Probability Questions in English

(D) When three unbiased coins are tossed,the sample space $S$ is given by:
$S = \{TTT, TTH, THT, HTT, THH, HTH, HHT, HHH\}$
Total number of outcomes $N(S) = 8$.
Let $E$ be the event of getting at most two heads.
This means we can have $0, 1,$ or $2$ heads.
The outcomes corresponding to $E$ are:
$E = \{TTT, TTH, THT, HTT, THH, HTH, HHT\}$
Number of favorable outcomes $N(E) = 7$.
The probability $P(E)$ is given by:
$P(E) = \frac{N(E)}{N(S)} = \frac{7}{8}$.

(B) When two dice are thrown,the total number of possible outcomes is $n(S) = 6 \times 6 = 36$.
The product of two numbers is odd only if both numbers are odd. The odd numbers on a die are $\{1, 3, 5\}$.
The number of outcomes where both dice show an odd number is $3 \times 3 = 9$.
These outcomes are: $(1,1), (1,3), (1,5), (3,1), (3,3), (3,5), (5,1), (5,3), (5,5)$.
The product is even if it is not odd. Therefore,the number of outcomes with an even product is $n(E) = n(S) - n(\text{odd product}) = 36 - 9 = 27$.
The probability of getting an even product is $P(E) = \frac{n(E)}{n(S)} = \frac{27}{36} = \frac{3}{4}$.

(A) Total number of students $= 15 + 10 = 25$.
Number of ways to select $3$ students out of $25$ is given by ${}^{25}C_{3} = \frac{25 \times 24 \times 23}{3 \times 2 \times 1} = 2300$.
Number of ways to select $1$ girl out of $10$ is ${}^{10}C_{1} = 10$.
Number of ways to select $2$ boys out of $15$ is ${}^{15}C_{2} = \frac{15 \times 14}{2 \times 1} = 105$.
Required probability $= \frac{{}^{10}C_{1} \times {}^{15}C_{2}}{{}^{25}C_{3}} = \frac{10 \times 105}{2300} = \frac{1050}{2300} = \frac{21}{46}$.

(B) Total number of balls $= 7 + 5 = 12$.
Probability of drawing a white ball in the first draw $P(W_1) = \frac{5}{12}$.
Since the ball is replaced,the total number of balls remains $12$ and the number of white balls remains $5$.
Probability of drawing a white ball in the second draw $P(W_2) = \frac{5}{12}$.
Since the events are independent,the probability that both balls drawn are white is $P(W_1 \cap W_2) = P(W_1) \times P(W_2) = \frac{5}{12} \times \frac{5}{12} = \frac{25}{144}$.

(C) Total number of balls $= 7 + 5 = 12$.
The probability of drawing a black ball in the first draw is $P(B_1) = \frac{7}{12}$.
Since the ball is replaced,the total number of balls remains $12$ for the second draw.
The probability of drawing a black ball in the second draw is $P(B_2) = \frac{7}{12}$.
Since the events are independent,the probability that both balls are black is $P(B_1 \cap B_2) = P(B_1) \times P(B_2) = \frac{7}{12} \times \frac{7}{12} = \frac{49}{144}$.

(A) Total number of balls in the bag $= 7 + 5 = 12$.
Since the ball is replaced after the first draw,the total number of balls remains $12$ for the second draw.
The probability of drawing a black ball in the first draw is $P(B_1) = \frac{7}{12}$.
The probability of drawing a white ball in the second draw is $P(W_2) = \frac{5}{12}$.
Since these are independent events,the probability that the first ball is black and the second is white is $P(B_1 \cap W_2) = P(B_1) \times P(W_2) = \frac{7}{12} \times \frac{5}{12} = \frac{35}{144}$.

(D) Total number of balls in the bag $= 7 + 5 = 12$.
Since the ball is replaced after the first draw,the total number of balls remains $12$ for the second draw.
The probability of drawing a white ball in the first draw is $P(W) = \frac{5}{12}$.
The probability of drawing a black ball in the second draw is $P(B) = \frac{7}{12}$.
Since these are independent events,the probability that the first ball is white and the second ball is black is $P(W \cap B) = P(W) \times P(B) = \frac{5}{12} \times \frac{7}{12} = \frac{35}{144}$.

(B) Total number of balls $= 5 + 4 + 3 = 12$.
The total number of ways to draw $3$ balls from $12$ is given by $n(S) = {}^{12}C_{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220$.
Let $E$ be the event that all $3$ balls are of the same colour.
This can happen if all $3$ are red,all $3$ are green,or all $3$ are black.
Number of ways to choose $3$ red balls $= {}^{5}C_{3} = 10$.
Number of ways to choose $3$ green balls $= {}^{4}C_{3} = 4$.
Number of ways to choose $3$ black balls $= {}^{3}C_{3} = 1$.
Total ways for same colour $= 10 + 4 + 1 = 15$.
Probability that all balls are of the same colour $P(E) = \frac{15}{220} = \frac{3}{44}$.
The probability that they are not of the same colour $= 1 - P(E) = 1 - \frac{3}{44} = \frac{41}{44}$.

(D) When a positive integer is divided by $9$,the possible remainders are the set of integers $\{0, 1, 2, 3, 4, 5, 6, 7, 8\}$.
There are $9$ possible outcomes in total,so $n(S) = 9$.
The event $E$ is that the remainder is equal to $1$. This corresponds to the outcome $\{1\}$.
Thus,the number of favorable outcomes is $n(E) = 1$.
The probability $P(E)$ is given by the ratio of favorable outcomes to total outcomes:
$P(E) = \frac{n(E)}{n(S)} = \frac{1}{9}$.

(C) When a positive integer is divided by $9$,the possible remainders are the set $S = \{0, 1, 2, 3, 4, 5, 6, 7, 8\}$.
There are $9$ possible outcomes in total,so $n(S) = 9$.
We want the probability that the remainder is not $1$.
The set of favorable outcomes is $E = \{0, 2, 3, 4, 5, 6, 7, 8\}$.
The number of favorable outcomes is $n(E) = 8$.
The probability $P(E)$ is given by the ratio of the number of favorable outcomes to the total number of outcomes.
$P(E) = \frac{n(E)}{n(S)} = \frac{8}{9}$.

(C) When two coins are tossed,the sample space $S$ is given by: $S = \{(H, H), (H, T), (T, H), (T, T)\}$.
Total number of outcomes $n(S) = 4$.
Let $E$ be the event of getting at most one head. This means getting zero heads or one head.
The favorable outcomes are: $E = \{(H, T), (T, H), (T, T)\}$.
Number of favorable outcomes $n(E) = 3$.
The probability $P(E)$ is given by the formula $P(E) = \frac{n(E)}{n(S)}$.
Therefore,$P(E) = \frac{3}{4}$.

(D) When two coins are tossed,the sample space $S$ is given by:
$S = \{(H, H), (H, T), (T, H), (T, T)\}$
Thus,the total number of outcomes is $n(S) = 4$.
Let $E$ be the event of getting at most two heads.
'At most two heads' means getting $0, 1,$ or $2$ heads.
All outcomes in the sample space satisfy this condition:
$E = \{(H, H), (H, T), (T, H), (T, T)\}$
Thus,the number of favorable outcomes is $n(E) = 4$.
The probability $P(E)$ is given by:
$P(E) = \frac{n(E)}{n(S)} = \frac{4}{4} = 1$.

(B) leap year has $366$ days,which consists of $52$ full weeks and $2$ additional days.
The possible pairs for these $2$ extra days are:
$\{(\text{Sunday, Monday}), (\text{Monday, Tuesday}), (\text{Tuesday, Wednesday}), (\text{Wednesday, Thursday}), (\text{Thursday, Friday}), (\text{Friday, Saturday}), (\text{Saturday, Sunday})\}$
Therefore,the total number of outcomes is $n(S) = 7$.
For the year to have $53$ Wednesdays,one of the two extra days must be a Wednesday. The favorable outcomes are:
$\{(\text{Tuesday, Wednesday}), (\text{Wednesday, Thursday})\}$
Thus,the number of favorable outcomes is $n(E) = 2$.
The probability $P(E)$ is given by $\frac{n(E)}{n(S)} = \frac{2}{7}$.

(C) non-leap year consists of $365$ days.
Dividing $365$ by $7$,we get $52$ weeks and $1$ extra day.
This extra day can be any one of the following: {Sunday,Monday,Tuesday,Wednesday,Thursday,Friday,Saturday}.
Thus,the total number of outcomes is $n(S) = 7$.
For the year to have $53$ Wednesdays,the extra day must be a Wednesday.
There is only $1$ favorable outcome,which is {Wednesday}.
Thus,$n(E) = 1$.
The probability $P(E) = \frac{n(E)}{n(S)} = \frac{1}{7}$.

(B) When two dice are thrown,the total number of possible outcomes is $n(S) = 6 \times 6 = 36$.
The sum of the numbers on the two dice can range from $2$ to $12$.
For the sum to be an odd number,it must be $3, 5, 7, 9,$ or $11$.
Let us list the favorable outcomes for each sum:
- Sum $= 3$: $(1, 2), (2, 1)$ ($2$ outcomes)
- Sum $= 5$: $(1, 4), (4, 1), (2, 3), (3, 2)$ ($4$ outcomes)
- Sum $= 7$: $(1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3)$ ($6$ outcomes)
- Sum $= 9$: $(3, 6), (6, 3), (4, 5), (5, 4)$ ($4$ outcomes)
- Sum $= 11$: $(5, 6), (6, 5)$ ($2$ outcomes)
Total number of favorable outcomes $n(E) = 2 + 4 + 6 + 4 + 2 = 18$.
The probability $P(E) = \frac{n(E)}{n(S)} = \frac{18}{36} = \frac{1}{2}$.

(C) When two dice are thrown,the total number of possible outcomes is $n(S) = 6 \times 6 = 36$.
The sum of the numbers on the two dice is even if both numbers are even or both numbers are odd.
Case $1$: Both numbers are odd. The possible outcomes are $(1,1), (1,3), (1,5), (3,1), (3,3), (3,5), (5,1), (5,3), (5,5)$. There are $9$ such outcomes.
Case $2$: Both numbers are even. The possible outcomes are $(2,2), (2,4), (2,6), (4,2), (4,4), (4,6), (6,2), (6,4), (6,6)$. There are $9$ such outcomes.
Total favorable outcomes $n(E) = 9 + 9 = 18$.
Therefore,the required probability $P(E) = \frac{n(E)}{n(S)} = \frac{18}{36} = \frac{1}{2}$.

(D) When two dice are rolled,the total number of outcomes is $n(S) = 6 \times 6 = 36$.
The sum of the numbers on the two dice can range from $2$ to $12$.
The sums that are multiples of $3$ are $3, 6, 9,$ and $12$.
Let $E$ be the event that the sum is a multiple of $3$. The favorable outcomes are:
- Sum $= 3$: $(1, 2), (2, 1)$
- Sum $= 6$: $(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)$
- Sum $= 9$: $(3, 6), (4, 5), (5, 4), (6, 3)$
- Sum $= 12$: $(6, 6)$
Counting these outcomes,we get $n(E) = 2 + 5 + 4 + 1 = 12$.
The required probability is $P(E) = \frac{n(E)}{n(S)} = \frac{12}{36} = \frac{1}{3}$.

(C) When two dice are rolled,the total number of outcomes is $n(S) = 6 \times 6 = 36$.
We want the probability that the sum of the numbers is $\leq 10$.
It is easier to calculate the complement: the probability that the sum is $> 10$.
The sums greater than $10$ are $11$ and $12$.
The outcomes for sum $= 11$ are $(5, 6)$ and $(6, 5)$.
The outcome for sum $= 12$ is $(6, 6)$.
Thus,the number of outcomes with sum $> 10$ is $n(E') = 3$.
The probability of the sum being $> 10$ is $P(E') = \frac{n(E')}{n(S)} = \frac{3}{36} = \frac{1}{12}$.
The probability of the sum being $\leq 10$ is $P(E) = 1 - P(E') = 1 - \frac{1}{12} = \frac{11}{12}$.

(A) When two dice are thrown,the total number of possible outcomes is $n(S) = 6 \times 6 = 36$.
Let $E_1$ be the event that the sum is $6$. The outcomes are $(1,5), (5,1), (2,4), (4,2), (3,3)$. Thus,$n(E_1) = 5$.
Let $E_2$ be the event that the sum is $7$. The outcomes are $(1,6), (6,1), (2,5), (5,2), (3,4), (4,3)$. Thus,$n(E_2) = 6$.
Since these events are mutually exclusive,the total number of favorable outcomes is $n(E) = n(E_1) + n(E_2) = 5 + 6 = 11$.
The required probability is $P(E) = \frac{n(E)}{n(S)} = \frac{11}{36}$.

(B) When two dice are thrown,the total number of possible outcomes is $n(S) = 6 \times 6 = 36$.
The favorable outcomes where the numbers on both dice are equal are $(1, 1), (2, 2), (3, 3), (4, 4), (5, 5),$ and $(6, 6)$.
Therefore,the number of favorable outcomes is $n(E) = 6$.
The probability of an event is given by $P(E) = \frac{n(E)}{n(S)}$.
Substituting the values,we get $P(E) = \frac{6}{36} = \frac{1}{6}$.

(C) The total number of ways to draw $2$ balls from a total of $(6 + 8) = 14$ balls is given by the combination formula ${}^{14}C_{2} = \frac{14 \times 13}{2 \times 1} = 91$.
The number of ways to choose $1$ white ball from $6$ white balls is ${}^{6}C_{1} = 6$.
The number of ways to choose $1$ red ball from $8$ red balls is ${}^{8}C_{1} = 8$.
The number of favorable outcomes where one ball is white and the other is red is ${}^{6}C_{1} \times {}^{8}C_{1} = 6 \times 8 = 48$.
Therefore,the probability is $\frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{48}{91}$.

(C) Total number of students = $4$ girls + $6$ boys = $10$ students.
We need to form a group of $4$ students.
The total number of ways to select $4$ students from $10$ is $n(S) = {}^{10}C_{4} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$.
We want the number of boys to be less than the number of girls in a group of $4$.
Possible cases for (boys,girls) such that boys < girls:
Case $1$: $1$ boy and $3$ girls. Number of ways = ${}^{6}C_{1} \times {}^{4}C_{3} = 6 \times 4 = 24$.
Case $2$: $0$ boys and $4$ girls. Number of ways = ${}^{6}C_{0} \times {}^{4}C_{4} = 1 \times 1 = 1$.
Total favorable outcomes $n(E) = 24 + 1 = 25$.
Probability $P(E) = \frac{n(E)}{n(S)} = \frac{25}{210} = \frac{5}{42}$.

(B) The given digits are $2, 3, 4, 6,$ and $7$. We need to form a $3$-digit number (let the positions be $H, T, U$ for Hundreds,Tens,and Units).
For the number to be greater than $300$,the hundreds digit $(H)$ must be $3, 4, 6,$ or $7$.
For the number to be odd,the units digit $(U)$ must be $3$ or $7$.
We analyze based on the choice of $H$:
Case $1$: $H = 3$. Then $U$ must be $7$ ($1$ way). The tens digit $(T)$ can be any of the remaining $3$ digits $(2, 4, 6)$. Total $= 1 \times 3 \times 1 = 3$.
Case $2$: $H = 4$. Then $U$ can be $3$ or $7$ ($2$ ways). The tens digit $(T)$ can be any of the remaining $3$ digits. Total $= 1 \times 3 \times 2 = 6$.
Case $3$: $H = 6$. Then $U$ can be $3$ or $7$ ($2$ ways). The tens digit $(T)$ can be any of the remaining $3$ digits. Total $= 1 \times 3 \times 2 = 6$.
Case $4$: $H = 7$. Then $U$ must be $3$ ($1$ way). The tens digit $(T)$ can be any of the remaining $3$ digits $(2, 4, 6)$. Total $= 1 \times 3 \times 1 = 3$.
Total numbers $= 3 + 6 + 6 + 3 = 18$.

(D) Total number of ways to draw $4$ cards from $52$ cards is $n(S) = {}^{52}C_4 = \frac{52 \times 51 \times 50 \times 49}{4 \times 3 \times 2 \times 1} = 270725$.
There are $4$ queens in a pack of $52$ cards. If we draw $4$ cards such that there is no queen,we must choose $4$ cards from the remaining $52 - 4 = 48$ cards.
Number of favorable outcomes is $n(E) = {}^{48}C_4 = \frac{48 \times 47 \times 46 \times 45}{4 \times 3 \times 2 \times 1} = 194580$.
The probability $P(E) = \frac{n(E)}{n(S)} = \frac{194580}{270725}$.
Dividing both numerator and denominator by $5$,we get $P(E) = \frac{38916}{54145}$.

(D) The total number of ways to draw $4$ cards from $52$ cards is given by $n(S) = {}^{52}C_{4}$.
$n(S) = \frac{52 \times 51 \times 50 \times 49}{4 \times 3 \times 2 \times 1} = 13 \times 17 \times 25 \times 49 = 270725$.
To get one ace,one king,one queen,and one jack,we choose $1$ card from each of the $4$ suits of these ranks.
There are $4$ aces,$4$ kings,$4$ queens,and $4$ jacks in a deck.
The number of favorable outcomes is $n(E) = {}^{4}C_{1} \times {}^{4}C_{1} \times {}^{4}C_{1} \times {}^{4}C_{1} = 4 \times 4 \times 4 \times 4 = 256$.
The probability $P(E) = \frac{n(E)}{n(S)} = \frac{256}{270725}$.

(C) Total number of ways to draw $4$ cards from $52$ is $n(S) = {}^{52}C_4 = \frac{52 \times 51 \times 50 \times 49}{4 \times 3 \times 2 \times 1} = 270725$.
Digit cards are cards numbered $2$ to $10$,so there are $9$ cards per suit,total $36$ digit cards.
Honours cards are Ace,King,Queen,Jack,so there are $4$ cards per suit.
We need $2$ digit cards,$1$ black honours card,and $1$ red honours card.
Number of ways to choose $2$ digit cards from $36$ is ${}^{36}C_2 = \frac{36 \times 35}{2} = 630$.
Number of ways to choose $1$ black honours card from $8$ (spades and clubs) is ${}^{8}C_1 = 8$.
Number of ways to choose $1$ red honours card from $8$ (hearts and diamonds) is ${}^{8}C_1 = 8$.
Total favorable outcomes $n(E) = 630 \times 8 \times 8 = 40320$.
Probability $P(E) = \frac{40320}{270725} = \frac{1152}{7735}$.

(A) Total number of balls in the box $= 8 + 7 + 6 = 21$.
Let $S$ be the sample space,so the total number of possible outcomes is $n(S) = 21$.
We need to find the probability that the ball is neither red nor green.
If the ball is neither red nor green,it must be blue.
The number of blue balls is $7$.
Let $E$ be the event of picking a blue ball,so $n(E) = 7$.
The probability $P(E)$ is given by $\frac{n(E)}{n(S)} = \frac{7}{21} = \frac{1}{3}$.

(A) Total number of balls $= 2 + 3 + 2 = 7$.
The number of ways to draw $2$ balls out of $7$ is given by ${}^{7}C_{2} = \frac{7 \times 6}{1 \times 2} = 21$.
So,$n(S) = 21$.
Let $E$ be the event that none of the balls drawn is blue. This means the $2$ balls must be chosen from the $2$ red and $3$ green balls,which makes a total of $5$ non-blue balls.
The number of ways to draw $2$ balls from these $5$ balls is ${}^{5}C_{2} = \frac{5 \times 4}{1 \times 2} = 10$.
So,$n(E) = 10$.
The probability $P(E) = \frac{n(E)}{n(S)} = \frac{10}{21}$.

(D) Total number of marbles $= 5 + 4 + 3 = 12$.
The number of ways to choose $3$ marbles out of $12$ is given by $^{12}C_{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220$.
The number of ways to choose $3$ green marbles is $^{5}C_{3} = ^{5}C_{2} = \frac{5 \times 4}{2 \times 1} = 10$.
The number of ways to choose $3$ yellow marbles is $^{4}C_{3} = ^{4}C_{1} = 4$.
The number of ways to choose $3$ white marbles is $^{3}C_{3} = 1$.
The total number of ways to choose $3$ marbles of the same colour is $10 + 4 + 1 = 15$.
The probability of choosing $3$ marbles of the same colour is $\frac{15}{220} = \frac{3}{44}$.
The probability that they are not of the same colour is $1 - \frac{3}{44} = \frac{41}{44}$.

(C) The total number of possible outcomes when a dice is thrown twice is $6 \times 6 = 36$.
The favorable outcomes to get a sum of $9$ when a dice is thrown twice are:
$(3, 6), (4, 5), (5, 4), (6, 3)$.
There are $4$ such favorable outcomes.
Let $E$ be the event of getting a sum of $9$.
Therefore,the probability $P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{4}{36} = \frac{1}{9}$.

(D) The total number of ways to draw $2$ cards from a pack of $52$ cards is given by the combination formula ${}^{n}C_{r} = \frac{n!}{r!(n-r)!}$.
Total outcomes $= {}^{52}C_{2} = \frac{52 \times 51}{2 \times 1} = 26 \times 51 = 1326$.
There are $4$ kings in a standard pack of $52$ cards. The number of ways to draw $2$ kings from $4$ is:
Favorable outcomes $= {}^{4}C_{2} = \frac{4 \times 3}{2 \times 1} = 6$.
The probability $P(E)$ of drawing $2$ kings is the ratio of favorable outcomes to total outcomes:
$P(E) = \frac{6}{1326} = \frac{1}{221}$.

(D) The total number of ways to draw $2$ cards from a pack of $52$ cards is given by $n(S) = {}^{52}C_{2}$.
$n(S) = \frac{52 \times 51}{2 \times 1} = 26 \times 51 = 1326$.
We need to draw $1$ spade and $1$ heart. There are $13$ spades and $13$ hearts in a pack.
The number of ways to choose $1$ spade is ${}^{13}C_{1} = 13$.
The number of ways to choose $1$ heart is ${}^{13}C_{1} = 13$.
Let $E$ be the event of getting $1$ spade and $1$ heart.
The number of favorable outcomes is $n(E) = 13 \times 13 = 169$.
The probability $P(E)$ is given by $\frac{n(E)}{n(S)} = \frac{169}{1326}$.
Dividing both numerator and denominator by $13$,we get $P(E) = \frac{13}{102}$.

(C) Total number of balls $= 4 + 5 + 6 = 15$.
The total number of ways to draw $3$ balls from $15$ balls is given by ${}^{15}C_{3}$:
${}^{15}C_{3} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 5 \times 7 \times 13 = 455$.
The number of ways to draw $3$ red balls from $5$ red balls is given by ${}^{5}C_{3}$:
${}^{5}C_{3} = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10$.
The probability $P(E)$ of drawing $3$ red balls is the ratio of the number of favourable outcomes to the total number of outcomes:
$P(E) = \frac{10}{455} = \frac{2}{91}$.

(A) Let $E$ be the event that a student passed English and $F$ be the event that a student passed French.
Given:
$P(E) = \frac{30}{100} = \frac{3}{10}$
$P(F) = \frac{20}{100} = \frac{2}{10} = \frac{1}{5}$
$P(E \cap F) = \frac{10}{100} = \frac{1}{10}$
We need to find the probability that the student passed English or French,which is $P(E \cup F)$.
Using the addition theorem of probability:
$P(E \cup F) = P(E) + P(F) - P(E \cap F)$
$P(E \cup F) = \frac{3}{10} + \frac{2}{10} - \frac{1}{10}$
$P(E \cup F) = \frac{3 + 2 - 1}{10} = \frac{4}{10} = \frac{2}{5}$

(D) The total number of persons is $3 + 2 + 4 = 9$.
The total number of ways to select $4$ persons from $9$ is given by ${}^{9}C_{4} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126$.
We want to select exactly $2$ children from $4$ children,which can be done in ${}^{4}C_{2} = \frac{4 \times 3}{2 \times 1} = 6$ ways.
To complete the selection of $4$ persons,the remaining $2$ persons must be selected from the $5$ adults ($3$ men + $2$ women). This can be done in ${}^{5}C_{2} = \frac{5 \times 4}{2 \times 1} = 10$ ways.
The number of favorable outcomes is ${}^{4}C_{2} \times {}^{5}C_{2} = 6 \times 10 = 60$.
The probability $P(E)$ is $\frac{60}{126} = \frac{10}{21}$.

(B) The total number of balls in the box $= 4 + 5 + 6 = 15$.
The total number of possible outcomes $n(S) = 15$.
We want to find the probability of drawing either a red or a green ball.
The number of red balls $= 4$.
The number of green balls $= 5$.
The number of favorable outcomes $n(E) = 4 + 5 = 9$.
The probability $P(E)$ is given by the ratio of favorable outcomes to total outcomes:
$P(E) = \frac{n(E)}{n(S)} = \frac{9}{15}$.
Simplifying the fraction by dividing both numerator and denominator by $3$,we get:
$P(E) = \frac{3}{5}$.

(A) Total number of balls = $6 + 4 = 10$.
Number of ways to draw $3$ balls from $10$ balls is given by ${}^{10}C_3 = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$.
Number of ways to select $1$ red ball from $4$ red balls is ${}^{4}C_1 = 4$.
Number of ways to select $2$ black balls from $6$ black balls is ${}^{6}C_2 = \frac{6 \times 5}{2 \times 1} = 15$.
Number of favourable outcomes = ${}^{4}C_1 \times {}^{6}C_2 = 4 \times 15 = 60$.
Probability $P(E) = \frac{\text{Favourable outcomes}}{\text{Total outcomes}} = \frac{60}{120} = \frac{1}{2}$.

(B) The total number of possible outcomes when two dice are thrown is $6 \times 6 = 36$.
The product of two numbers is odd only if both numbers are odd. The odd numbers on a die are ${1, 3, 5}$.
The number of outcomes where both numbers are odd is $3 \times 3 = 9$.
The product is even if it is not odd. Therefore,the number of favourable outcomes where the product is even is $36 - 9 = 27$.
The probability $P(E)$ is given by $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} = \frac{27}{36} = \frac{3}{4}$.

(B) The sample space $S$ for tossing $2$ coins simultaneously is given by $S = \{HH, HT, TH, TT\}$.
The total number of exhaustive cases is $n(S) = 4$.
The favorable outcome for getting $2$ tails is the event $E = \{TT\}$.
The number of favorable cases is $n(E) = 1$.
Therefore,the probability $P(E)$ is given by the formula $P(E) = \frac{n(E)}{n(S)} = \frac{1}{4}$.

(A) When $2$ coins are tossed simultaneously,the sample space $S$ is given by:
$S = \{HH, HT, TH, TT\}$
The total number of possible outcomes is $n(S) = 4$.
We are looking for the probability of getting exactly $1$ tail. The favorable outcomes are $HT$ and $TH$.
The number of favorable outcomes is $n(E) = 2$.
The probability $P(E)$ is given by the formula:
$P(E) = \frac{n(E)}{n(S)} = \frac{2}{4} = \frac{1}{2}$.
Therefore,the probability of getting exactly $1$ tail is $1/2$.

(C) When $2$ coins are tossed simultaneously,the sample space $S$ is given by:
$S = \{HH, HT, TH, TT\}$
The total number of possible outcomes is $n(S) = 4$.
We want to find the probability of getting 'no tail'. This corresponds to the outcome where both coins show heads,which is $HH$.
The number of favourable outcomes is $n(E) = 1$.
The probability $P(E)$ is given by the formula:
$P(E) = \frac{n(E)}{n(S)} = \frac{1}{4}$
Therefore,the probability of getting no tail is $\frac{1}{4}$.

(B) When $3$ coins are tossed,the total number of possible outcomes in the sample space $S$ is $2^3 = 8$.
The sample space is $S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$.
The event of getting all heads is $E = \{HHH\}$.
The number of favorable outcomes is $n(E) = 1$.
The probability $P(E)$ is given by the ratio of favorable outcomes to the total number of outcomes:
$P(E) = \frac{n(E)}{n(S)} = \frac{1}{8}$.

(A) When $3$ coins are tossed,the sample space $S$ is given by:
$S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$
The total number of possible outcomes is $n(S) = 8$.
We are looking for the probability of getting exactly $2$ heads. The favorable outcomes are:
$E = \{HHT, HTH, THH\}$
The number of favorable outcomes is $n(E) = 3$.
The probability $P(E)$ is given by the formula:
$P(E) = \frac{n(E)}{n(S)} = \frac{3}{8}$

(A) When $3$ coins are tossed,the sample space $S$ is given by:
$S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$
Total number of exhaustive cases $= 8$.
We need to find the probability of getting at least $2$ heads. The favorable outcomes are:
$E = \{HHH, HHT, HTH, THH\}$
Number of favorable cases $= 4$.
Therefore,the probability $P$ is:
$P(\text{at least } 2 \text{ heads}) = \frac{\text{Number of favorable cases}}{\text{Total number of cases}} = \frac{4}{8} = \frac{1}{2}$.

(C) When $3$ coins are tossed,the sample space $S$ is given by:
$S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$
Total number of outcomes $= 8$.
Let $E$ be the event of getting at most $2$ heads.
The complement event $E'$ is getting $3$ heads.
The only outcome with $3$ heads is $\{HHH\}$,so $n(E') = 1$.
The probability of getting $3$ heads is $P(E') = \frac{1}{8}$.
Using the complement rule,the probability of getting at most $2$ heads is:
$P(E) = 1 - P(E') = 1 - \frac{1}{8} = \frac{7}{8}$.

(B) The sample space $S$ for tossing $3$ coins is given by:
$S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$
The total number of exhaustive outcomes is $n(S) = 8$.
We need to find the probability of getting no heads,which is equivalent to getting all tails $(TTT)$.
The favorable outcome is $E = \{TTT\}$.
The number of favorable outcomes is $n(E) = 1$.
The probability $P(E)$ is given by the formula:
$P(E) = \frac{n(E)}{n(S)} = \frac{1}{8}$.

(C) When $3$ coins are tossed,the total number of possible outcomes is $2^3 = 8$. The sample space is $S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$.
We need to find the probability of getting at least $1$ head and $1$ tail. This means we exclude the cases where all outcomes are heads $(HHH)$ or all outcomes are tails $(TTT)$.
The favourable outcomes are $\{HHT, HTH, HTT, THH, THT, TTH\}$.
The number of favourable outcomes is $6$.
Required probability $= \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} = \frac{6}{8} = \frac{3}{4}$.

(B) The sample space $S$ for tossing a coin $3$ times is given by:
$S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$
The total number of exhaustive cases is $8$.
We are looking for outcomes where heads and tails appear alternately. These outcomes are:
$HTH$ (Head,Tail,Head)
$THT$ (Tail,Head,Tail)
Thus,the number of favourable cases is $2$.
Therefore,the required probability is $\frac{\text{Number of favourable cases}}{\text{Total number of cases}} = \frac{2}{8} = \frac{1}{4}$.

(A) When $4$ coins are tossed,the total number of possible outcomes is $2^n$,where $n$ is the number of coins.
Here,$n = 4$,so the total outcomes = $2^4 = 16$.
The sample space $S$ consists of all possible combinations of Heads $(H)$ and Tails $(T)$:
$S = \{HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, HTTH, THTH, THHT, TTHH, TTHT, TTTH, THTT, HTTT, TTTT\}$.
There is only $1$ favorable outcome for getting $4$ tails,which is ${TTTT}$.
Therefore,the probability of getting $4$ tails is given by:
$P(4 \text{ tails}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{1}{16}$.

(B) When $4$ coins are tossed,the total number of possible outcomes is $2^4 = 16$.
The favorable outcomes for getting exactly $3$ tails are: $(T, T, T, H), (T, T, H, T), (T, H, T, T), (H, T, T, T)$.
There are $4$ such favorable outcomes.
Therefore,the probability of getting exactly $3$ tails is $P = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{4}{16} = \frac{1}{4}$.