Probability Questions in English

(D) The total number of ways to choose two distinct numbers $a$ and $b$ from the set ${1, 2, 3, \dots, 30}$ is given by ${}^{30}C_2 = \frac{30 \times 29}{2} = 435$.
We know that $a^2 - b^2$ is divisible by $3$ if $a^2 \equiv b^2 \pmod{3}$.
For any integer $n$, $n^2 \pmod{3}$ can be $0$ (if $n$ is a multiple of $3$) or $1$ (if $n$ is not a multiple of $3$).
Let $S_0$ be the set of numbers divisible by $3$ $(|S_0| = 10)$ and $S_1$ be the set of numbers not divisible by $3$ $(|S_1| = 20)$.
$a^2 - b^2$ is divisible by $3$ in two cases:
Case $1$: Both $a$ and $b$ are in $S_0$. The number of ways is ${}^{10}C_2 = \frac{10 \times 9}{2} = 45$.
Case $2$: Both $a$ and $b$ are in $S_1$. The number of ways is ${}^{20}C_2 = \frac{20 \times 19}{2} = 190$.
Total favorable cases $= 45 + 190 = 235$.
Therefore, the required probability $= \frac{235}{435} = \frac{47}{87}$.

(D) Let the number of daughters each friend has be $x$.
Total number of daughters = $x + x = 2x$.
We need to select $3$ tickets out of $2x$ daughters. The total number of ways to distribute $3$ tickets is ${}^{2x}C_3$.
The number of ways to select $3$ tickets such that all go to the daughters of $A$ is ${}^xC_3$.
The probability is given by $\frac{{}^xC_3}{{}^{2x}C_3} = \frac{1}{20}$.
Expanding the combinations: $\frac{x(x-1)(x-2)}{3!} / \frac{2x(2x-1)(2x-2)}{3!} = \frac{1}{20}$.
$\frac{x(x-1)(x-2)}{2x(2x-1)2(x-1)} = \frac{1}{20}$.
$\frac{x-2}{4(2x-1)} = \frac{1}{20}$.
$20(x-2) = 4(2x-1)$.
$5(x-2) = 2x-1$.
$5x - 10 = 2x - 1$.
$3x = 9$.
$x = 3$.

(B) There are $10$ possible digits: $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$.
Since the last two digits are different,the total number of ways to choose and arrange these two digits is given by the permutation formula ${}^{10}P_2$.
${}^{10}P_2 = 10 \times 9 = 90$.
Out of these $90$ possible outcomes,only $1$ outcome is the correct sequence of the last two digits.
Therefore,the required probability is $\frac{1}{90}$.

(C) The total number of balls in the box is $2 + 3 + 4 = 9$.
The total number of ways to draw $3$ balls out of $9$ is given by ${}^9C_3 = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$.
For the balls to be of the same color,they must be either all black or all white (since there are only $2$ red balls,it is impossible to draw $3$ red balls).
The number of ways to draw $3$ black balls from $3$ is ${}^3C_3 = 1$.
The number of ways to draw $3$ white balls from $4$ is ${}^4C_3 = 4$.
Therefore,the total number of favorable ways is $1 + 4 = 5$.
The required probability is $\frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{5}{84}$.

(B) Total number of ways to arrange $6$ boys and $6$ girls in a row is $12!$.
To find the number of ways where all $6$ girls sit together,we treat the $6$ girls as a single unit. Now,we have $6$ boys and $1$ unit of girls,totaling $7$ entities,which can be arranged in $7!$ ways.
Within the unit,the $6$ girls can be arranged among themselves in $6!$ ways.
Thus,the number of favorable arrangements is $7! \times 6!$.
The required probability is $\frac{7! \times 6!}{12!} = \frac{7! \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{12 \times 11 \times 10 \times 9 \times 8 \times 7!} = \frac{720}{95040} = \frac{1}{132}$.

(B) Total number of ways to select $6$ persons out of $11$ is given by ${}^{11}C_6 = \frac{11 \times 10 \times 9 \times 8 \times 7 \times 6}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 462$.
To have exactly $2$ ladies in a committee of $6$,we must select $2$ ladies from $4$ and $4$ men from $7$.
Number of ways to select $2$ ladies from $4$ is ${}^4C_2 = \frac{4 \times 3}{2 \times 1} = 6$.
Number of ways to select $4$ men from $7$ is ${}^7C_4 = {}^7C_3 = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$.
Total favorable ways = ${}^4C_2 \times {}^7C_4 = 6 \times 35 = 210$.
Probability = $\frac{\text{Favorable ways}}{\text{Total ways}} = \frac{210}{462} = \frac{5}{11}$.

(A) Total number of balls = $4$ (white) + $3$ (red) = $7$ balls.
Probability of drawing the first red ball = $\frac{3}{7}$.
Since the ball is not replaced,the remaining number of balls is $6$,and the remaining number of red balls is $2$.
Probability of drawing the second red ball = $\frac{2}{6} = \frac{1}{3}$.
Therefore,the probability that both balls are red = $\frac{3}{7} \times \frac{1}{3} = \frac{1}{7}$.

(D) Total number of balls in the bag $= 5 + 7 + 4 = 16$.
We need to draw $3$ balls out of $16$. The total number of ways to select $3$ balls is given by $^{16}C_3 = \frac{16 \times 15 \times 14}{3 \times 2 \times 1} = 16 \times 5 \times 7 = 560$.
The number of ways to select $3$ white balls out of $5$ white balls is given by $^{5}C_3 = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10$.
The probability of drawing $3$ white balls is $P = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{10}{560} = \frac{1}{56}$.

(B) In a set of $40$ consecutive natural numbers,there are exactly $20$ odd numbers and $20$ even numbers.
For the sum of two chosen numbers to be odd,one number must be odd and the other must be even.
The total number of ways to choose $2$ numbers out of $40$ is given by ${}^{40}C_2 = \frac{40 \times 39}{2} = 780$.
The number of ways to choose one odd number and one even number is ${}^{20}C_1 \times {}^{20}C_1 = 20 \times 20 = 400$.
Therefore,the required probability is $P = \frac{400}{780} = \frac{40}{78} = \frac{20}{39}$.

(D) Total number of cards in a pack = $52$.
Number of red cards in a pack = $26$.
We need to draw $3$ cards from the $26$ red cards.
The number of ways to choose $3$ red cards out of $26$ is given by ${}^{26}C_3$.
The total number of ways to choose $3$ cards out of $52$ is given by ${}^{52}C_3$.
The probability $P$ is given by:
$P = \frac{{}^{26}C_3}{{}^{52}C_3} = \frac{\frac{26 \times 25 \times 24}{3 \times 2 \times 1}}{\frac{52 \times 51 \times 50}{3 \times 2 \times 1}} = \frac{26 \times 25 \times 24}{52 \times 51 \times 50}$
Simplifying the expression:
$P = \frac{26}{52} \times \frac{25}{50} \times \frac{24}{51} = \frac{1}{2} \times \frac{1}{2} \times \frac{8}{17} = \frac{8}{68} = \frac{2}{17}$.
Thus,the correct option is $D$.

(D) Total number of experts $= 2 + 3 + 4 = 9$.
We need to select $3$ experts out of $9$ to resign. The total number of ways to choose $3$ experts is given by ${}^9C_3 = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$.
We want the probability that the $3$ experts belong to different institutions. This means we must select $1$ expert from institution $A$,$1$ from $B$,and $1$ from $C$.
The number of ways to select $1$ expert from each is ${}^2C_1 \times {}^3C_1 \times {}^4C_1 = 2 \times 3 \times 4 = 24$.
The required probability is $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{24}{84}$.
Simplifying the fraction,$\frac{24}{84} = \frac{2}{7}$.

(A) Total number of ways to select two numbers from $100$ is ${}^{100}C_2 = \frac{100 \times 99}{2} = 4950$.
Let $S$ be the set $\{1, 2, 3, \dots, 100\}$. The numbers in $S$ divisible by $3$ are $\{3, 6, 9, \dots, 99\}$, which are $33$ in total. The numbers not divisible by $3$ are $100 - 33 = 67$.
A product of two numbers is divisible by $3$ if at least one of the selected numbers is divisible by $3$.
It is easier to calculate the complement: the product is $NOT$ divisible by $3$ if both selected numbers are $NOT$ divisible by $3$.
The number of ways to select two numbers such that neither is divisible by $3$ is ${}^{67}C_2 = \frac{67 \times 66}{2} = 2211$.
The number of ways the product is divisible by $3$ is $4950 - 2211 = 2739$.
The required probability is $\frac{2739}{4950} = 0.5533\dots$, which is $0.55$ correct to two decimal places.

(A) The set of available digits is ${1, 2, 3, 4, 5, 6, 8}$. There are $7$ digits in total.
Total number of $5$-digit numbers that can be formed using these $7$ digits (assuming repetition is not allowed) is $^7P_5 = 7 \times 6 \times 5 \times 4 \times 3 = 2520$.
Even digits available in the set are ${2, 4, 6, 8}$,which are $4$ in total.
For a $5$-digit number to have even digits at both ends,we must choose $2$ even digits for the two ends from the $4$ available even digits. The number of ways to arrange these is $^4P_2 = 4 \times 3 = 12$.
After placing $2$ even digits at the ends,we are left with $5$ digits (from the original $7$ digits minus the $2$ used). We need to fill the remaining $3$ positions using these $5$ digits. The number of ways to do this is $^5P_3 = 5 \times 4 \times 3 = 60$.
Total favorable outcomes = $12 \times 60 = 720$.
Probability = $\frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{720}{2520} = \frac{72}{252} = \frac{2}{7}$.

(A) Total number of balls $= 3 + 4 + 5 = 12$.
Number of ways to draw $3$ balls out of $12$ is given by $^{12}C_3 = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220$.
For the balls to be of different colours,we must draw $1$ red,$1$ white,and $1$ black ball.
Number of favorable outcomes $= ^3C_1 \times ^4C_1 \times ^5C_1 = 3 \times 4 \times 5 = 60$.
Probability $= \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{60}{220} = \frac{6}{22} = \frac{3}{11}$.

(D) regular octagon has $8$ vertices. The total number of ways to choose $4$ vertices out of $8$ is given by the combination formula $^nC_r = \frac{n!}{r!(n-r)!}$.
Total number of cases $= ^8C_4 = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70$.
In a regular octagon,a rectangle is formed by choosing $4$ vertices such that the diagonals of the rectangle are diameters of the circumscribed circle. There are only $2$ such rectangles possible in a regular octagon (as shown in the figure: $ADEH$ and $GFCB$).
Number of favourable cases $= 2$.
Therefore,the required probability $= \frac{\text{Number of favourable cases}}{\text{Total number of cases}} = \frac{2}{70} = \frac{1}{35}$.

(D) Total number of socks $= 5 + 4 = 9$.
The total number of ways to select $2$ socks out of $9$ is given by $^9C_2 = \frac{9 \times 8}{2 \times 1} = 36$.
The number of ways to select $2$ brown socks from $5$ is $^5C_2 = \frac{5 \times 4}{2 \times 1} = 10$.
The number of ways to select $2$ white socks from $4$ is $^4C_2 = \frac{4 \times 3}{2 \times 1} = 6$.
The number of ways to select $2$ socks of the same colour is $10 + 6 = 16$.
Therefore,the probability $P$ that both socks are of the same colour is $P = \frac{16}{36}$.
To match the options,we multiply the numerator and denominator by $3$: $P = \frac{16 \times 3}{36 \times 3} = \frac{48}{108}$.

(C) Total number of ways to choose a committee of $3$ people from $38$ is $\binom{38}{3}$.
If you are on the committee,then you have already been selected,and we need to choose the remaining $2$ members from the remaining $37$ people.
The number of ways to choose the remaining $2$ members is $\binom{37}{2}$.
Therefore,the probability that you will be on the committee is the ratio of the number of favorable outcomes to the total number of outcomes:
$P = \frac{\binom{37}{2}}{\binom{38}{3}}$.

(B) Total number of people $= 4 \text{ boys} + 3 \text{ girls} = 7 \text{ people}$.
Total ways to arrange $7$ people in a queue $= 7!$.
For the boys and girls to stand in alternate positions,the arrangement must be $B, G, B, G, B, G, B$.
Number of ways to arrange $4$ boys in the $4$ designated positions $= 4!$.
Number of ways to arrange $3$ girls in the $3$ designated positions $= 3!$.
Total favourable arrangements $= 4! \times 3!$.
Probability $= \frac{4! \times 3!}{7!} = \frac{24 \times 6}{5040} = \frac{144}{5040} = \frac{1}{35}$.

(A) Total number of ways to draw $5$ tickets from $90$ is $^{90}C_5$.
We need to draw $2$ specific tickets ($15$ and $89$) and $3$ other tickets from the remaining $88$ tickets.
The number of favorable outcomes is $^{2}C_2 \times ^{88}C_3 = 1 \times ^{88}C_3$.
The probability $P = \frac{^{88}C_3}{^{90}C_5}$.
$P = \frac{88 \times 87 \times 86}{3 \times 2 \times 1} \div \frac{90 \times 89 \times 88 \times 87 \times 86}{5 \times 4 \times 3 \times 2 \times 1}$.
$P = \frac{88 \times 87 \times 86}{6} \times \frac{120}{90 \times 89 \times 88 \times 87 \times 86}$.
$P = \frac{120}{90 \times 89 \times 6} = \frac{20}{89 \times 6} = \frac{10}{89 \times 3} = \frac{10}{267}$.
Wait,re-calculating: $P = \frac{88! / (3! 85!)}{90! / (5! 85!)} = \frac{88!}{3! 85!} \times \frac{5! 85!}{90!} = \frac{5 \times 4}{90 \times 89} = \frac{20}{8010} = \frac{2}{801}$.

(A) Total number of ways to select $11$ players from $15$ players is given by ${^{15}C_{11}}$.
Number of ways to select $6$ batsmen from $8$ batsmen is ${^8C_6}$.
Number of ways to select $5$ bowlers from $7$ bowlers is ${^7C_5}$.
Therefore,the number of favourable outcomes is ${^8C_6} \times {^7C_5}$.
The required probability is the ratio of favourable outcomes to total outcomes:
$P = \frac{{^8C_6} \times {^7C_5}}{{^{15}C_{11}}}$.

(D) Total number of balls in the bag = $5$ (black) + $4$ (white) + $3$ (red) = $12$ balls.
We need to find the probability of selecting a black or red ball.
Number of favorable outcomes (black or red) = $5 + 3 = 8$.
Total number of possible outcomes = $12$.
Probability $P(\text{Black or Red}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{8}{12} = \frac{2}{3}$.

(C) The total number of ways to choose $2$ integers from $30$ consecutive integers is given by $^{30}C_2 = \frac{30 \times 29}{2} = 435$.
For the sum of two numbers to be odd,one number must be even and the other must be odd.
In $30$ consecutive integers,there are $15$ even numbers and $15$ odd numbers.
The number of ways to choose one even number and one odd number is $^{15}C_1 \times ^{15}C_1 = 15 \times 15 = 225$.
The required probability is $\frac{\text{Favourable outcomes}}{\text{Total outcomes}} = \frac{225}{435}$.
Dividing both numerator and denominator by $15$,we get $\frac{225 \div 15}{435 \div 15} = \frac{15}{29}$.

(C) The total number of ways to choose $3$ integers from the first $20$ integers is $^{20}C_3 = \frac{20 \times 19 \times 18}{3 \times 2 \times 1} = 1140$.
The product of three integers is even if at least one of the chosen integers is even.
It is easier to calculate the probability of the complement event,which is that the product is odd. The product is odd if and only if all three chosen integers are odd.
There are $10$ odd integers and $10$ even integers in the first $20$ integers.
The number of ways to choose $3$ odd integers is $^{10}C_3 = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$.
The probability that the product is odd is $P(\text{odd}) = \frac{^{10}C_3}{^{20}C_3} = \frac{120}{1140} = \frac{12}{114} = \frac{2}{19}$.
Therefore,the probability that the product is even is $P(\text{even}) = 1 - P(\text{odd}) = 1 - \frac{2}{19} = \frac{17}{19}$.

(D) The total number of ways to select $2$ numbers from $6$ numbers without replacement where order matters is $P(6, 2) = 6 \times 5 = 30$.
Let $X$ and $Y$ be the two selected numbers. We want to find the probability that $\min(X, Y) < 4$.
It is easier to calculate the complement: the probability that $\min(X, Y) \ge 4$.
If $\min(X, Y) \ge 4$,then both numbers must be chosen from the set $\{4, 5, 6\}$.
The number of ways to choose $2$ numbers from $\{4, 5, 6\}$ without replacement is $P(3, 2) = 3 \times 2 = 6$.
The probability that $\min(X, Y) \ge 4$ is $\frac{6}{30} = \frac{1}{5}$.
Therefore,the probability that $\min(X, Y) < 4$ is $1 - \frac{1}{5} = \frac{4}{5}$.

(B) Total number of balls in the bag $= 6 + 7 + 5 = 18$.
The number of ways to draw $3$ balls from $18$ balls is given by ${}^{18}C_3 = \frac{18 \times 17 \times 16}{3 \times 2 \times 1} = 3 \times 17 \times 16 = 816$.
The number of ways to draw $3$ white balls from $6$ white balls is given by ${}^{6}C_3 = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$.
The required probability is the ratio of favourable outcomes to total outcomes:
$P = \frac{{}^{6}C_3}{{}^{18}C_3} = \frac{20}{816}$.
Simplifying the fraction by dividing both numerator and denominator by $4$:
$P = \frac{5}{204}$.

(D) Total number of balls = $4 + 5 + 6 = 15$.
We need to select $3$ balls out of $15$. The total number of ways to select $3$ balls is given by ${}^{15}C_3 = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 5 \times 7 \times 13 = 455$.
The number of ways to select $1$ white,$1$ red,and $1$ green ball is ${}^4C_1 \times {}^5C_1 \times {}^6C_1 = 4 \times 5 \times 6 = 120$.
The required probability is $\frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{120}{455}$.
Dividing both numerator and denominator by $5$,we get $\frac{24}{91}$.

(B) Total number of balls $= 10 + 15 = 25$.
We need to draw one red ball and one green ball in two draws.
The number of ways to choose $1$ red ball from $10$ and $1$ green ball from $15$ is given by $10 \times 15 = 150$.
The total number of ways to choose $2$ balls from $25$ is given by $^{25}C_2 = \frac{25 \times 24}{2} = 300$.
The probability is $\frac{150}{300} = \frac{1}{2}$.
Alternatively,considering the order: (Red then Green) or (Green then Red).
$P(RG) + P(GR) = (\frac{10}{25} \times \frac{15}{24}) + (\frac{15}{25} \times \frac{10}{24}) = \frac{150}{600} + \frac{150}{600} = \frac{300}{600} = \frac{1}{2}$.

(C) The total number of ways to draw $3$ cards from a pack of $52$ cards is given by ${}^{52}C_3$.
${}^{52}C_3 = \frac{52 \times 51 \times 50}{3 \times 2 \times 1} = 52 \times 17 \times 25 = 22100$.
There are $4$ aces in a pack of $52$ cards. The number of ways to draw $3$ aces from $4$ is given by ${}^{4}C_3$.
${}^{4}C_3 = \frac{4 \times 3 \times 2}{3 \times 2 \times 1} = 4$.
The probability of drawing $3$ aces is $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{{}^{4}C_3}{{}^{52}C_3} = \frac{4}{22100} = \frac{1}{5525}$.

(A) Total number of cards = $4 + 4 + 4 + 4 = 16$.
Number of cards drawn = $2$.
Total ways to draw $2$ cards from $16$ is ${}^{16}C_2 = \frac{16 \times 15}{2 \times 1} = 120$.
Number of non-ace cards = $16 - 4 = 12$.
Ways to draw $2$ cards such that none of them is an ace = ${}^{12}C_2 = \frac{12 \times 11}{2 \times 1} = 66$.
Probability of drawing no ace = $\frac{66}{120} = \frac{11}{20}$.
Probability of drawing at least one ace = $1 - P(\text{No Ace}) = 1 - \frac{11}{20} = \frac{9}{20}$.

(A) The total number of possible outcomes when a coin is tossed $100$ times is $2^{100}$.
The number of ways to get tails an odd number of times is given by the sum of combinations: ${}^{100}C_1 + {}^{100}C_3 + {}^{100}C_5 + \dots + {}^{100}C_{99}$.
Using the binomial identity,the sum of odd-indexed binomial coefficients is ${}^{n}C_1 + {}^{n}C_3 + \dots = 2^{n-1}$.
For $n = 100$,the number of favourable outcomes is $2^{100-1} = 2^{99}$.
Therefore,the required probability is $\frac{2^{99}}{2^{100}} = \frac{1}{2}$.

(A) Total number of balls $= 3 + 4 + 5 = 12$.
Number of ways to draw $2$ balls from $12$ is ${}^{12}C_2 = \frac{12 \times 11}{2} = 66$.
To find the probability that the balls are of different colours,it is easier to calculate the probability that they are of the same colour and subtract it from $1$.
Number of ways to draw $2$ balls of the same colour:
- Both red: ${}^{3}C_2 = 3$
- Both white: ${}^{4}C_2 = 6$
- Both blue: ${}^{5}C_2 = 10$
Total ways for same colour $= 3 + 6 + 10 = 19$.
Probability of same colour $= \frac{19}{66}$.
Probability of different colours $= 1 - \frac{19}{66} = \frac{66 - 19}{66} = \frac{47}{66}$.

(A) Total number of ways to arrange $10$ students in a row is $10!$.
To find the probability that two particular students sit together,we treat them as a single unit. Now,we have $9$ units (the pair + $8$ other students),which can be arranged in $9!$ ways. Within the pair,the two students can be arranged in $2! = 2$ ways.
Therefore,the number of ways they sit together is $2 \times 9!$.
The probability that they sit together is $P(\text{together}) = \frac{2 \times 9!}{10!} = \frac{2 \times 9!}{10 \times 9!} = \frac{2}{10} = \frac{1}{5}$.
The probability that they are not seated side by side is $P(\text{not together}) = 1 - P(\text{together}) = 1 - \frac{1}{5} = \frac{4}{5}$.

(A) Total number of socks = $5 + 4 = 9$.
The total number of ways to choose $2$ socks out of $9$ is given by ${}^9C_2 = \frac{9 \times 8}{2 \times 1} = 36$.
The socks will match if either both are brown or both are blue.
Number of ways to choose $2$ brown socks from $5$ is ${}^5C_2 = \frac{5 \times 4}{2 \times 1} = 10$.
Number of ways to choose $2$ blue socks from $4$ is ${}^4C_2 = \frac{4 \times 3}{2 \times 1} = 6$.
Total favorable outcomes = $10 + 6 = 16$.
Therefore,the required probability = $\frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{16}{36} = \frac{4}{9}$.

(C) Total number of ways to arrange $5$ persons in a queue is $5! = 120$.
To find the number of favourable ways where $A$ and $E$ are always together,we treat $(AE)$ as a single unit.
Now,we have $4$ units to arrange: $(AE), B, C, D$. These can be arranged in $4!$ ways.
Within the unit $(AE)$,$A$ and $E$ can be arranged in $2! = 2$ ways.
So,the total number of favourable ways is $2 \times 4! = 2 \times 24 = 48$.
The required probability is $\frac{\text{Favourable outcomes}}{\text{Total outcomes}} = \frac{2 \times 4!}{5!} = \frac{2 \times 24}{120} = \frac{48}{120} = \frac{2}{5}$.

(C) Total number of balls = $8 + 7 = 15$.
Number of ways to draw $2$ balls from $15$ is $^{15}C_2 = \frac{15 \times 14}{2 \times 1} = 105$.
For both balls to be of the same colour,they must either both be red or both be black.
Number of ways to draw $2$ red balls from $8$ is $^8C_2 = \frac{8 \times 7}{2 \times 1} = 28$.
Number of ways to draw $2$ black balls from $7$ is $^7C_2 = \frac{7 \times 6}{2 \times 1} = 21$.
Total favorable outcomes = $28 + 21 = 49$.
Required probability = $\frac{49}{105} = \frac{7}{15}$.

(B) Total number of cards = $80$.
Numbers divisible by $4$ between $1$ and $80$ are $4, 8, 12, \dots, 80$. This is an arithmetic progression where $a = 4$, $d = 4$, and $l = 80$.
Using $l = a + (n - 1)d$, we get $80 = 4 + (n - 1)4$, which implies $n = 20$.
Total ways to select $2$ cards from $80$ is $^{80}C_2 = \frac{80 \times 79}{2} = 3160$.
Favorable ways to select $2$ cards from the $20$ cards divisible by $4$ is $^{20}C_2 = \frac{20 \times 19}{2} = 190$.
Required probability $P = \frac{190}{3160} = \frac{19}{316}$.

(D) Let $B_1$ be the first basket and $B_2$ be the second basket.
Total fruits in $B_1 = 5 + 7 = 12$.
Total fruits in $B_2 = 4 + 8 = 12$.
Probability of picking an apple from $B_1 = \frac{5}{12}$.
Probability of picking an apple from $B_2 = \frac{4}{12}$.
Probability of picking an orange from $B_1 = \frac{7}{12}$.
Probability of picking an orange from $B_2 = \frac{8}{12}$.
Since the events are independent,the probability that both are apples is $\frac{5}{12} \times \frac{4}{12} = \frac{20}{144}$.
The probability that both are oranges is $\frac{7}{12} \times \frac{8}{12} = \frac{56}{144}$.
Since these events are mutually exclusive,the total probability is $\frac{20}{144} + \frac{56}{144} = \frac{76}{144}$.

(B) The total number of functions (mappings) from a set $A$ with $m$ elements to a set $B$ with $n$ elements is given by $n^m$.
An injection (one-to-one function) exists only if $m \le n$. The number of ways to choose $m$ distinct elements from $B$ and arrange them is given by the permutation formula $P(n, m) = \frac{n!}{(n - m)!}$.
Therefore,the number of injective mappings is $\frac{n!}{(n - m)!}$.
The probability $P$ of selecting an injection is the ratio of the number of injective mappings to the total number of mappings:
$P = \frac{\text{Number of injective mappings}}{\text{Total number of mappings}} = \frac{\frac{n!}{(n - m)!}}{n^m} = \frac{n!}{(n - m)! n^m}$.

(A) Total number of ways to select $2$ persons out of $n$ is given by $^nC_2 = \frac{n(n-1)}{2}$.
To find the number of ways they are together,we treat the $2$ adjacent persons as a single unit. There are $(n-1)$ such pairs in a row of $n$ persons.
So,the number of ways they are together is $(n-1)$.
The probability that they are together is $P(\text{together}) = \frac{n-1}{^nC_2} = \frac{n-1}{\frac{n(n-1)}{2}} = \frac{2}{n}$.
The probability that they are not together is $P(\text{not together}) = 1 - P(\text{together}) = 1 - \frac{2}{n}$.

(D) Let $A$ occupy any seat at the round table. There are $14$ remaining seats available for $B$.
For there to be $4$ persons between $A$ and $B$,$B$ must occupy one of the two specific seats (one on the left side of $A$ and one on the right side of $A$) such that there are exactly $4$ people in the gap between them.
Since there are $14$ total possible positions for $B$ and $2$ favorable positions,the required probability is $P = \frac{2}{14} = \frac{1}{7}$.

(B) The total number of ways to arrange $10$ people ( $5$ boys and $5$ girls) in a row is $10!$.
For the boys and girls to sit alternatively,there are two possible patterns:
$1$. Boy,Girl,Boy,Girl,Boy,Girl,Boy,Girl,Boy,Girl
$2$. Girl,Boy,Girl,Boy,Girl,Boy,Girl,Boy,Girl,Boy
In each pattern,the $5$ boys can be arranged in $5!$ ways and the $5$ girls can be arranged in $5!$ ways.
Therefore,the number of favorable ways is $2 \times (5! \times 5!)$.
The probability is given by the ratio of favorable outcomes to total outcomes:
$P = \frac{2 \times 5! \times 5!}{10!}$
$P = \frac{2 \times 120 \times 120}{3628800} = \frac{2 \times 120}{10 \times 9 \times 8 \times 7 \times 6} = \frac{2}{30240} \times 120 = \frac{2}{252} = \frac{1}{126}$.

(B) Let $A$ be the event that the sum of the numbers is $11$. The possible outcomes for a sum of $11$ are $(5, 6)$ and $(6, 5)$.
Let $B$ be the event that $5$ appears on the first die. The possible outcomes for $B$ are $(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)$.
We need to find the conditional probability $P(A|B) = \frac{P(A \cap B)}{P(B)}$.
The intersection $A \cap B$ is the event where the sum is $11$ $AND$ $5$ is on the first die,which is only the outcome $(5, 6)$.
Thus,$n(A \cap B) = 1$ and $n(B) = 6$.
The required probability is $\frac{n(A \cap B)}{n(B)} = \frac{1}{6}$.

(C) The formula for conditional probability is given by $P(B/A) = \frac{P(A \cap B)}{P(A)}$.
Given values are $P(A) = \frac{1}{2}$ and $P(A \cap B) = \frac{1}{4}$.
Substituting these values into the formula:
$P(B/A) = \frac{1/4}{1/2} = \frac{1}{4} \times \frac{2}{1} = \frac{2}{4} = \frac{1}{2}$.
Therefore,the correct option is $C$.

(C) By the definition of conditional probability,$P\left( \frac{\overline{A}}{\overline{B}} \right) = \frac{P(\overline{A} \cap \overline{B})}{P(\overline{B})}$.
Using De Morgan's Law,we know that $\overline{A} \cap \overline{B} = \overline{A \cup B}$.
Therefore,$P(\overline{A} \cap \overline{B}) = P(\overline{A \cup B}) = 1 - P(A \cup B)$.
Substituting this into the expression,we get $P\left( \frac{\overline{A}}{\overline{B}} \right) = \frac{1 - P(A \cup B)}{P(\overline{B})}$.

(B) Let $A$ be the event that $4$ appears on the first die.
The sample space for the first die showing $4$ is: $(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)$.
There are $6$ possible outcomes in this reduced sample space.
Let $B$ be the event that the sum of the numbers is greater than $7$.
We look for outcomes $(4, x)$ such that $4 + x > 7$,which implies $x > 3$.
The favorable outcomes are $(4, 4), (4, 5), (4, 6)$.
There are $3$ favorable outcomes.
The required probability is $\frac{\text{Number of favorable outcomes}}{\text{Total outcomes in reduced sample space}} = \frac{3}{6} = \frac{1}{2}$.

(A) We are given $P(A) = \frac{1}{3}$,$P(B) = \frac{1}{4}$,and $P(A \cap B) = \frac{1}{5}$.
First,calculate $P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{1}{3} + \frac{1}{4} - \frac{1}{5} = \frac{20 + 15 - 12}{60} = \frac{23}{60}$.
We need to find $P\left( \frac{\overline{B}}{\overline{A}} \right)$. By the definition of conditional probability,$P\left( \frac{\overline{B}}{\overline{A}} \right) = \frac{P(\overline{B} \cap \overline{A})}{P(\overline{A})}$.
Using De Morgan's Law,$\overline{B} \cap \overline{A} = \overline{A \cup B}$,so $P(\overline{B} \cap \overline{A}) = 1 - P(A \cup B) = 1 - \frac{23}{60} = \frac{37}{60}$.
Also,$P(\overline{A}) = 1 - P(A) = 1 - \frac{1}{3} = \frac{2}{3}$.
Therefore,$P\left( \frac{\overline{B}}{\overline{A}} \right) = \frac{37/60}{2/3} = \frac{37}{60} \times \frac{3}{2} = \frac{37}{40}$.

(A) We are given the probabilities $P(A) = \frac{3}{8}$,$P(B) = \frac{5}{8}$,and $P(A \cup B) = \frac{3}{4}$.
Using the addition rule of probability: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Substituting the values: $\frac{3}{4} = \frac{3}{8} + \frac{5}{8} - P(A \cap B)$.
$\frac{3}{4} = \frac{8}{8} - P(A \cap B) = 1 - P(A \cap B)$.
Therefore,$P(A \cap B) = 1 - \frac{3}{4} = \frac{1}{4}$.
Now,the conditional probability $P(A|B)$ is given by the formula: $P(A|B) = \frac{P(A \cap B)}{P(B)}$.
Substituting the values: $P(A|B) = \frac{1/4}{5/8} = \frac{1}{4} \times \frac{8}{5} = \frac{2}{5}$.

(A) The conditional probability is defined as $P\left( \frac{A}{B} \right) = \frac{P(A \cap B)}{P(B)}$.
Since the events $A$ and $B$ are mutually exclusive,they cannot occur simultaneously.
Therefore,the probability of their intersection is $P(A \cap B) = 0$.
Substituting this value into the formula,we get $P\left( \frac{A}{B} \right) = \frac{0}{P(B)} = 0$ (provided $P(B) \neq 0$).

(B) Given that $A \subseteq B$.
This implies that whenever event $A$ occurs,event $B$ must also occur.
Therefore,the intersection of $A$ and $B$ is $A \cap B = A$.
By the definition of conditional probability,$P\left( \frac{B}{A} \right) = \frac{P(A \cap B)}{P(A)}$.
Substituting $A \cap B = A$ into the formula,we get $P\left( \frac{B}{A} \right) = \frac{P(A)}{P(A)} = 1$ (provided $P(A) > 0$).

(C) By the definition of conditional probability,$P(A/B) = \frac{P(A \cap B)}{P(B)}$.
Since $A$ and $B$ are independent events,the probability of their intersection is $P(A \cap B) = P(A) \cdot P(B)$.
Substituting this into the formula,we get $P(A/B) = \frac{P(A) \cdot P(B)}{P(B)}$.
Canceling $P(B)$ from the numerator and denominator,we obtain $P(A/B) = P(A)$.