Set Based probability Questions in English

(A) The given inequality is $x + \frac{336}{x} \le 50$.
Since $x$ is a positive integer from $1$ to $50$,we can multiply by $x$ without changing the inequality sign:
$x^2 + 336 \le 50x$
$x^2 - 50x + 336 \le 0$
Factoring the quadratic expression,we get:
$(x - 8)(x - 42) \le 0$
This inequality holds for $x \in [8, 42]$.
Since $x$ is an integer,the possible values for $x$ are $\{8, 9, 10, \dots, 42\}$.
The number of such integers is $42 - 8 + 1 = 35$.
The total number of integers from $1$ to $50$ is $50$.
Therefore,the probability is $P = \frac{35}{50} = \frac{7}{10}$.

(A) Let $P(P), P(Q), P(R)$ be the probabilities of hitting the target by $P, Q, R$ respectively.
$P(P) = \frac{3}{4}, P(Q) = \frac{1}{2}, P(R) = \frac{5}{8}$.
The probabilities of missing the target are $P(P') = 1 - \frac{3}{4} = \frac{1}{4}$,$P(Q') = 1 - \frac{1}{2} = \frac{1}{2}$,and $P(R') = 1 - \frac{5}{8} = \frac{3}{8}$.
The event that the target is hit by $P$ or $Q$ but not by $R$ is $(P \cap Q' \cap R') \cup (P' \cap Q \cap R') \cup (P \cap Q \cap R')$.
Required probability $= P(P)P(Q')P(R') + P(P')P(Q)P(R') + P(P)P(Q)P(R')$.
$= (\frac{3}{4} \times \frac{1}{2} \times \frac{3}{8}) + (\frac{1}{4} \times \frac{1}{2} \times \frac{3}{8}) + (\frac{3}{4} \times \frac{1}{2} \times \frac{3}{8})$.
$= \frac{9}{64} + \frac{3}{64} + \frac{9}{64} = \frac{21}{64}$.

(D) Given $P(A \cup B) = P(A \cap B)$.
We know that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Substituting the given condition,we get $P(A \cap B) = P(A) + P(B) - P(A \cap B)$,which implies $P(A) + P(B) = 2P(A \cap B)$.
Also,we know $P(A \cap B') = P(A) - P(A \cap B)$ and $P(A' \cap B) = P(B) - P(A \cap B)$.
Since $P(A \cap B) \le P(A)$ and $P(A \cap B) \le P(B)$,and $P(A \cup B) = P(A \cap B)$ implies that $A \subseteq B$ and $B \subseteq A$ (i.e.,$A = B$),we have $P(A) = P(B) = P(A \cap B) = P(A \cup B)$.
Thus,$P(A \cap B') = P(A) - P(A) = 0$ and $P(A' \cap B) = P(B) - P(B) = 0$.
Also,$P(A) = P(B)$ implies $A$ and $B$ are equally likely.
However,$P(A) + P(B) = 1$ is not necessarily true,as $P(A) + P(B) = 2P(A \cap B)$,which is only $1$ if $P(A \cap B) = 0.5$.

(B) Let $S = \{x_1, x_2, x_3, x_4, x_5, x_6, x_7\}$.
The total number of non-empty subsets of $S$ is $2^7 - 1 = 127$.
Let the chosen element be $x_i$. We want to find the number of non-empty subsets $A$ such that $x_i \in A$.
For any subset $A$,each of the $7$ elements can either be included or excluded,giving $2^7$ total subsets.
If we fix $x_i$ to be in the subset,the remaining $6$ elements can either be included or excluded,giving $2^6 = 64$ such subsets.
Since $64$ is non-zero,all these $64$ subsets are non-empty.
Thus,the probability that $x \in A$ is $\frac{64}{127}$.

(A) We are given the inequality $\frac{(x-10)(x-50)}{(x-30)} \ge 0$.
Using the wavy curve method (sign scheme) for the expression $f(x) = \frac{(x-10)(x-50)}{(x-30)}$,the critical points are $x = 10, 30, 50$.
The sign of $f(x)$ in different intervals is:
- For $x < 10$: $f(x) < 0$
- For $10 \le x < 30$: $f(x) \ge 0$
- For $30 < x < 50$: $f(x) < 0$
- For $x \ge 50$: $f(x) \ge 0$
Since $x \in \{1, 2, \dots, 100\}$,we look for integers $x$ in the intervals $[10, 30)$ and $[50, 100]$.
In the interval $[10, 30)$,the integers are $\{10, 11, \dots, 29\}$. The number of such integers is $29 - 10 + 1 = 20$.
In the interval $[50, 100]$,the integers are $\{50, 51, \dots, 100\}$. The number of such integers is $100 - 50 + 1 = 51$.
The total number of favorable outcomes is $20 + 51 = 71$.
The total number of possible outcomes is $100$.
Therefore,$P(A) = \frac{71}{100} = 0.71$.

(C) For any event $E$,the probability $P(E)$ must satisfy $0 \le P(E) \le 1$.
For event $A$: $0 \le \frac{3x+1}{3} \le 1 \Rightarrow 0 \le 3x+1 \le 3 \Rightarrow -1 \le 3x \le 2 \Rightarrow -\frac{1}{3} \le x \le \frac{2}{3}$.
For event $B$: $0 \le \frac{1-x}{4} \le 1 \Rightarrow 0 \le 1-x \le 4 \Rightarrow -1 \le -x \le 3 \Rightarrow -3 \le x \le 1$.
Since $A$ and $B$ are mutually exclusive,$P(A \cup B) = P(A) + P(B) \le 1$.
$\frac{3x+1}{3} + \frac{1-x}{4} \le 1 \Rightarrow \frac{4(3x+1) + 3(1-x)}{12} \le 1 \Rightarrow 12x + 4 + 3 - 3x \le 12 \Rightarrow 9x + 7 \le 12 \Rightarrow 9x \le 5 \Rightarrow x \le \frac{5}{9}$.
Combining all conditions: $x \ge -\frac{1}{3}$,$x \le \frac{2}{3}$,$x \ge -3$,$x \le 1$,and $x \le \frac{5}{9}$.
The intersection of these intervals is $[-\frac{1}{3}, \frac{5}{9}]$.

(A) Let the probabilities of hitting the target by four persons be $P(A) = \frac{1}{2}$,$P(B) = \frac{1}{3}$,$P(C) = \frac{1}{4}$,and $P(D) = \frac{1}{8}$.
The probability that the target is hit is $1 - P(\text{none of them hits the target})$.
Since the events are independent,$P(\text{none hits}) = P(\overline{A}) \cdot P(\overline{B}) \cdot P(\overline{C}) \cdot P(\overline{D})$.
$P(\overline{A}) = 1 - \frac{1}{2} = \frac{1}{2}$,$P(\overline{B}) = 1 - \frac{1}{3} = \frac{2}{3}$,$P(\overline{C}) = 1 - \frac{1}{4} = \frac{3}{4}$,and $P(\overline{D}) = 1 - \frac{1}{8} = \frac{7}{8}$.
$P(\text{none hits}) = \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdot \frac{7}{8} = \frac{1 \cdot 2 \cdot 3 \cdot 7}{2 \cdot 3 \cdot 4 \cdot 8} = \frac{42}{192} = \frac{7}{32}$.
Therefore,the probability that the target is hit is $1 - \frac{7}{32} = \frac{25}{32}$.

(D) Let $P(A)$ and $P(B)$ be the probabilities of events $A$ and $B$ respectively.
The probability that exactly one of them occurs is given by $P(A) + P(B) - 2P(A \cap B) = \frac{2}{5}$.
The probability that $A$ or $B$ occurs is given by $P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{1}{2}$.
Subtracting the first equation from the second equation:
$(P(A) + P(B) - P(A \cap B)) - (P(A) + P(B) - 2P(A \cap B)) = \frac{1}{2} - \frac{2}{5}$.
$P(A \cap B) = \frac{5 - 4}{10} = \frac{1}{10} = 0.10$.

(A) The sample space $S$ for tossing three coins is:
$S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$,so $n(S) = 8$.
The events are:
$E = \{HHH, TTT\} \implies P(E) = \frac{2}{8} = \frac{1}{4}$
$F = \{HHH, HHT, HTH, THH\} \implies P(F) = \frac{4}{8} = \frac{1}{2}$
$G = \{HHT, HTH, THH, HTT, THT, TTH, TTT\} \implies P(G) = \frac{7}{8}$
Intersections:
$E \cap F = \{HHH\} \implies P(E \cap F) = \frac{1}{8}$
$E \cap G = \{TTT\} \implies P(E \cap G) = \frac{1}{8}$
$F \cap G = \{HHT, HTH, THH\} \implies P(F \cap G) = \frac{3}{8}$
Checking independence $(P(A \cap B) = P(A) \cdot P(B))$:
$1$. For $(E, F)$: $P(E) \cdot P(F) = \frac{1}{4} \cdot \frac{1}{2} = \frac{1}{8} = P(E \cap F)$. Thus,$(E, F)$ are independent.
$2$. For $(E, G)$: $P(E) \cdot P(G) = \frac{1}{4} \cdot \frac{7}{8} = \frac{7}{32} \neq P(E \cap G) = \frac{1}{8}$. Thus,$(E, G)$ are dependent.
$3$. For $(F, G)$: $P(F) \cdot P(G) = \frac{1}{2} \cdot \frac{7}{8} = \frac{7}{16} \neq P(F \cap G) = \frac{3}{8}$. Thus,$(F, G)$ are dependent.

(N/A) The sample space $S$ for tossing a fair coin and an unbiased die is given by:
$S = \{(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)\}$
The total number of outcomes is $n(S) = 12$.
Let $A$ be the event 'head appears on the coin':
$A = \{(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6)\}$
$P(A) = \frac{n(A)}{n(S)} = \frac{6}{12} = \frac{1}{2}$.
Let $B$ be the event '$3$ on the die':
$B = \{(H, 3), (T, 3)\}$
$P(B) = \frac{n(B)}{n(S)} = \frac{2}{12} = \frac{1}{6}$.
The intersection $A \cap B$ is the event 'head appears on the coin and $3$ appears on the die':
$A \cap B = \{(H, 3)\}$
$P(A \cap B) = \frac{n(A \cap B)}{n(S)} = \frac{1}{12}$.
Now,check for independence:
$P(A) \times P(B) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12}$.
Since $P(A \cap B) = P(A) \times P(B)$,the events $A$ and $B$ are independent.

(N/A) When a die is thrown,the sample space $S$ is $S = \{1, 2, 3, 4, 5, 6\}$.
Let $A$ be the event that the number is even,so $A = \{2, 4, 6\}$.
Thus,$P(A) = \frac{3}{6} = \frac{1}{2}$.
Let $B$ be the event that the number is red,so $B = \{1, 2, 3\}$.
Thus,$P(B) = \frac{3}{6} = \frac{1}{2}$.
The intersection $A \cap B$ is the set of numbers that are both even and red,so $A \cap B = \{2\}$.
Thus,$P(A \cap B) = \frac{1}{6}$.
Now,calculate $P(A) \times P(B) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$.
Since $P(A \cap B) = \frac{1}{6}$ and $P(A) \times P(B) = \frac{1}{4}$,we see that $P(A \cap B) \neq P(A) \times P(B)$.
Therefore,the events $A$ and $B$ are not independent.

(A) We are given $P(A) = \frac{1}{4}$,$P(B) = \frac{1}{2}$,and $P(A \cap B) = \frac{1}{8}$.
We need to find $P(A' \cap B')$,where $A'$ and $B'$ are the complements of events $A$ and $B$ respectively.
By De Morgan's Law,$A' \cap B' = (A \cup B)'$.
Therefore,$P(A' \cap B') = P((A \cup B)') = 1 - P(A \cup B)$.
Using the addition theorem,$P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
$P(A \cup B) = \frac{1}{4} + \frac{1}{2} - \frac{1}{8} = \frac{2+4-1}{8} = \frac{5}{8}$.
Thus,$P(A' \cap B') = 1 - \frac{5}{8} = \frac{3}{8}$.

(C) The probability of getting an odd number in a single throw of a die is $P(\text{Odd}) = \frac{3}{6} = \frac{1}{2}$.
The probability of getting an even number in a single throw is $P(\text{Even}) = \frac{3}{6} = \frac{1}{2}$.
When a die is tossed thrice,the probability of getting an even number in all three throws is $P(\text{Even, Even, Even}) = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}$.
The probability of getting an odd number at least once is given by $1 - P(\text{No odd number in any throw})$.
This is equivalent to $1 - P(\text{Even in all three throws})$.
Therefore,the probability is $1 - \frac{1}{8} = \frac{7}{8}$.

(C) Let $P(A)$ be the probability that $A$ solves the problem,$P(A) = \frac{1}{2}$.
Let $P(B)$ be the probability that $B$ solves the problem,$P(B) = \frac{1}{3}$.
The problem is solved if at least one of them solves it.
The probability that the problem is not solved by either is $P(A') \times P(B')$.
$P(A') = 1 - P(A) = 1 - \frac{1}{2} = \frac{1}{2}$.
$P(B') = 1 - P(B) = 1 - \frac{1}{3} = \frac{2}{3}$.
Probability that the problem is not solved $= P(A') \times P(B') = \frac{1}{2} \times \frac{2}{3} = \frac{1}{3}$.
Probability that the problem is solved $= 1 - P(\text{not solved}) = 1 - \frac{1}{3} = \frac{2}{3}$.

(B) Let $P(A) = \frac{1}{2}$ be the probability that $A$ solves the problem.
Let $P(B) = \frac{1}{3}$ be the probability that $B$ solves the problem.
Then,$P(A') = 1 - P(A) = 1 - \frac{1}{2} = \frac{1}{2}$ and $P(B') = 1 - P(B) = 1 - \frac{1}{3} = \frac{2}{3}$.
The probability that exactly one of them solves the problem is given by $P(A \cap B') + P(B \cap A')$.
Since the events are independent,this is equal to $P(A) \cdot P(B') + P(B) \cdot P(A')$.
$= (\frac{1}{2} \times \frac{2}{3}) + (\frac{1}{3} \times \frac{1}{2})$
$= \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}$.

(A) When two dice are rolled,the total number of possible outcomes is $6 \times 6 = 36$.
The only even prime number is $2$.
Let $E$ be the event of getting an even prime number on each die.
Since each die must show $2$,the favorable outcome is $(2, 2)$.
Therefore,the number of favorable outcomes is $1$.
The probability $P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{1}{36}$.
Thus,the correct answer is $A$.

(A) Since the two coins are distinguishable (a one rupee coin and a two rupee coin),we can denote the outcomes as ordered pairs $(C_1, C_2)$,where $C_1$ is the outcome of the one rupee coin and $C_2$ is the outcome of the two rupee coin.
Each coin can result in either Head $(H)$ or Tail $(T)$.
The possible outcomes are:
$1$. Head on both coins: $(H, H) = HH$
$2$. Head on the first coin and Tail on the second: $(H, T) = HT$
$3$. Tail on the first coin and Head on the second: $(T, H) = TH$
$4$. Tail on both coins: $(T, T) = TT$
Therefore,the sample space is $S = \{ HH, HT, TH, TT \}$.

(A) Let the coins be represented by their values: $1, 2, 5$.
Since he takes out two coins one after the other,the order matters.
The first coin can be any of the three: $1, 2, 5$.
If the first coin is $1$,the second can be $2$ or $5$. Outcomes: $(1, 2), (1, 5)$.
If the first coin is $2$,the second can be $1$ or $5$. Outcomes: $(2, 1), (2, 5)$.
If the first coin is $5$,the second can be $1$ or $2$. Outcomes: $(5, 1), (5, 2)$.
Thus,the sample space is $S = \{ (1, 2), (1, 5), (2, 1), (2, 5), (5, 1), (5, 2) \}$.

(N/A) The number of accidents along a busy highway during the year of observation can be $0$ (for no accident),$1$,$2$,or any other positive integer. Thus,the sample space associated with this experiment is $S = \{0, 1, 2, 3, \ldots\}$.

(N/A) Let us denote the blue balls by $B_1, B_2, B_3$ and the white balls by $W_1, W_2, W_3, W_4$.
The sample space $S$ of the experiment consists of all possible outcomes.
If the coin shows head $(H)$,we draw one of the $7$ balls. The outcomes are ${HB_1, HB_2, HB_3, HW_1, HW_2, HW_3, HW_4}$.
If the coin shows tail $(T)$,we roll a die. The outcomes are ${T1, T2, T3, T4, T5, T6}$.
Thus,the sample space is:
$S = \{HB_1, HB_2, HB_3, HW_1, HW_2, HW_3, HW_4, T1, T2, T3, T4, T5, T6\}$.

(A) In the experiment,the coin is tossed until a head $(H)$ appears.
If the head appears on the $1^{st}$ toss,the outcome is $H$.
If the head appears on the $2^{nd}$ toss,the outcome is $TH$.
If the head appears on the $3^{rd}$ toss,the outcome is $TTH$.
If the head appears on the $4^{th}$ toss,the outcome is $TTTH$,and so on.
Thus,the sample space $S$ is the set of all possible outcomes:
$S = \{ H, TH, TTH, TTTH, TTTTH, \dots \}$

(A) coin has two faces: head $(H)$ and tail $(T)$.
When a coin is tossed three times,the total number of possible outcomes is $2^{3} = 8$.
Thus,the sample space $S$ is the set of all possible outcomes:
$S = \{ HHH, HHT, HTH, HTT, THH, THT, TTH, TTT \}$

When a die is thrown,the possible outcomes are $1, 2, 3, 4, 5,$ or $6$.
When a die is thrown two times,the sample space $S$ is given by the set of all ordered pairs $(x, y)$ where $x$ and $y$ represent the outcomes of the first and second throw respectively,such that $x, y \in \{1, 2, 3, 4, 5, 6\}$.
The total number of elements in this sample space is $6 \times 6 = 36$.
The sample space $S$ is:
$S = \{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}$

When a coin is tossed once,there are two possible outcomes: head $(H)$ and tail $(T)$.
When a coin is tossed four times,the total number of possible outcomes is $2^{4} = 16$.
Thus,when a coin is tossed four times,the sample space $S$ is given by:
$S = \{ HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT \}$

(A) coin has two faces: head $(H)$ and tail $(T)$.
$A$ die has six faces that are numbered from $1$ to $6$,with one number on each face.
Thus,when a coin is tossed and a die is thrown,the sample space is given by:
$S = \{H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6\}$

(N/A) coin has two faces: head $(H)$ and tail $(T)$.
$A$ die has six faces numbered from $1$ to $6$.
According to the experiment,if a tail $(T)$ appears,the die is not rolled. If a head $(H)$ appears,the die is rolled.
Therefore,the sample space $S$ is the set of all possible outcomes:
$S = \{H1, H2, H3, H4, H5, H6, T\}$

(N/A) Let us denote the $2$ boys and $2$ girls in Room $X$ as $B_{1}, B_{2}$ and $G_{1}, G_{2}$ respectively.
Let us denote the $1$ boy and $3$ girls in Room $Y$ as $B_{3}$ and $G_{3}, G_{4}, G_{5}$ respectively.
When a room is selected first and then a person,the possible outcomes are the combinations of the room and the individuals within that room.
Thus,the required sample space $S$ is given by:
$S = \{X B_{1}, X B_{2}, X G_{1}, X G_{2}, Y B_{3}, Y G_{3}, Y G_{4}, Y G_{5}\}$

$A$ die has six faces that are numbered from $1$ to $6,$ with one number on each face. Let us denote the red,white,and blue dice as $R$,$W$,and $B$ respectively.
Accordingly,when a die is selected and then rolled,the sample space $S$ is given by the set of all possible outcomes of the form (colour,number):
$S = \{R1, R2, R3, R4, R5, R6, W1, W2, W3, W4, W5, W6, B1, B2, B3, B4, B5, B6\}$

(A) In an experiment with $2$ children,each child can be either a boy $(B)$ or a girl $(G)$.
Since we are interested in the order of birth,we consider all possible sequences of length $2$.
The possible outcomes are:
$1$. First child is a boy,second is a boy: $BB$
$2$. First child is a boy,second is a girl: $BG$
$3$. First child is a girl,second is a boy: $GB$
$4$. First child is a girl,second is a girl: $GG$
Therefore,the sample space $S$ is given by $S = \{ BB, BG, GB, GG \}$.

(N/A) Since the maximum number of children in each family is $2$,a family can have $2$ girls,$1$ girl,or $0$ girls.
Therefore,the sample space $S$ for the number of girls is $S = \{0, 1, 2\}$.

(B) The box contains $1$ red ball $(R)$ and $3$ identical white balls $(W)$.
Since the balls are drawn without replacement,the first ball can be red or white.
If the first ball is red,the second must be white (as there is only $1$ red ball).
If the first ball is white,the second can be red or white.
Thus,the possible outcomes are:
$1$. First is red,second is white: $(R, W)$
$2$. First is white,second is red: $(W, R)$
$3$. First is white,second is white: $(W, W)$
Therefore,the sample space is $S = \{ RW, WR, WW \}$.

(N/A) coin has two faces: head $(H)$ and tail $(T)$.
$A$ die has six faces that are numbered from $1$ to $6,$ with one number on each face.
If the first toss is a head $(H)$,the coin is tossed again,resulting in $(HH)$ or $(HT)$.
If the first toss is a tail $(T)$,a die is rolled,resulting in $(T1), (T2), (T3), (T4), (T5),$ or $(T6)$.
Thus,the sample space $S$ is given by:
$S = \{HH, HT, T1, T2, T3, T4, T5, T6\}$

(N/A) Since each bulb can be either defective $(D)$ or non-defective $(N)$,and there are $3$ bulbs,each bulb has $2$ possible outcomes.
Total number of outcomes $= 2^3 = 8$.
The sample space $S$ is the set of all possible outcomes:
$S = \{ DDD, DDN, DND, DNN, NDD, NDN, NND, NNN \}$

(N/A) When a coin is tossed,the possible outcomes are head $(H)$ and tail $(T)$.
When a die is thrown,the possible outcomes are $1, 2, 3, 4, 5,$ or $6$.
If the coin shows $T$,the experiment ends.
If the coin shows $H$,a die is thrown. If the result is odd $(1, 3, 5)$,the experiment ends.
If the result is even $(2, 4, 6)$,the die is thrown again.
Thus,the sample space $S$ is:
$S = \{T, H1, H3, H5, H21, H22, H23, H24, H25, H26, H41, H42, H43, H44, H45, H46, H61, H62, H63, H64, H65, H66\}$

(N/A) die has six faces numbered from $1$ to $6$. Among these,$2, 4,$ and $6$ are even numbers,while $1, 3,$ and $5$ are odd numbers.
$A$ coin has two faces: head $(H)$ and tail $(T)$.
If the number is even,the outcome is (number,coin face). If the number is odd,the outcome is (number,coin face $1$,coin face $2$).
The sample space $S$ is:
$S = \{(2, H), (2, T), (4, H), (4, T), (6, H), (6, T), (1, H, H), (1, H, T), (1, T, H), (1, T, T), (3, H, H), (3, H, T), (3, T, H), (3, T, T), (5, H, H), (5, H, T), (5, T, H), (5, T, T)\}$

The box contains $2$ red balls and $3$ black balls. Let us denote the $2$ red balls as $R_{1}, R_{2}$ and the $3$ black balls as $B_{1}, B_{2}, B_{3}$.
The sample space $S$ of this experiment is the set of all possible outcomes.
If the coin shows a tail $(T)$,we draw a ball from the box,resulting in outcomes: $TR_{1}, TR_{2}, TB_{1}, TB_{2}, TB_{3}$.
If the coin shows a head $(H)$,we throw a die,resulting in outcomes: $H1, H2, H3, H4, H5, H6$.
Therefore,the sample space is $S = \{TR_{1}, TR_{2}, TB_{1}, TB_{2}, TB_{3}, H1, H2, H3, H4, H5, H6\}$.

(D) The sample space $S$ contains $36$ outcomes: $S = \{(x, y) : x, y \in \{1, 2, 3, 4, 5, 6\}\}$.
$A = \{(1,1), (1,3), (1,5), (2,2), (2,4), (2,6), (3,1), (3,3), (3,5), (4,2), (4,4), (4,6), (5,1), (5,3), (5,5), (6,2), (6,4), (6,6)\}$
$B = \{(1,2), (2,1), (1,5), (5,1), (3,3), (2,4), (4,2), (3,6), (6,3), (4,5), (5,4), (6,6)\}$
$C = \{(1,1), (1,2), (2,1)\}$
$D = \{(6,6)\}$
Two events are mutually exclusive if their intersection is the empty set $(\phi)$.
Checking the intersections:
$A \cap B = \{(1,5), (2,4), (3,3), (4,2), (5,1), (6,6)\} \neq \phi$
$A \cap D = \{(6,6)\} \neq \phi$
$B \cap D = \{(6,6)\} \neq \phi$
$C \cap D = \phi$
Since $C \cap D = \phi$,the events $C$ and $D$ are mutually exclusive.

(A) The sample space of the experiment is:
$S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$
The events are defined as:
$A = \{TTT\}$
$B = \{HTT, THT, TTH\}$
$C = \{HHT, HTH, THH, HHH\}$
Check for exhaustive events:
$A \cup B \cup C = \{TTT, HTT, THT, TTH, HHT, HTH, THH, HHH\} = S$
Since the union of the events is the sample space $S$,the events are exhaustive.
Check for mutually exclusive events:
$A \cap B = \phi$
$A \cap C = \phi$
$B \cap C = \phi$
Since the intersection of any two events is the empty set $\phi$,the events are pair-wise disjoint,i.e.,mutually exclusive.
Conclusion:
Yes,$A, B,$ and $C$ form a set of mutually exclusive and exhaustive events.

(B) When a die is rolled,the sample space $S$ is given by:
$S = \{1, 2, 3, 4, 5, 6\}$
The event $E$ (die shows $4$) is:
$E = \{4\}$
The event $F$ (die shows an even number) is:
$F = \{2, 4, 6\}$
To check if $E$ and $F$ are mutually exclusive,we find their intersection:
$E \cap F = \{4\}$
Since $E \cap F \neq \phi$ (where $\phi$ is the empty set),the events $E$ and $F$ are not mutually exclusive.

(D) When a die is thrown,the sample space is $S = \{1, 2, 3, 4, 5, 6\}$.
Event $A$ is a number less than $7$,so $A = \{1, 2, 3, 4, 5, 6\}$.
Event $B$ is a number greater than $7$,so $B = \emptyset$ (impossible event).
Event $C$ is a multiple of $3$,so $C = \{3, 6\}$.
Therefore,$B \cup C = \emptyset \cup \{3, 6\} = \{3, 6\}$.

(A) When a pair of dice is rolled,the sample space $S$ contains $36$ outcomes.
$A = \{(3,6), (4,5), (4,6), (5,4), (5,5), (5,6), (6,3), (6,4), (6,5), (6,6)\}$
$B = \{(2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (1,2), (3,2), (4,2), (5,2), (6,2)\}$
$C = \{(1,6), (2,5), (3,4), (4,3), (5,2), (6,1), (3,6), (4,5), (5,4), (6,3), (6,6)\}$
Two events are mutually exclusive if their intersection is empty $(\phi)$.
$A \cap B = \phi$ (No common elements).
$B \cap C = \{(2,5), (5,2)\} \neq \phi$.
$A \cap C = \{(3,6), (4,5), (5,4), (6,3), (6,6)\} \neq \phi$.
Therefore,only events $A$ and $B$ are mutually exclusive.

(N/A) When three coins are tossed,the sample space is given by
$S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$
Accordingly,
$A = \{HHH\}$
$B = \{HHT, HTH, THH\}$
$C = \{TTT\}$
$D = \{HHH, HHT, HTH, HTT\}$
Two events are mutually exclusive if their intersection is the empty set $(\phi)$.
$A \cap B = \phi$
$A \cap C = \phi$
$A \cap D = \{HHH\} \neq \phi$
$B \cap C = \phi$
$B \cap D = \{HHT, HTH\} \neq \phi$
$C \cap D = \phi$
Thus,the mutually exclusive pairs are $(A, B)$,$(A, C)$,$(B, C)$,and $(C, D)$.

(B) When three coins are tossed,the sample space $S$ is given by:
$S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$
An event is called a simple event if it contains only one sample point of the sample space.
$A = \{HHH\}$ (contains $1$ sample point,so it is a simple event)
$B = \{HHT, HTH, THH\}$ (contains $3$ sample points,so it is a compound event)
$C = \{TTT\}$ (contains $1$ sample point,so it is a simple event)
$D = \{HHH, HHT, HTH, HTT\}$ (contains $4$ sample points,so it is a compound event)
Thus,$A$ and $C$ are simple events.

(B) When three coins are tossed,the sample space $S$ is given by:
$S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$
Accordingly,the events are:
$A = \{HHH\}$
$B = \{HHT, HTH, THH\}$
$C = \{TTT\}$
$D = \{HHH, HHT, HTH, HTT\}$
An event is called a compound event if it contains more than one sample point.
- Event $A$ has $1$ sample point (simple event).
- Event $B$ has $3$ sample points (compound event).
- Event $C$ has $1$ sample point (simple event).
- Event $D$ has $4$ sample points (compound event).
Thus,$B$ and $D$ are compound events.

(N/A) When three coins are tossed,the sample space $S$ is given by:
$S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$
Two events $A$ and $B$ are mutually exclusive if they have no common outcomes,i.e.,$A \cap B = \emptyset$.
Let $A$ be the event of getting no heads: $A = \{TTT\}$.
Let $B$ be the event of getting no tails: $B = \{HHH\}$.
Since $A \cap B = \emptyset$,these two events are mutually exclusive.

(N/A) When three coins are tossed,the sample space $S$ is given by:
$S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$
Three events that are mutually exclusive and exhaustive can be defined as:
$A$: Getting no heads.
$B$: Getting exactly one head.
$C$: Getting at least two heads.
These events are represented as:
$A = \{TTT\}$
$B = \{HTT, THT, TTH\}$
$C = \{HHH, HHT, HTH, THH\}$
These events are mutually exclusive because $A \cap B = B \cap C = C \cap A = \phi$.
They are exhaustive because $A \cup B \cup C = S$.

(N/A) When three coins are tossed,the sample space $S$ is given by:
$S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$
Two events that are not mutually exclusive can be defined as:
$A$: Getting three heads
$B$: Getting at least $2$ heads
Here,the sets are:
$A = \{HHH\}$
$B = \{HHH, HHT, HTH, THH\}$
Since $A \cap B = \{HHH\} \neq \phi$,the events $A$ and $B$ are not mutually exclusive.

(N/A) When three coins are tossed,the sample space $S$ is given by:
$S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$
Two events which are mutually exclusive but not exhaustive can be defined as:
$A$: Getting exactly one head.
$B$: Getting exactly one tail.
Here,the sets are:
$A = \{HTT, THT, TTH\}$
$B = \{HHT, HTH, THH\}$
These events are mutually exclusive because $A \cap B = \phi$.
They are not exhaustive because $A \cup B = \{HTT, THT, TTH, HHT, HTH, THH\} \neq S$.

(N/A) When three coins are tossed,the sample space $S$ is given by:
$S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$
Three events that are mutually exclusive but not exhaustive can be defined as:
$A$: Getting exactly three heads,i.e.,$A = \{HHH\}$
$B$: Getting exactly one head,i.e.,$B = \{HTT, THT, TTH\}$
$C$: Getting exactly two heads,i.e.,$C = \{HHT, HTH, THH\}$
These events are mutually exclusive because $A \cap B = \phi$,$B \cap C = \phi$,and $C \cap A = \phi$.
They are not exhaustive because $A \cup B \cup C = \{HHH, HTT, THT, TTH, HHT, HTH, THH\} \neq S$ (since $TTT \notin A \cup B \cup C$).