Permutation and Combination based Probability Questions in English

(A) Total number of ways to draw $2$ cards from $52$ is $^{52}C_2 = \frac{52 \times 51}{2} = 1326$.
Number of ways to draw $2$ cards such that none of them is an ace (i.e.,choosing from $48$ non-ace cards) is $^{48}C_2 = \frac{48 \times 47}{2} = 1128$.
Probability of drawing no ace is $P(\text{no ace}) = \frac{1128}{1326} = \frac{47 \times 24}{51 \times 13} = \frac{47 \times 8}{17 \times 13} = \frac{376}{663} = \frac{48}{52} \times \frac{47}{51} = \frac{12}{13} \times \frac{47}{51} = \frac{4}{13} \times \frac{47}{17} = \frac{188}{221}$.
Therefore,the probability that at least one card is an ace is $1 - P(\text{no ace}) = 1 - \frac{188}{221} = \frac{221 - 188}{221} = \frac{33}{221}$.

(A) The total number of ways to place $n$ letters into $n$ envelopes is $n!$.
There is only $1$ way to place each letter into its correct corresponding envelope.
Therefore,the probability $P$ is given by the ratio of the number of favorable outcomes to the total number of outcomes.
$P = \frac{1}{n!}$

(A) Total number of balls $= 5 + 7 + 8 = 20$.
Since four balls are drawn one by one without replacement,the probability that all four are white is given by:
$P = \frac{5}{20} \times \frac{4}{19} \times \frac{3}{18} \times \frac{2}{17}$
$P = \frac{1}{4} \times \frac{4}{19} \times \frac{1}{6} \times \frac{2}{17}$
$P = \frac{1 \times 1 \times 1 \times 2}{19 \times 6 \times 17} = \frac{2}{1938} = \frac{1}{969}$.

(A) The total number of two-digit numbers that can be formed using the digits $\{1, 2, 3, 4, 5\}$ with repetition allowed is $5 \times 5 = 25$.
$A$ number is divisible by $4$ if the number formed by its last two digits is divisible by $4$. For a two-digit number,the number itself must be divisible by $4$.
The possible two-digit numbers formed by these digits are:
$12, 16$ (not possible),$24, 32, 44, 52$.
Checking divisibility by $4$:
$12 \div 4 = 3$
$24 \div 4 = 6$
$32 \div 4 = 8$
$44 \div 4 = 11$
$52 \div 4 = 13$
The favourable outcomes are $\{12, 24, 32, 44, 52\}$.
Number of favourable outcomes $= 5$.
Required probability $= \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} = \frac{5}{25} = \frac{1}{5}$.

(B) Let $A$ be the event that one card is an ace of hearts and the other is not.
There are two mutually exclusive cases:
$(i)$ The first card is an ace of hearts and the second is not: $P(E_1) = \frac{1}{52} \times \frac{51}{51} = \frac{1}{52}$.
$(ii)$ The first card is not an ace of hearts and the second is an ace of hearts: $P(E_2) = \frac{51}{52} \times \frac{1}{51} = \frac{1}{52}$.
Therefore,the required probability is $P(E_1) + P(E_2) = \frac{1}{52} + \frac{1}{52} = \frac{2}{52} = \frac{1}{26}$.

(D) Total number of ways to select $2$ horses out of $5$ is given by $^5C_2 = \frac{5 \times 4}{2 \times 1} = 10$.
There is only $1$ winning horse. The number of ways to select $2$ horses such that the winning horse is included is equivalent to choosing the winning horse and one other horse from the remaining $4$ horses,which is $^4C_1 = 4$.
Therefore,the probability that $Mr. A$ selected the winning horse is $\frac{4}{10} = \frac{2}{5}$.

(C) The total number of ways to draw $2$ cards from $52$ cards is given by $^{52}C_2$.
The number of ways to draw $2$ cards of spade from $13$ spade cards is given by $^{13}C_2$.
The required probability is $P = \frac{^{13}C_2}{^{52}C_2}$.
$P = \frac{\frac{13 \times 12}{2 \times 1}}{\frac{52 \times 51}{2 \times 1}} = \frac{13 \times 12}{52 \times 51} = \frac{1 \times 12}{4 \times 51} = \frac{12}{204} = \frac{1}{17}$.
Thus,the correct option is $C$.

(B) The total number of ways to choose $2$ integers from $20$ consecutive integers is given by ${}^{20}C_2 = \frac{20 \times 19}{2} = 190$.
The sum of two numbers is odd if and only if one number is even and the other is odd.
In any set of $20$ consecutive integers,there are exactly $10$ even numbers and $10$ odd numbers.
The number of ways to choose one even number and one odd number is ${}^{10}C_1 \times {}^{10}C_1 = 10 \times 10 = 100$.
Therefore,the required probability is $P = \frac{100}{190} = \frac{10}{19}$.

(C) The total number of ways to draw $2$ tickets from $25$ is $^{25}C_2 = \frac{25 \times 24}{2} = 300$.
For the product of two numbers to be even,at least one of the numbers must be even.
It is easier to calculate the complement: the product is odd only if both numbers drawn are odd.
There are $13$ odd numbers $(1, 3, 5, \dots, 25)$ and $12$ even numbers $(2, 4, 6, \dots, 24)$ in the set.
The number of ways to choose $2$ odd tickets is $^{13}C_2 = \frac{13 \times 12}{2} = 78$.
The probability that both are odd is $P(\text{odd}) = \frac{78}{300} = \frac{13}{50}$.
Therefore,the probability that the product is even is $1 - P(\text{odd}) = 1 - \frac{13}{50} = \frac{37}{50}$.

(A) Total number of students $= 12 + 18 = 30$.
Number of ways to choose $2$ students out of $30$ is given by $^{30}C_2 = \frac{30 \times 29}{2 \times 1} = 435$.
Number of ways to choose $2$ girls out of $12$ is given by $^{12}C_2 = \frac{12 \times 11}{2 \times 1} = 66$.
Required probability $= \frac{^{12}C_2}{^{30}C_2} = \frac{66}{435} = \frac{22}{145}$.

(B) Total number of letters = $11$.
Number of consonants = $7$.
Number of ways to choose $2$ letters out of $11$ is given by $^{11}C_2 = \frac{11 \times 10}{2 \times 1} = 55$.
Number of ways to choose $2$ consonants out of $7$ is given by $^{7}C_2 = \frac{7 \times 6}{2 \times 1} = 21$.
Therefore,the required probability is $P = \frac{^{7}C_2}{^{11}C_2} = \frac{21}{55}$.

(A) The total number of ways to draw $3$ tickets out of $20$ is given by $^{20}C_3 = \frac{20 \times 19 \times 18}{3 \times 2 \times 1} = 1140$.
We want to find the probability that the tickets marked $7$ and $11$ are included in the selection.
If $7$ and $11$ are already chosen,we only need to choose $1$ more ticket from the remaining $18$ tickets $(20 - 2 = 18)$.
The number of ways to choose this $3^{rd}$ ticket is $^{18}C_1 = 18$.
Therefore,the required probability is $\frac{18}{1140} = \frac{18}{20 \times 19 \times 18 / 6} = \frac{18}{1140} = \frac{3}{190}$.

(A) Total number of balls $= 4 + 5 + 6 = 15$.
Number of ways to draw $2$ balls from $15$ is $^{15}C_2 = \frac{15 \times 14}{2} = 105$.
We want the probability that exactly one ball is white. This means one ball is white and the other is non-white (red or black).
Number of white balls $= 4$.
Number of non-white balls $= 5 + 6 = 11$.
Number of ways to choose $1$ white ball and $1$ non-white ball $= ^4C_1 \times ^{11}C_1 = 4 \times 11 = 44$.
Required probability $= \frac{44}{105}$.

(A) Total number of tickets $= 50$.
Number of prize-winning tickets $= 14$.
Number of blank tickets $= 50 - 14 = 36$.
The man buys $2$ tickets.
The total number of ways to choose $2$ tickets from $50$ is ${}^{50}C_2 = \frac{50 \times 49}{2 \times 1} = 1225$.
The number of ways to choose $2$ tickets such that none of them are prize-winning (both are blank) is ${}^{36}C_2 = \frac{36 \times 35}{2 \times 1} = 630$.
The probability of not winning any prize is $P(\text{no prize}) = \frac{630}{1225} = \frac{18}{35}$.
The probability of winning at least one prize is $P(\text{at least one prize}) = 1 - P(\text{no prize}) = 1 - \frac{18}{35} = \frac{17}{35}$.

(C) The total number of three-digit numbers that can be formed using the digits $1, 2, 3,$ and $4$ without repetition is given by $^4P_3 = 4 \times 3 \times 2 = 24$.
$A$ number is divisible by $3$ if the sum of its digits is divisible by $3$.
Possible sets of three digits from ${1, 2, 3, 4}$ whose sum is divisible by $3$ are:
$1 + 2 + 3 = 6$ (divisible by $3$)
$2 + 3 + 4 = 9$ (divisible by $3$)
For the set ${1, 2, 3}$,the number of arrangements is $3! = 6$.
For the set ${2, 3, 4}$,the number of arrangements is $3! = 6$.
Total favorable outcomes = $6 + 6 = 12$.
Required probability = $\frac{12}{24} = \frac{1}{2}$.

(A) The total number of tickets is $20$. The number of ways to draw $2$ tickets from $20$ is given by $^{20}C_2 = \frac{20 \times 19}{2 \times 1} = 190$.
The prime numbers between $1$ and $20$ are $2, 3, 5, 7, 11, 13, 17, 19$. There are $8$ such prime numbers.
The number of ways to choose $2$ prime numbers from these $8$ is given by $^8C_2 = \frac{8 \times 7}{2 \times 1} = 28$.
The probability that both numbers are prime is $\frac{28}{190} = \frac{14}{95}$.

(A) Total number of balls $= 6 + 5 + 4 = 15$.
Number of non-red balls $= 5 + 4 = 9$.
We need to draw $2$ balls from the $9$ non-red balls.
The number of ways to choose $2$ balls from $15$ is $^{15}C_2 = \frac{15 \times 14}{2} = 105$.
The number of ways to choose $2$ balls from $9$ non-red balls is $^{9}C_2 = \frac{9 \times 8}{2} = 36$.
Required probability $= \frac{36}{105} = \frac{12}{35}$.

(B) Total number of ways to arrange $m$ rupee coins and $n$ ten paise coins in a line is given by the number of permutations of $m+n$ objects where $m$ are of one type and $n$ are of another type: $\frac{(m + n)!}{m! n!}$.
If the extreme coins are ten paise coins,we fix two ten paise coins at the two ends. The remaining $(m + n - 2)$ coins consist of $m$ rupee coins and $(n - 2)$ ten paise coins.
The number of ways to arrange these remaining coins is $\frac{(m + n - 2)!}{m! (n - 2)!}$.
Therefore,the required probability is:
$P = \frac{\frac{(m + n - 2)!}{m! (n - 2)!}}{\frac{(m + n)!}{m! n!}}$
$P = \frac{(m + n - 2)!}{m! (n - 2)!} \times \frac{m! n!}{(m + n)!}$
$P = \frac{n(n - 1)}{(m + n)(m + n - 1)}$.

(A) Total number of balls $= 4 + 3 = 7$.
Probability of drawing the first red ball $= \frac{3}{7}$.
Since the draw is without replacement,the number of red balls remaining is $2$ and the total number of balls remaining is $6$.
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 $= 5 + 7 + 4 = 16$.
Number of ways to draw $3$ balls out of $16$ is given by $^{16}C_3 = \frac{16 \times 15 \times 14}{3 \times 2 \times 1} = 16 \times 5 \times 7 = 560$.
Number of ways to draw $3$ white balls out of $5$ is given by $^{5}C_3 = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10$.
Required probability $= \frac{^{5}C_3}{^{16}C_3} = \frac{10}{560} = \frac{1}{56}$.

(A) Total number of ways to select two distinct numbers from $100$ is ${}^{100}C_2 = \frac{100 \times 99}{2} = 4950$.
Let $S$ be the set ${1, 2, \dots, 100}$. The numbers in $S$ divisible by $3$ are ${3, 6, \dots, 99}$,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 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 $0.55$ when rounded to two decimal places.

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

(C) The total number of ways to choose a committee of $3$ people from $38$ is $\binom{38}{3}$.
If you are on the committee,the remaining $2$ members must be chosen from the remaining $37$ people.
The number of ways to choose the committee such that you are included is $\binom{37}{2}$.
Therefore,the probability is $P = \frac{\binom{37}{2}}{\binom{38}{3}}$.

(B) Total number of ways to arrange $4$ boys and $3$ girls in a queue is $7!$.
For them to be 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 is $4!$.
Number of ways to arrange $3$ girls in the $3$ designated positions is $3!$.
Favourable cases $= 4! \times 3!$.
Required probability $= \frac{4! \times 3!}{7!} = \frac{24 \times 6}{5040} = \frac{144}{5040} = \frac{1}{35}$.

(C) The total number of ways to choose $2$ integers from $30$ consecutive integers is given by $^{30}C_2 = \frac{30 \times 29}{2} = 435$.
In $30$ consecutive integers,there are $15$ even numbers and $15$ odd numbers.
The sum of two numbers is odd if and only if one number is even and the other is odd.
The number of ways to choose one even number and one odd number is $^{15}C_1 \times ^{15}C_1 = 15 \times 15 = 225$.
Therefore,the required probability is $P = \frac{225}{435} = \frac{15}{29}$.

(D) The total number of ways to select $2$ numbers from $6$ numbers without replacement is $P(6, 2) = 6 \times 5 = 30$.
Let the two numbers be $x$ and $y$. We want the probability that $\min(x, y) < 4$.
It is easier to calculate the complement: the probability that $\min(x, y) \geq 4$.
This means both numbers must be chosen from the set $\{4, 5, 6\}$.
The number of ways to choose $2$ numbers from $\{4, 5, 6\}$ is $P(3, 2) = 3 \times 2 = 6$.
The probability that $\min(x, y) \geq 4$ 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 $= 6 + 7 + 5 = 18$.
Number of ways to draw $3$ balls from $18$ balls is given by ${}^{18}C_3 = \frac{18 \times 17 \times 16}{3 \times 2 \times 1} = 3 \times 17 \times 16 = 816$.
Number of ways to draw $3$ white balls from $6$ white balls is given by ${}^{6}C_3 = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$.
Required probability $= \frac{\text{Favourable cases}}{\text{Total cases}} = \frac{20}{816}$.
Simplifying the fraction: $\frac{20}{816} = \frac{5}{204}$.

(B) Total number of balls $= 10 + 15 = 25$.
We need to draw one red ball and one green ball in two draws.
The possible cases are (Red,Green) or (Green,Red).
Probability $= P(RG) + P(GR) = \left( \frac{10}{25} \times \frac{15}{24} \right) + \left( \frac{15}{25} \times \frac{10}{24} \right)$.
Probability $= \left( \frac{150}{600} \right) + \left( \frac{150}{600} \right) = \frac{300}{600} = \frac{1}{2}$.

(C) The 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 $4$ units can be arranged in $4!$ ways.
Within the unit $(AE)$,$A$ and $E$ can be arranged in $2!$ ways.
So,the total favourable ways $= 2! \times 4! = 2 \times 24 = 48$.
The required probability $= \frac{\text{Favourable ways}}{\text{Total ways}} = \frac{48}{120} = \frac{2}{5}$.

(A) Total number of ways to select $2$ persons out of $n$ is $^nC_2 = \frac{n(n-1)}{2}$.
To find the number of ways they are together,we treat the $2$ persons as a single unit. There are $(n-1)$ such units,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 $P(\text{not together}) = 1 - P(\text{together}) = 1 - \frac{2}{n}$.

(B) Total number of ways to arrange $10$ people in a row is $n = 10!$.
For boys and girls to sit alternatively,they must occupy positions like $(B, G, B, G, B, G, B, G, B, G)$ or $(G, B, G, B, G, B, G, B, G, B)$.
Case $1$: Starting with a boy,the number of arrangements is $5! \times 5!$.
Case $2$: Starting with a girl,the number of arrangements is $5! \times 5!$.
Total favourable ways $m = 2 \times (5! \times 5!)$.
Required probability $P = \frac{m}{n} = \frac{2 \times 5! \times 5!}{10!}$.
$P = \frac{2 \times 120 \times 120}{3628800} = \frac{2 \times 120}{10 \times 9 \times 8 \times 7 \times 6} = \frac{2}{30240} \times 120 = \frac{2}{252} = \frac{1}{126}$.

(B) Let the four machines be $M_1, M_2, F_1, F_2$,where $F$ denotes a faulty machine and $M$ denotes a working machine.
We need to find the probability that both faulty machines are identified in exactly two tests.
This means the first test must result in a faulty machine,and the second test must also result in a faulty machine.
The total number of ways to choose $2$ machines out of $4$ for the first two tests is $^4P_2 = 4 \times 3 = 12$.
The number of ways to choose $2$ faulty machines out of $2$ for the first two tests is $^2P_2 = 2 \times 1 = 2$.
Therefore,the required probability is $\frac{2}{12} = \frac{1}{6}$.
Alternatively,the probability that the first machine tested is faulty is $\frac{2}{4}$.
Given that the first machine was faulty,the probability that the second machine tested is also faulty is $\frac{1}{3}$.
Thus,the required probability is $\frac{2}{4} \times \frac{1}{3} = \frac{1}{6}$.

(A) The set of available digits is $S = \{1, 2, 3, 4, 5, 6, 8\}$,which contains $7$ digits in total.
Among these,the even digits are $\{2, 4, 6, 8\}$,so there are $4$ even digits and $3$ odd digits.
The total number of $5$-digit numbers that can be formed using these $7$ digits is $P(7, 5) = \frac{7!}{2!} = 7 \times 6 \times 5 \times 4 \times 3 = 2520$.
For the number to have even digits at both ends,we must choose $2$ even digits out of $4$ for the two ends and arrange them in $P(4, 2) = 4 \times 3 = 12$ ways.
The remaining $3$ positions can be filled by the remaining $5$ digits in $P(5, 3) = 5 \times 4 \times 3 = 60$ ways.
Thus,the number of favorable outcomes is $12 \times 60 = 720$.
The probability is $\frac{720}{2520} = \frac{72}{252} = \frac{2}{7}$.

(A) Let the letters be $L_1, L_2, L_3$ and their corresponding envelopes be $E_1, E_2, E_3$.
The total number of ways to place $3$ letters in $3$ envelopes is $3! = 6$.
The possible arrangements are: $(1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), (3,2,1)$.
We want exactly one letter in the correct envelope.
If $L_1$ is in $E_1$,then $L_2$ and $L_3$ must be swapped ($L_2$ in $E_3$,$L_3$ in $E_2$). This is $1$ way.
If $L_2$ is in $E_2$,then $L_1$ and $L_3$ must be swapped. This is $1$ way.
If $L_3$ is in $E_3$,then $L_1$ and $L_2$ must be swapped. This is $1$ way.
Total favorable ways $= 1 + 1 + 1 = 3$.
The probability is $\frac{\text{Favorable ways}}{\text{Total ways}} = \frac{3}{6} = \frac{1}{2}$.

(A) Total number of balls $= 5 + 3 = 8$.
Total ways to draw $2$ balls from $8$ balls is given by $^8C_2 = \frac{8 \times 7}{2 \times 1} = 28$.
Number of ways to select $1$ white ball from $5$ and $1$ black ball from $3$ is $^5C_1 \times ^3C_1 = 5 \times 3 = 15$.
Therefore,the probability $P = \frac{15}{28}$.

(B) Total number of ways to arrange $7$ people in a row is $7!$.
For the boys and girls to stand in alternating positions,the arrangement must be $B G B G B G B$.
Number of ways to arrange $4$ boys in $4$ positions is $4!$.
Number of ways to arrange $3$ girls in $3$ positions is $3!$.
Total favorable outcomes $= 4! \times 3!$.
Probability $= \frac{4! \times 3!}{7!} = \frac{24 \times 6}{5040} = \frac{144}{5040} = \frac{1}{35}$.

(A) The total number of ways to choose $2$ numbers from $100$ is $^{100}C_2 = \frac{100 \times 99}{2} = 4950$.
Among the numbers from $1$ to $100$,the numbers divisible by $3$ are $3, 6, 9, \dots, 99$. There are $33$ such numbers.
The numbers not divisible by $3$ are $100 - 33 = 67$.
The product of two numbers is divisible by $3$ if at least one of the chosen numbers is a multiple of $3$.
This can happen in two cases:
Case $1$: One number is a multiple of $3$ and the other is not. Number of ways = $^{33}C_1 \times ^{67}C_1 = 33 \times 67 = 2211$.
Case $2$: Both numbers are multiples of $3$. Number of ways = $^{33}C_2 = \frac{33 \times 32}{2} = 528$.
Total favorable outcomes = $2211 + 528 = 2739$.
Probability = $\frac{2739}{4950} = 0.55$.

(A) The total number of ways to select $2$ cards from a deck of $52$ cards is given by $n(S) = \binom{52}{2} = \frac{52 \times 51}{2 \times 1} = 1326$.
There are $4$ kings in a deck. The number of ways to select $2$ kings from $4$ is $n(E) = \binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6$.
The probability $P(E)$ that both cards are kings is $\frac{n(E)}{n(S)} = \frac{6}{1326} = \frac{1}{221}$.

(D) Total number of chalks = $5 + 4 = 9$.
Total ways to select $2$ chalks from $9$ is $n(S) = _9C_2 = \frac{9 \times 8}{2 \times 1} = 36$.
Let $A$ be the event that both chalks are white. The number of ways to select $2$ white chalks from $4$ is $n(A) = _4C_2 = \frac{4 \times 3}{2 \times 1} = 6$.
Let $B$ be the event that both chalks are blue. The number of ways to select $2$ blue chalks from $5$ is $n(B) = _5C_2 = \frac{5 \times 4}{2 \times 1} = 10$.
Since $A$ and $B$ are mutually exclusive events,the probability that both chalks are of the same color is $P(A \cup B) = P(A) + P(B)$.
$P(A) = \frac{6}{36} = \frac{1}{6}$.
$P(B) = \frac{10}{36} = \frac{5}{18}$.
$P(A \cup B) = \frac{6}{36} + \frac{10}{36} = \frac{16}{36} = \frac{4}{9}$.

(A) Total number of cards = $16$. Number of ways to draw $2$ cards from $16$ is $\binom{16}{2} = \frac{16 \times 15}{2} = 120$.
Number of Aces = $4$. Number of non-Ace cards = $12$.
The probability of getting no Ace is the probability of drawing $2$ cards from the $12$ non-Ace cards.
$P(\text{no Ace}) = \frac{\binom{12}{2}}{\binom{16}{2}} = \frac{\frac{12 \times 11}{2}}{120} = \frac{66}{120} = \frac{11}{20}$.
The probability of getting at least one Ace is $1 - P(\text{no Ace})$.
$P(\text{at least one Ace}) = 1 - \frac{11}{20} = \frac{9}{20}$.

(B) The word $MISSISSIPPI$ has $11$ letters,consisting of $4$ $S$,$4$ $I$,$2$ $P$,and $1$ $M$.
The total number of arrangements is $\frac{11!}{4!4!2!1!} = \frac{39916800}{24 \times 24 \times 2} = 34650$.
To find the number of arrangements where all $4$ $S$ come together,treat the block $(SSSS)$ as a single unit.
Now we have $8$ units to arrange: $(SSSS), I, I, I, I, P, P, M$.
The number of arrangements of these $8$ units is $\frac{8!}{4!2!1!} = \frac{40320}{24 \times 2} = 840$.
The probability is $\frac{840}{34650} = \frac{84}{3465} = \frac{4}{165}$.

(A) The total number of $5$-digit numbers formed using the digits $1, 2, 3, 4, 5$ is $5! = 120$.
$A$ number is divisible by $4$ if the number formed by its last two digits is divisible by $4$.
The possible pairs from the given digits that are divisible by $4$ are $12, 24, 32, 52$.
For each of these $4$ pairs,the remaining $3$ digits can be arranged in $3! = 6$ ways.
Total favorable outcomes $= 4 \times 6 = 24$.
Probability $= \frac{24}{120} = \frac{1}{5}$.

(C) The total number of ways to choose $2$ numbers from $30$ consecutive integers is $n = \binom{30}{2} = \frac{30 \times 29}{2} = 15 \times 29$.
The sum of two numbers is odd if and only if one number is even and the other is odd.
In $30$ consecutive integers,there are $15$ even and $15$ odd numbers.
The number of ways to choose one even and one odd number is $r = \binom{15}{1} \times \binom{15}{1} = 15 \times 15$.
The required probability is $P = \frac{15 \times 15}{15 \times 29} = \frac{15}{29}$.

(A) Total number of ways to select $2$ people from $13$ is given by $n(S) = {}_{13}C_2 = \frac{13 \times 12}{2 \times 1} = 78$.
Let $A$ be the event that at least one woman is selected.
The number of ways to select at least one woman is:
$n(A) = ({}_5C_1 \times {}_8C_1) + ({}_5C_2 \times {}_8C_0) = (5 \times 8) + (10 \times 1) = 40 + 10 = 50$.
Therefore,the probability $P(A) = \frac{n(A)}{n(S)} = \frac{50}{78} = \frac{25}{39}$.

(D) Let $A$ be the event of drawing $2$ brown socks and $B$ be the event of drawing $2$ white socks.
Total number of ways to draw $2$ socks from $9$ is $^{9}C_{2} = \frac{9 \times 8}{2 \times 1} = 36$.
Probability of drawing $2$ brown socks: $P(A) = \frac{^{5}C_{2}}{^{9}C_{2}} = \frac{10}{36} = \frac{5}{18}$.
Probability of drawing $2$ white socks: $P(B) = \frac{^{4}C_{2}}{^{9}C_{2}} = \frac{6}{36} = \frac{3}{18}$.
Since $A$ and $B$ are mutually exclusive events,the required probability is $P(A \cup B) = P(A) + P(B) = \frac{5}{18} + \frac{3}{18} = \frac{8}{18} = \frac{4}{9}$.

(D) The word $ASSASSIN$ contains $8$ letters: $A(2), S(4), I(2), N(1)$.
Total arrangements = $\frac{8!}{2!4!2!} = \frac{40320}{2 \times 24 \times 2} = 420$.
To ensure no two $S$s occur together,we arrange the remaining letters $A, A, I, N$ first.
Number of ways to arrange $A, A, I, N$ = $\frac{4!}{2!} = 12$.
These $4$ letters create $5$ gaps: $\_ L \_ L \_ L \_ L \_$.
We need to place $4$ $S$s in these $5$ gaps,which can be done in $^5C_4 = 5$ ways.
Favorable outcomes = $12 \times 5 = 60$.
Probability = $\frac{60}{420} = \frac{1}{7}$.
Note: The provided option $\frac{1}{14}$ is incorrect based on the standard calculation.

(B) In $20$ consecutive integers,there are $10$ even numbers and $10$ odd numbers.
The total number of ways to select $2$ numbers from $20$ is $^{20}C_2 = \frac{20 \times 19}{2} = 190$.
The sum of two numbers is odd if and only if one number is even and the other is odd.
The number of favorable outcomes is $^{10}C_1 \times ^{10}C_1 = 10 \times 10 = 100$.
The required probability is $P = \frac{100}{190} = \frac{10}{19}$.

(A) The total number of ways to select $2$ tickets from $20$ is given by $n = \binom{20}{2} = \frac{20 \times 19}{2 \times 1} = 190$.
The prime numbers between $1$ and $20$ are $2, 3, 5, 7, 11, 13, 17, 19$. There are $8$ such numbers.
The number of ways to select $2$ prime numbers from these $8$ is $r = \binom{8}{2} = \frac{8 \times 7}{2 \times 1} = 28$.
The probability of the event is $P = \frac{r}{n} = \frac{28}{190} = \frac{14}{95}$.

(D) Total number of balls $= 5 + 7 + 4 = 16$.
Number of ways to select $3$ balls out of $16$ is given by $^{16}C_3 = \frac{16 \times 15 \times 14}{3 \times 2 \times 1} = 560$.
Number of ways to select $3$ white balls out of $5$ is given by $^{5}C_3 = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10$.
Probability $= \frac{^{5}C_3}{^{16}C_3} = \frac{10}{560} = \frac{1}{56}$.

(C) Total number of people = $3 + 2 + 4 = 9$.
We need to select a group of $3$ people from $9$.
The total number of ways to select $3$ people is $n = \binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$.
We want exactly $2$ children in the group of $3$.
This means we select $2$ children from $4$ children and $1$ person from the remaining $5$ people ($3$ men + $2$ women).
The number of favorable ways is $r = \binom{4}{2} \times \binom{5}{1} = 6 \times 5 = 30$.
The probability $P(A) = \frac{r}{n} = \frac{30}{84} = \frac{5}{14}$.