Let the sample space for a single throw of a fair die be \[ S=\{1,2,3,4,5,6\} \]

Given \[ E=\{\text{die shows }4\}=\{4\} \] and \[ F=\{\text{die shows an even number}\}=\{2,4,6\} \]

To check whether two events are mutually exclusive, we examine their intersection. If \[ E\cap F=\varnothing \] then the events are mutually exclusive

Here, \[ E\cap F=\{4\} \]

Since the intersection is not empty and the outcome 4 is common to both events, the occurrence of event \(E\) automatically implies the occurrence of event \(F\)

Therefore, \(E\) and \(F\) are not mutually exclusive because they can occur simultaneously

For a single throw of a fair die, the sample space is \[ S=\{1,2,3,4,5,6\} \]

Event \(A\) represents numbers less than 7, so \[ A=\{1,2,3,4,5,6\} \] Event \(B\) represents numbers greater than 7, which is impossible on a die, hence \[ B=\varnothing \] Event \(C\) represents multiples of 3, therefore \[ C=\{3,6\} \] Event \(D\) represents numbers less than 4, so \[ D=\{1,2,3\} \] Event \(E\) represents even numbers greater than 4, giving \[ E=\{6\} \] Event \(F\) represents numbers not less than 3, hence \[ F=\{3,4,5,6\} \]

Now, using set operations,

\[ \begin{aligned} A\cup B&=\{1,2,3,4,5,6\} \\ A\cap B&=\varnothing \\ B\cup C&=\{3,6\} \\ E\cap F&=\{6\} \\ D\cap E&=\varnothing \\ A-C&=\{1,2,4,5\} \\ D-E&=\{1,2,3\} \\ F'&=\{1,2\} \\ E\cap F'&=\varnothing \end{aligned} \]

Here, \(F'\) denotes the complement of \(F\) with respect to the sample space \(S\). Each result follows directly from the definitions of union, intersection, difference, and complement of sets.

Let the outcome of throwing two dice be written as an ordered pair \((x,y)\), where \(x\) is the number on the first die and \(y\) on the second die.

For event \(A\), the sum is greater than 8, that is \(x+y\ge 9\). Hence,

\[ \begin{aligned} A=\{&(3,6),(4,5),(5,4),(6,3),(4,6),(5,5),(6,4),(5,6),(6,5),(6,6)\} \end{aligned} \]

For event \(B\), the number 2 appears on either die. Therefore,

\[ \begin{aligned} B=\{&(1,2),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,2),(4,2),(5,2),(6,2)\} \end{aligned} \]

For event \(C\), the sum is at least 7 and also a multiple of 3. The possible sums satisfying these conditions are 9 and 12. Hence,

\[ \begin{aligned} C=\{&(3,6),(4,5),(5,4),(6,3),(6,6)\} \end{aligned} \]

Observing the sets, there is no common ordered pair between \(A\) and \(B\), so \(A\cap B=\varnothing\). Also, none of the outcomes in \(C\) contains a 2, so \(B\cap C=\varnothing\). Therefore, \(A\) and \(B\), as well as \(B\) and \(C\), are mutually exclusive events.

Let the outcomes of tossing three coins be written as ordered triples using \(H\) for head and \(T\) for tail.

Event \(A\) represents three heads, so

\[ \begin{aligned} A&=\{(H,H,H)\} \\ B&=\{(H,H,T),(H,T,H),(T,H,H)\} \\ C&=\{(T,T,T)\} \\ D&=\{(H,H,H),(H,H,T),(H,T,H),(H,T,T)\} \end{aligned} \]

Here, \(A\) contains only the outcome with all heads, \(B\) contains all outcomes with exactly two heads and one tail, \(C\) contains only the outcome with all tails, and \(D\) contains all outcomes in which the first coin shows head.

For mutual exclusiveness, two events must have no common outcome. From the sets above, \(A\cap B=\varnothing\), \(A\cap C=\varnothing\), \(B\cap C=\varnothing\), and \(C\cap D=\varnothing\). Hence, the mutually exclusive pairs are \(A\) and \(B\), \(A\) and \(C\), \(B\) and \(C\), and \(C\) and \(D\).

A simple event contains only one outcome. Therefore, \(A\) and \(C\) are simple events.

A compound event contains more than one outcome. Hence, \(B\) and \(D\) are compound events.

Let the sample space for tossing three coins be \[ S=\{HHH,HHT,HTH,THH,HTT,THT,TTH,TTT\} \]

(i) Two mutually exclusive events can be taken as “getting at least two heads” and “getting at least two tails”. These events cannot occur together because an outcome cannot simultaneously contain at least two heads and at least two tails when only three coins are tossed.

(ii) Three events which are mutually exclusive and exhaustive may be chosen as “getting no head”, “getting exactly one head”, and “getting at least two heads”. These three events have no common outcomes among themselves and together they cover the entire sample space.

(iii) Two events which are not mutually exclusive can be “getting at least one head” and “getting exactly two heads”. These are not mutually exclusive because outcomes with exactly two heads are included in both events.

(iv) Two events which are mutually exclusive but not exhaustive can be “getting three heads” and “getting three tails”. They are mutually exclusive since they have no common outcome, but they are not exhaustive because several outcomes such as HHT or HTT are left out.

(v) Three events which are mutually exclusive but not exhaustive can be “getting three heads”, “getting exactly two heads”, and “getting three tails”. These events are mutually exclusive, but they are not exhaustive since outcomes with exactly one head are not included.

Let the sample space be all ordered pairs \((x,y)\) where \(x\) is the score on the first die and \(y\) on the second die.

Event \(A\) denotes getting an even number on the first die, event \(B\) denotes getting an odd number on the first die, and event \(C\) denotes getting a total not exceeding 5.

\[ \begin{aligned} A=\{&(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),\\ &(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),\\ &(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\} \end{aligned} \]

\[ \begin{aligned} B=\{&(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),\\ &(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),\\ &(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)\} \end{aligned} \]

\[ \begin{aligned} C=\{&(1,1),(1,2),(1,3),(1,4),\\&(2,1),(2,2),(2,3),\\&(3,1),(3,2),(4,1)\} \end{aligned} \]

Since the first die must be either even or odd, \(A'=B\) and \(B'=A\).

\[ A\cup B=S \qquad A\cap B=\varnothing \]

Now,

\[ \begin{aligned} A-C=\{&(2,4),(2,5),(2,6),\\&(4,2),(4,3),(4,4),(4,5),(4,6),\\ &(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\}\\\\ B\cup C=\{&(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),\\&(2,1),(2,2),(2,3),\\ &(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),\\&(4,1),\\ &(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)\}\\\\ B\cap C=\{&(1,1),(1,2),(1,3),(1,4),\\&(3,1),(3,2)\} \end{aligned} \]

Finally, \(A\cap B'\cap C'\) represents outcomes where the first die is even, not odd, and the sum exceeds 5. Since \(B'=A\), this reduces to \(A\cap C'\).

\[ \begin{aligned} A\cap B'\cap C'=\{&(2,4),(2,5),(2,6),\\&(4,2),(4,3),(4,4),(4,5),(4,6),\\ &(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\} \end{aligned} \]

Each result follows directly from the definitions of complement, union, intersection, and difference of events.

(i) True. Events \(A\) and \(B\) are mutually exclusive because the first die cannot be even and odd simultaneously. Hence, \[ A\cap B=\varnothing \]

(ii) True. From Question 6, we have \(A\cap B=\varnothing\) and \[ A\cup B=S \] Therefore, \(A\) and \(B\) are both mutually exclusive and exhaustive.

(iii) True. Every outcome has the first die either even or odd. Hence the complement of \(B\) consists exactly of outcomes where the first die is even. Therefore, \[ A=B' \]

(iv) False. There are common outcomes in \(A\) and \(C\), such as \((2,1)\) and \((2,2)\). Thus, \[ A\cap C\neq\varnothing \] So \(A\) and \(C\) are not mutually exclusive.

(v) False. Since \(B'=A\), the events \(A\) and \(B'\) are identical. Hence their intersection is not empty, and they are not mutually exclusive.

(vi) False. Here \(A'=B\). Although \(A'\) and \(B'\) are mutually exclusive, the three events \(A'\), \(B'\), and \(C\) are not mutually exclusive because \[ B'\cap C\neq\varnothing \] Also, their union does not cover the entire sample space. Therefore, they are neither mutually exclusive nor exhaustive.