## Chapter 14 Probability EXERCISE 14.1

### EXERCISE 14.1

**1.** A die is rolled. Let $E$ be the event “die shows 4 " and $F$ be the event “die shows even number”. Are $E$ and $F$ mutually exclusive?

## Show Answer

**Answer :**

When a die is rolled, the sample space is given by

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

Accordingly, $E=\{4\}$ and $F=\{2,4,6\}$

It is observed that $E \cap$ F={4}

Therefore, $E$ and $F$ are not mutually exclusive events.

**2.** A die is thrown. Describe the following events:

(i) A: a number less than 7

(ii) $B$ : a number greater than 7

(iii) $C$ : a multiple of 3

(iv) D: a number less than 4

(v) $E$ : an even number greater than 4

(vi) $F$ : a number not less than 3

Also find $A \cup B, A \cap B, B \cup C, E \cap F, D \cap E, A-C, D-E, E \cap F^{\prime}, F^{\prime}$

## Show Answer

**Answer :**

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

(ii) $\phi$

(iii) $\{3,6\}$

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

(v) $\{6\}$

(vi) $\{3,4,5,6\}, \mathrm{A} \cup \mathrm{B}=\{1,2,3,4,5,6\}, \mathrm{A} \cap \mathrm{B}=\phi, \mathrm{B} \cup \mathrm{C}=\{3,6\}, \mathrm{E} \cap \mathrm{F}=\{6\}$, $\mathrm{D} \cap \mathrm{E}=\phi$,

$ \mathrm{A}-\mathrm{C}=\{1,2,4,5\}, \mathrm{D}-\mathrm{E}=\{1,2,3\}, \mathrm{E} \cap \mathrm{F}^{\prime}=\phi, \mathrm{F}^{\prime}=\{1,2\} $

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

A: the sum is greater than 8,

B: 2 occurs on either die

$C$ : the sum is at least 7 and a multiple of 3 .

Which pairs of these events are mutually exclusive?

## Show Answer

**Answer :**

When a pair of dice is rolled, the sample space is given by

S= {(x, y): x, y=1,2,3,4,5,6}

= $ \begin{cases} (1,1), (1,2), (1,3), (1,4), (1,5), (1,6) \\ (2,1), (2,2), (2,3), (2,4), (2,5), (2,6) \\ (3,1), (3,2), (3,3), (3,4), (3,5), (3,6) \\ (4,1), (4,2), (4,3), (4,4), (4,5), (4,6) \\ (5,1), (5,2), (5,3), (5,4), (5,5), (5,6) \\ (6,1), (6,2), (6,3), (6,4), (6,5), (6,6) \end{cases} $

Accordingly,

A={(3,6),(4,5),(4,6),(5,4),(5,5),(5,6),(6,3),(6,4),(6,5),(6,6)}

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

C={(3,6),(4,5),(5,4),(6,3),(6,6)}

It is observed that

$A \cap B=\Phi$ and $C \cap B=\Phi$

$C \cap$ A={(3,6),(4,5),(5,4),(6,3),(6,6)} $\neq \phi$

Hence, events $A$ and $B$ and events $B$ and $C$ are mutually exclusive.

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

(i) mutually exclusive?

(ii) simple?

(iii) Compound?

## Show Answer

**Answer :**

When three coins are tossed, the sample space is given by

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

Accordingly,

A={HHH}

B={HHT, HTH, THH}

C={TTT}

D={HHH, HHT, HTH, HTT}

We now observe that

$ \begin{aligned} & \mathrm{A} \cap \mathrm{B}=\phi \\ & \mathrm{A} \cap \mathrm{C}=\phi \\ & \mathrm{A} \cap \mathrm{D}=\{\mathrm{HHH}\} \neq \phi \\ & \mathrm{B} \cap \mathrm{C}=\phi \\ & \mathrm{B} \cap \mathrm{D}=\{\mathrm{HHT}, \mathrm{HTH}\} \neq \phi \\ & \mathrm{C} \cap \mathrm{D}=\phi \end{aligned} $

(i) Event A and B; event A and C; event B and C; and event C and D are all mutually exclusive.

(ii) If an event has only one sample point of a sample space, it is called a simple event. Thus, $A$ and $D$ are simple events.

(iii) If an event has more than one sample point of a sample space, it is called a compound event. Thus, B and D are compound events.

**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.

## Show Answer

**Answer :**

When three coins are tossed, the sample space is given by

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

(i) Two events that are mutually exclusive can be

A : getting no heads and $B$ : getting no tails

This is because sets A={TTT} and {B}={HHH} are disjoint

(ii) Three events that are mutually exclusive and exhaustive can

$\mathrm{A}:$ getting no heads $\Rightarrow \mathrm{A}=\{\mathrm{TTT}\}$

$\mathrm{B}$ : getting exactly one head {B}={HTT},{THT},{TTH}

$\mathrm{C}$ : getting at least two heads {C}={HHH},{HHT}, {HTH}, {THH}

This is because $\mathrm{A} \cap \mathrm{B}=\mathrm{B} \cap \mathrm{C}=\mathrm{C} \cap \mathrm{A}=\phi$

and $\mathrm{A} \cup \mathrm{B} \cup \mathrm{C}=\mathrm{S}$

(iii) Two events that are not mutually exclusive can be

$A$ : getting three heads $\Rightarrow A=\{\mathrm{HHH}\}$

$B$ : getting at least 2 heads B={HHH}, {HHT}, {HTH}, {THH}

This is because $\mathrm{A} \cap \mathrm{B}=\{\mathrm{HHH}\} \neq \phi$

(iv) Two events which are mutually exclusive but not exhaustive can be

A : getting exactly one head $\Rightarrow \mathrm{A}=\{\mathrm{HTT}, \mathrm{THT}, \mathrm{TTH}\}$

$B$ : getting exactly one tail $\Rightarrow B=\{\mathrm{HHT}, \mathrm{HTH}, \mathrm{THH}\}$

This is because $A \cap B=\phi$ but $A \cup B \neq S$

(v) Three events that are mutually exclusive but not exhaustive can be

A : getting exactly three heads $\Rightarrow \mathrm{A}=\{\mathrm{HHH}\}$

$\mathrm{B}$ : getting one head and two tails $\Rightarrow \mathrm{B}=\{\mathrm{HTT}, \mathrm{THT}, \mathrm{TTH}\}$

$\mathrm{C}$ : getting one tail and two heads $\Rightarrow \mathrm{C}=\{\mathrm{HHT}, \mathrm{HTH}, \mathrm{THH}\}$

This is because $\mathrm{A} \cap \mathrm{B}=\mathrm{B} \cap \mathrm{C}=\mathrm{C} \cap \mathrm{A}=\phi$ but $\mathrm{A} \cup \mathrm{B} \cup \mathrm{C} \neq \mathrm{S}$

**6.** Two dice are thrown. The events A, B and C are as follows:

A: getting an even number on the first die.

B: getting an odd number on the first die.

$C$ : getting the sum of the numbers on the dice $\leq 5$.

Describe the events

(i) $A^{\prime}$

(ii) not B

(iii) $A$ or $B$

(iv) $A$ and $B$

(v) A but not $C$

(vi) $B$ or $C$

(vii) B and C

(viii) $A \cap B^{\prime} \cap C^{\prime}$

## Show Answer

**Answer :**

When two dice are thrown, the sample space is given by

S={(x, y): x, y=1,2,3,4,5,6}

= $ \begin{cases} (1,1), (1,2), (1,3), (1,4), (1,5), (1,6) \\ (2,1), (2,2), (2,3), (2,4), (2,5), (2,6) \\ (3,1), (3,2), (3,3), (3,4), (3,5), (3,6) \\ (4,1), (4,2), (4,3), (4,4), (4,5), (4,6) \\ (5,1), (5,2), (5,3), (5,4), (5,5), (5,6) \\ (6,1), (6,2), (6,3), (6,4), (6,5), (6,6) \end{cases} $

Accordingly,

A: Getting an even number on the first die

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

B: getting an odd number on the first die

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

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

(i) $ A^{\prime}$=$\{ (1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(3,1),(3,2),(3,3), \\ (3,4),(3,5),(3,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6) \}$

(ii)Not $ B= \{ (2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(4,1),(4,2),(4,3), \\ (4,4),(4,5),(4,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\}=A$

(iii)$ A \text{ or } B= A \cup B =$

$ \begin{array}{l} = \\ \{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6) \\ (2,1), (2,2), (2,3), (2,4), (2,5), (2,6) \\ (3,1), (3,2), (3,3), (3,4), (3,5), (3,6) \\ (4,1), (4,2), (4,3), (4,4), (4,5), (4,6) \\ (5,1), (5,2), (5,3), (5,4), (5,5), (5,6) \\ (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\} \\ =\text{S} \end{array} $

(iv) $\mathrm{A} \cap \mathrm{B}=\phi$

(v) $A-C=\{(2,4),(2,5),(2,6),(4,2),(4,3),(4,4),(4,5),(4,6),(6,1),(6,2),(6,3), \\ (6,4),(6,5),(6,6)\}$

(vi)

$ \begin{array}{l} & \mathrm{B} \cup \mathrm{C}=\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(3,1),(3,2), \\ &(3,3),(3,4),(3,5),(3,6),(4,1),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)\} \end{array} $

(vii) $\mathrm{B} \cap \mathrm{C}=\{(1,1),(1,2),(1,3),(1,4),(3,1),(3,2)\}$

(viii) $ \begin{array}{l} {A} \cap {B}^{\prime} \cap {C}^{\prime}=\{(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)\} \end{array} $

**7.** Refer to question 6 above, state true or false: (give reason for your answer)

(i) A and B are mutually exclusive

(ii) $A$ and $B$ are mutually exclusive and exhaustive

(iii) $A=B^{\prime}$

(iv) $A$ and $C$ are mutually exclusive

(v) $A$ and $B^{\prime}$ are mutually exclusive.

(vi) $A^{\prime}, B^{\prime}, C$ are mutually exclusive and exhaustive.

## Show Answer

**Answer :**

A= $ \begin{cases} (2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(4,1),(4,2),(4,3), \\ (4,4),(4,5),(4,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6) \end{cases} $

B= $ \begin{cases} (1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(3,1),(3,2),(3,3), \\ (3,4),(3,5),(3,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6) \end{cases} $

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

(i) It is observed that $A \cap B= \phi $

$\therefore A$ and $B$ are mutually exclusive.

Thus, the given statement is true.

(ii) It is observed that $A \cap B=\phi $ and $A \cup B=S$

$\therefore A$ and $B$ are mutually exclusive and exhaustive.

Thus, the given statement is true.

(iii) It is observed that

$B^{\prime}=$ $ \begin{cases} (2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(4,1),(4,2),(4,3), \\ (4,4),(4,5),(4,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6) \end{cases} $ =A

Thus, the given statement is true.

(iv) It is observed that $A \cap C=\{(2,1),(2,2),(2,3),(4,1)\} = \phi $

$\therefore A$ and $C$ are not mutually exclusive.

Thus, the given statement is false.

(v) $A \cap B^{\prime}=A \cap A=A$

$\therefore A \cap B^{\prime} \neq \phi$

$\therefore A$ and $B^{\prime}$ are not mutually exclusive.

Thus, the given statement is false.

(vi) It is observed that $A^{\prime} \cup B^{\prime} \cup C=S$;

However,

$ B^{\prime} \cap C=\{(2,1),(2,2),(2,3),(4,1)\} \neq \phi $

Therefore, events $A^{\prime}, B^{\prime}$ and $C$ are not mutually exclusive and exhaustive.

Thus, the given statement is false.