphi$ ✓
• $\mathrm A \cap \mathrm D = \{\mathrm{HHH}\} \ne \varphi$ ✗
• $\mathrm B \cap \mathrm D = \{\mathrm{HHT, HTH}\} \ne \varphi$ ✗

(ii) Simple events contain exactly one outcome: $\mathrm A$ and $\mathrm C$.

(iii) Compound events contain more than one: $\mathrm B$ and $\mathrm D$.

Question 5

Three coins are tossed. Describe

(i) two events which are mutually exclusive; (ii) three events which are mutually exclusive and exhaustive; (iii) two events which are not mutually exclusive; (iv) two events which are mutually exclusive but not exhaustive; (v) three events which are mutually exclusive but not exhaustive.

Solution. The sample space is the eight outcomes listed in question 4. More than one answer is correct in each part; the following are the simplest.

(i) “Getting at least two heads” $= \{\mathrm{HHH, HHT, HTH, THH}\}$ and “getting at least two tails” $= \{\mathrm{TTT, TTH, THT, HTT}\}$. These cannot both happen, since three coins cannot give two heads and two tails.

(ii) Partition by the number of heads, grouping the top two together:

• “Getting no heads” $= \{\mathrm{TTT}\}$
• “Getting exactly one head” $= \{\mathrm{HTT, THT, TTH}\}$
• “Getting at least two heads” $= \{\mathrm{HHT, HTH, THH, HHH}\}$

They are pairwise disjoint and their union is all eight outcomes, so they are mutually exclusive and exhaustive.

(iii) “Getting at most two tails” and “getting exactly two tails” overlap — the second is contained in the first.

(iv) “Getting exactly one head” and “getting exactly two heads” are disjoint, but together they miss $\mathrm{HHH}$ and $\mathrm{TTT}$, so they are not exhaustive.

(v) “Getting exactly one tail”, “getting exactly two tails” and “getting exactly three tails” are pairwise disjoint, but they miss $\mathrm{HHH}$ (no tails), so they are not exhaustive.

Question 6

Two dice are thrown. The events $\mathrm A$, $\mathrm B$ and $\mathrm C$ are as follows:

$\mathrm A$: getting an even number on the first die. $\mathrm B$: getting an odd number on the first die. $\mathrm C$: getting the sum of the numbers on the dice $\le 5$.

Describe the events (i) $\mathrm A’$ (ii) not $\mathrm B$ (iii) $\mathrm A$ or $\mathrm B$ (iv) $\mathrm A$ and $\mathrm B$ (v) $\mathrm A$ but not $\mathrm C$ (vi) $\mathrm B$ or $\mathrm C$ (vii) $\mathrm B$ and $\mathrm C$ (viii) $\mathrm A \cap \mathrm B’ \cap \mathrm C’$.

Solution. The first die is either even or odd, never both and always one, so $\mathrm A$ and $\mathrm B$ partition the sample space — that single observation answers the first four parts.

$$\mathrm A = \{(2,y), (4,y), (6,y) : y = 1,\ldots,6\} \quad (18 \text{ outcomes})$$ $$\mathrm B = \{(1,y), (3,y), (5,y) : y = 1,\ldots,6\} \quad (18 \text{ outcomes})$$ $$\mathrm C = \{(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(3,1),(3,2),(4,1)\}$$

(i) $\mathrm A’ = \mathrm B$ — the complement of “even first” is “odd first”.

(ii) not $\mathrm B = \mathrm B’ = \mathrm A$.

(iii) $\mathrm A \cup \mathrm B = \mathrm S$, all $36$ outcomes.

(iv) $\mathrm A \cap \mathrm B = \varphi$.

(v) $\mathrm A – \mathrm C$: take $\mathrm A$’s eighteen outcomes and drop those with sum at most $5$, namely $(2,1)$, $(2,2)$, $(2,3)$ and $(4,1)$:

$$\{(2,4),(2,5),(2,6),(4,2),(4,3),(4,4),(4,5),(4,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\}$$

(vi) $\mathrm B \cup \mathrm C$: all eighteen of $\mathrm B$, plus the members of $\mathrm C$ with an even first entry — $(2,1), (2,2), (2,3), (4,1)$:

$$\{(1,1),\ldots,(1,6),(3,1),\ldots,(3,6),(5,1),\ldots,(5,6),(2,1),(2,2),(2,3),(4,1)\}$$

(vii) $\mathrm B \cap \mathrm C$: members of $\mathrm C$ with odd first entry:

$$\{(1,1),(1,2),(1,3),(1,4),(3,1),(3,2)\}$$

(viii) Since $\mathrm B’ = \mathrm A$, we have $\mathrm A \cap \mathrm B’ \cap \mathrm C’ = \mathrm A \cap \mathrm C’ = \mathrm A – \mathrm C$ — exactly the answer to part (v).

Question 7

Refer to question 6 above. State true or false, giving a reason:

(i) $\mathrm A$ and $\mathrm B$ are mutually exclusive (ii) $\mathrm A$ and $\mathrm B$ are mutually exclusive and exhaustive (iii) $\mathrm A = \mathrm B’$ (iv) $\mathrm A$ and $\mathrm C$ are mutually exclusive (v) $\mathrm A$ and $\mathrm B’$ are mutually exclusive (vi) $\mathrm A’$, $\mathrm B’$, $\mathrm C$ are mutually exclusive and exhaustive

Solution.

(i) True. The first die cannot be both even and odd, so $\mathrm A \cap \mathrm B = \varphi$.

(ii) True. In addition to (i), every outcome has a first die that is even or odd, so $\mathrm A \cup \mathrm B = \mathrm S$.

(iii) True. Given (i) and (ii), $\mathrm B$ is precisely the complement of $\mathrm A$.

(iv) False. $\mathrm A \cap \mathrm C = \{(2,1),(2,2),(2,3),(4,1)\} \ne \varphi$ — an even first die can still give a small sum.

(v) False. $\mathrm B’ = \mathrm A$, so $\mathrm A \cap \mathrm B’ = \mathrm A \cap \mathrm A = \mathrm A \ne \varphi$. An event is never mutually exclusive with itself unless it is impossible.

(vi) False. $\mathrm A’ = \mathrm B$ and $\mathrm B’ = \mathrm A$, and although these two are disjoint, $\mathrm A’ \cap \mathrm C = \mathrm B \cap \mathrm C \ne \varphi$ by question 6(vii). So the three are not pairwise disjoint.

Common mistakes

• Question 1, calling the events mutually exclusive because $\mathrm E$ is smaller. Containment is the opposite of exclusion: $\mathrm E \subset \mathrm F$ means they overlap as much as possible.
• Question 2(ii), writing $\{7, 8, \ldots\}$. A die shows only $1$ to $6$, so “greater than $7$” is impossible and the event is $\varphi$.
• Question 2(vi), reading “not less than $3$” as “less than $3$”. It means $\ge 3$, so the event is $\{3,4,5,6\}$ and its complement is $\{1,2\}$.
• Question 3, listing $\mathrm B$ with $(2,2)$ twice. It qualifies under both “$2$ on the first” and “$2$ on the second”, but a set lists it once — giving $11$ outcomes, not $12$.
• Question 3, reading $\mathrm C$ as “at least $7$ or a multiple of $3$”. Both conditions must hold, so the sums are $9$ and $12$ only.
• Question 4, calling $\mathrm A$ and $\mathrm D$ mutually exclusive. $\mathrm{HHH}$ satisfies both, so they overlap.
• Question 7(v), taking $\mathrm B’$ as something new. $\mathrm B’ = \mathrm A$, so the claim reduces to “$\mathrm A$ is mutually exclusive with itself”, which is false for any non-impossible event.
• Confusing mutually exclusive with exhaustive. In question 5, parts (iv) and (v) are exclusive without being exhaustive — the distinction is the whole point of those parts.

Practise next

• Exercise 14.2 — the axiomatic approach, where probabilities are attached to the events described here.
• Exercise 1.4 — worth revisiting for union, intersection and difference, since every event operation in this exercise is a set operation renamed.