Probability Questions in English

(D) Total number of horses = $5$.
Number of horses selected by $Mr. A$ = $2$.
Total 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 $NOT$ included is the number of ways to select $2$ horses from the $4$ losing horses,which is $^4C_2 = \frac{4 \times 3}{2 \times 1} = 6$.
The probability that the winning horse is $NOT$ selected = $\frac{6}{10} = \frac{3}{5}$.
The probability that the winning horse $IS$ selected = $1 - \frac{3}{5} = \frac{2}{5}$.

(C) Let $P(X)$ be the probability that $X$ speaks the truth and $P(Y)$ be the probability that $Y$ speaks the truth.
Given: $P(X) = 60\% = \frac{60}{100} = \frac{3}{5}$ and $P(Y) = 50\% = \frac{50}{100} = \frac{1}{2}$.
$X$ and $Y$ contradict each other if one speaks the truth and the other lies.
This can happen in two ways: ($X$ speaks the truth $AND$ $Y$ lies) $OR$ ($X$ lies $AND$ $Y$ speaks the truth).
Required probability $= P(X) \cdot P(\bar{Y}) + P(\bar{X}) \cdot P(Y)$.
Here,$P(\bar{X}) = 1 - P(X) = 1 - \frac{3}{5} = \frac{2}{5}$ and $P(\bar{Y}) = 1 - P(Y) = 1 - \frac{1}{2} = \frac{1}{2}$.
Required probability $= (\frac{3}{5} \cdot \frac{1}{2}) + (\frac{2}{5} \cdot \frac{1}{2}) = \frac{3}{10} + \frac{2}{10} = \frac{5}{10} = \frac{1}{2}$.

(C) Let $P(A)$ be the probability that $A$ speaks the truth and $P(B)$ be the probability that $B$ speaks the truth.
Given: $P(A) = \frac{4}{5}$ and $P(B) = \frac{3}{4}$.
The probability that $A$ lies is $P(\bar{A}) = 1 - P(A) = 1 - \frac{4}{5} = \frac{1}{5}$.
The probability that $B$ lies is $P(\bar{B}) = 1 - P(B) = 1 - \frac{3}{4} = \frac{1}{4}$.
They contradict each other if $A$ speaks the truth and $B$ lies,$OR$ $A$ lies and $B$ speaks the truth.
Required Probability $= P(A) \cdot P(\bar{B}) + P(\bar{A}) \cdot P(B)$
$= (\frac{4}{5} \cdot \frac{1}{4}) + (\frac{1}{5} \cdot \frac{3}{4})$
$= \frac{4}{20} + \frac{3}{20} = \frac{7}{20}$.

(A) The total number of outcomes when throwing three dice is $n(S) = 6 \times 6 \times 6 = 216$.
To get a sum of $16$,the possible combinations $(x, y, z)$ are:
$(6, 6, 4), (6, 4, 6), (4, 6, 6), (6, 5, 5), (5, 6, 5), (5, 5, 6)$.
The number of favorable outcomes is $n(E) = 6$.
The required probability is $P(E) = \frac{n(E)}{n(S)} = \frac{6}{216} = \frac{1}{36}$.

(D) Let $A$ be the set of numbers divisible by $6$ and $B$ be the set of numbers divisible by $8$ in the range $1$ to $90$.
Total numbers $n(S) = 90$.
Number of elements divisible by $6$ is $n(A) = \lfloor \frac{90}{6} \rfloor = 15$.
Number of elements divisible by $8$ is $n(B) = \lfloor \frac{90}{8} \rfloor = 11$.
Numbers divisible by both $6$ and $8$ are divisible by $\text{lcm}(6, 8) = 24$.
Number of elements divisible by $24$ is $n(A \cap B) = \lfloor \frac{90}{24} \rfloor = 3$.
Using the inclusion-exclusion principle,the number of elements divisible by $6$ or $8$ is $n(A \cup B) = n(A) + n(B) - n(A \cap B) = 15 + 11 - 3 = 23$.
The probability is $P(A \cup B) = \frac{n(A \cup B)}{n(S)} = \frac{23}{90}$.

(A) Let $A$ be the event that the first student solves the problem,so $P(A) = \frac{1}{2}$.
Let $B$ be the event that the second student solves the problem,so $P(B) = \frac{1}{3}$.
The problem is solved if at least one of the students solves it.
The probability that the problem is not solved by either student is $P(A^c) \times P(B^c) = (1 - \frac{1}{2}) \times (1 - \frac{1}{3}) = \frac{1}{2} \times \frac{2}{3} = \frac{1}{3}$.
The probability that the problem is solved is $1 - P(\text{not solved}) = 1 - \frac{1}{3} = \frac{2}{3}$.

(D) There are $3$ houses available and $3$ persons applying for them.
Each person has $3$ choices for selecting a house.
Since each person applies independently,the total number of ways the $3$ persons can apply for the $3$ houses is $3 \times 3 \times 3 = 27$.
For all $3$ persons to apply for the same house,they can either all choose the first house,all choose the second house,or all choose the third house.
There are $3$ such favorable outcomes.
Therefore,the probability that all $3$ persons apply for the same house is $\frac{3}{27} = \frac{1}{9}$.

(B) Let $p$ be the probability of getting a $1$ in a single throw,so $p = \frac{1}{6}$.
Let $q$ be the probability of not getting a $1$,so $q = 1 - \frac{1}{6} = \frac{5}{6}$.
The event of getting a $1$ in an even number of throws means getting it on the $2^{nd}, 4^{th}, 6^{th}, \dots$ throw.
This is represented by the sum of the probabilities: $q \cdot p + q^3 \cdot p + q^5 \cdot p + \dots$
This is an infinite geometric series with the first term $a = qp = \frac{5}{6} \cdot \frac{1}{6} = \frac{5}{36}$ and common ratio $r = q^2 = (\frac{5}{6})^2 = \frac{25}{36}$.
The sum $S$ of an infinite geometric series is given by $S = \frac{a}{1 - r}$.
$S = \frac{5/36}{1 - 25/36} = \frac{5/36}{11/36} = \frac{5}{11}$.

(D) Given that $A$ and $B$ are independent events,the probability of neither $A$ nor $B$ occurring is given by $P(\bar{A} \cap \bar{B})$.
Since $A$ and $B$ are independent,their complements $\bar{A}$ and $\bar{B}$ are also independent.
Therefore,$P(\bar{A} \cap \bar{B}) = P(\bar{A}) \times P(\bar{B})$.
We know that $P(\bar{A}) = 1 - P(A) = 1 - 1/2 = 1/2$.
Similarly,$P(\bar{B}) = 1 - P(B) = 1 - 1/3 = 2/3$.
Thus,$P(\bar{A} \cap \bar{B}) = (1/2) \times (2/3) = 1/3$.

(NONE) The circuit consists of two parallel branches connected between points $A$ and $B$.
$1$. The upper branch contains two switches $a$ and $b$ in series. Current flows through this branch only if both switches $a$ and $b$ are closed.
Since the probability of each switch being closed is $p$,the probability that the upper branch is closed is $P(\text{upper}) = p \cdot p = p^2$.
$2$. The lower branch contains a single switch $c$. Current flows through this branch if switch $c$ is closed.
The probability that the lower branch is closed is $P(\text{lower}) = p$.
$3$. Current flows from $A$ to $B$ if either the upper branch is closed $OR$ the lower branch is closed.
Using the inclusion-exclusion principle for probability: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Here,$P(\text{upper} \cup \text{lower}) = P(\text{upper}) + P(\text{lower}) - P(\text{upper} \cap \text{lower})$.
$P(\text{upper} \cap \text{lower}) = p^2 \cdot p = p^3$.
Therefore,the required probability is $p^2 + p - p^3$.

(C) Total number of ways to draw $2$ cards from $52$ cards is given by $^{52}C_2$.
Number of ways to draw $2$ spade cards from $13$ spade cards is given by $^{13}C_2$.
The required probability is $P = \frac{^{13}C_2}{^{52}C_2}$.
Calculating the combinations: $^{13}C_2 = \frac{13 \times 12}{2 \times 1} = 78$ and $^{52}C_2 = \frac{52 \times 51}{2 \times 1} = 1326$.
Thus,$P = \frac{78}{1326} = \frac{1}{17}$.

(C) Total number of cards in a pack is $52$.
There are $26$ red cards and $26$ black cards in a pack.
We need to draw $6$ cards simultaneously,which can be done in $^{52}C_6$ ways.
The number of ways to select $3$ red cards from $26$ red cards is $^{26}C_3$.
The number of ways to select $3$ black cards from $26$ black cards is $^{26}C_3$.
Therefore,the number of favorable outcomes is $^{26}C_3 \times ^{26}C_3$.
The probability is given by the ratio of favorable outcomes to total outcomes: $P = \frac{^{26}C_3 \times ^{26}C_3}{^{52}C_6}$.

(C) The total number of cards in a pack is $52$.
The number of spade cards is $13$.
The probability of drawing a spade card in a single trial is $P(S) = \frac{13}{52} = \frac{1}{4}$.
The probability of not drawing a spade card (failure) in a single trial is $P(F) = 1 - P(S) = 1 - \frac{1}{4} = \frac{3}{4}$.
Since the card is replaced and the pack is shuffled,each trial is independent.
The probability that he fails the first two times is the probability of getting a non-spade card in the first trial $AND$ a non-spade card in the second trial.
Required probability $= P(F) \times P(F) = \frac{3}{4} \times \frac{3}{4} = \frac{9}{64}$.

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

(C) Total number of balls $= 3 + 7 + 4 = 14$.
Number of ways to draw $3$ balls out of $14$ is given by $^{14}C_3 = \frac{14 \times 13 \times 12}{3 \times 2 \times 1} = 14 \times 13 \times 2 = 364$.
For all three balls to be of the same color,they must be either all red,all white,or all black.
Number of ways to choose $3$ red balls $= ^3C_3 = 1$.
Number of ways to choose $3$ white balls $= ^7C_3 = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$.
Number of ways to choose $3$ black balls $= ^4C_3 = 4$.
Total favorable outcomes $= 1 + 35 + 4 = 40$.
Required probability $= \frac{40}{364} = \frac{10}{91}$.

(A) Total number of persons $= 3 + 2 + 4 = 9$.
We need to choose $4$ persons out of $9$,which can be done in ${}^9C_4$ ways.
${}^9C_4 = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126$.
We want exactly $2$ children out of $4$ children,which can be chosen in ${}^4C_2$ ways.
${}^4C_2 = \frac{4 \times 3}{2 \times 1} = 6$.
The remaining $2$ persons must be chosen from the $5$ adults ($3$ men $+ 2$ women),which can be done in ${}^5C_2$ ways.
${}^5C_2 = \frac{5 \times 4}{2 \times 1} = 10$.
The number of favorable outcomes $= {}^4C_2 \times {}^5C_2 = 6 \times 10 = 60$.
The required probability $= \frac{60}{126} = \frac{10}{21}$.

(C) The total number of ways to draw $2$ tickets from $25$ is given by $^{25}C_2 = \frac{25 \times 24}{2} = 300$.
The product of two numbers is even if at least one of the numbers is even.
It is easier to calculate the probability of the complement event: the product is odd.
The product is odd only if both selected numbers are odd.
In the set ${1, 2, \dots, 25}$,there are $13$ odd numbers $(1, 3, 5, \dots, 25)$ and $12$ even numbers.
The number of ways to choose $2$ odd tickets is $^{13}C_2 = \frac{13 \times 12}{2} = 78$.
The probability that the product is 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 from $30$ is given by $^{30}C_2 = \frac{30 \times 29}{2 \times 1} = 435$.
Number of ways to choose $2$ girls from $12$ is given by $^{12}C_2 = \frac{12 \times 11}{2 \times 1} = 66$.
The probability that both chosen students are girls is $P = \frac{^{12}C_2}{^{30}C_2} = \frac{66}{435}$.
Dividing both numerator and denominator by $3$,we get $P = \frac{22}{145}$.

(B) Total number of letters = $11$.
Number of consonants = $7$.
Number of vowels = $4$.
We need to choose $2$ letters out of $11$. The total number of ways to choose $2$ letters is given by $^{11}C_2 = \frac{11 \times 10}{2 \times 1} = 55$.
The number of ways to choose $2$ consonants out of $7$ is given by $^7C_2 = \frac{7 \times 6}{2 \times 1} = 21$.
The probability that both chosen letters are consonants is the ratio of the number of favorable outcomes to the total number of outcomes.
Required probability = $\frac{^7C_2}{^{11}C_2} = \frac{21}{55}$.

(A) 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.
Since $2$ tickets ($7$ and $11$) are already fixed,we only need to choose $1$ more ticket from the remaining $18$ tickets $(20 - 2 = 18)$.
The number of ways to choose the $3^{rd}$ ticket is $^{18}C_1 = 18$.
Therefore,the required probability is $\frac{18}{1140} = \frac{18}{1140} = \frac{3}{190}$.

(A) Total number of tickets $= 3 + 9 = 12$.
Mohan selects $3$ tickets out of $12$. The total number of ways to select $3$ tickets is ${}^{12}C_3 = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220$.
The probability of winning at least one prize is $1 - P(\text{winning no prize})$.
Winning no prize means all $3$ tickets are selected from the $9$ blanks.
The number of ways to select $3$ blanks out of $9$ is ${}^9C_3 = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$.
The probability of winning no prize is $\frac{84}{220} = \frac{21}{55}$.
Therefore,the probability of winning at least one prize is $1 - \frac{21}{55} = \frac{34}{55}$.

(D) Total number of balls in the bag = $3$ (white) + $7$ (red) = $10$ balls.
Let $E$ be the event of drawing a ball that is either white or red.
Since all balls in the bag are either white or red,this is a certain event.
The number of favorable outcomes = $3 + 7 = 10$.
The total number of possible outcomes = $10$.
Probability $P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{10}{10} = 1$.

(A) Total number of balls $= 4 + 5 + 6 = 15$.
Number of ways to draw $2$ balls from $15$ is $^{15}C_2 = \frac{15 \times 14}{2 \times 1} = 105$.
We want the probability that 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 $= \frac{44}{105}$.

(C) Total number of balls = $6 + 4 + 8 = 18$.
Number of ways to draw $3$ balls from $18$ is given by ${}^{18}C_3 = \frac{18 \times 17 \times 16}{3 \times 2 \times 1} = 3 \times 17 \times 16 = 816$.
We need to draw $2$ white balls from $4$ and $1$ red ball from $6$.
Number of ways to choose $2$ white balls = ${}^4C_2 = \frac{4 \times 3}{2 \times 1} = 6$.
Number of ways to choose $1$ red ball = ${}^6C_1 = 6$.
Total favorable outcomes = ${}^4C_2 \times {}^6C_1 = 6 \times 6 = 36$.
Required probability = $\frac{36}{816} = \frac{3}{68}$.

(C) Total ways to choose a committee of $5$ from $9$ people is ${}^9C_5 = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126$.
Case $1$: The couple serves together.
If the couple is included,we need to choose $3$ more people from the remaining $7$ people. This can be done in ${}^7C_3 = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$ ways.
Case $2$: The couple does not serve at all.
If the couple is excluded,we need to choose $5$ people from the remaining $7$ people. This can be done in ${}^7C_5 = {}^7C_2 = \frac{7 \times 6}{2 \times 1} = 21$ ways.
Total favorable ways $= 35 + 21 = 56$.
Required probability $= \frac{56}{126} = \frac{4}{9}$.

(B) The word $ASSASSIN$ contains $8$ letters: $A(2), S(4), I(1), N(1)$.
Total number of arrangements $ = \frac{8!}{4! \times 2!} = \frac{8 \times 7 \times 6 \times 5 \times 4!}{4! \times 2} = 840$.
To ensure no two $S$ occur together,we use the gap method. First,arrange the remaining $4$ letters $(A, A, I, N)$.
The number of ways to arrange these $4$ letters is $\frac{4!}{2!} = 12$.
These $4$ letters create $5$ gaps (including ends): $\_ L_1 \_ L_2 \_ L_3 \_ L_4 \_$.
We need to place $4$ $S$'s in these $5$ gaps. The number of ways to choose $4$ gaps out of $5$ is $\binom{5}{4} = 5$.
Total favorable arrangements $ = 12 \times 5 = 60$.
Probability $ = \frac{\text{Favorable arrangements}}{\text{Total arrangements}} = \frac{60}{840} = \frac{1}{14}$.

(D) The probability of getting $k$ successes in $n$ independent Bernoulli trials is given by the binomial distribution formula: $P(X = k) = {}^nC_k \cdot p^k \cdot q^{n-k}$.
Here,$n = 8$ (total number of throws),$k = 4$ (number of heads required),$p = \frac{1}{2}$ (probability of getting a head),and $q = 1 - p = \frac{1}{2}$ (probability of getting a tail).
Substituting these values into the formula:
$P(X = 4) = {}^8C_4 \cdot (\frac{1}{2})^4 \cdot (\frac{1}{2})^{8-4}$
$P(X = 4) = {}^8C_4 \cdot (\frac{1}{2})^4 \cdot (\frac{1}{2})^4$
$P(X = 4) = {}^8C_4 \cdot (\frac{1}{2})^8$
$P(X = 4) = \frac{{}^8C_4}{2^8}$.
Thus,the correct option is $(d)$.

(A) Total number of tickets = $50$.
Number of prize-winning tickets = $14$.
Number of non-prize-winning (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 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 $P(\text{At least one prize}) = 1 - P(\text{No prize}) = 1 - \frac{18}{35} = \frac{17}{35}$.

(B) Total number of balls $= 8 + 7 = 15$.
Total ways to draw $2$ balls $= {}^{15}C_2 = \frac{15 \times 14}{2} = 105$.
$1$. Probability that both balls are white: $P(WW) = \frac{{}^7C_2}{{}^{15}C_2} = \frac{21}{105} = \frac{1}{5} = 0.2$.
$2$. Probability that both balls are black: $P(BB) = \frac{{}^8C_2}{{}^{15}C_2} = \frac{28}{105} = \frac{4}{15} \approx 0.266$.
$3$. Probability that one ball is white and one is black: $P(WB) = \frac{{}^7C_1 \times {}^8C_1}{{}^{15}C_2} = \frac{7 \times 8}{105} = \frac{56}{105} = \frac{8}{15} \approx 0.533$.
Comparing the probabilities: $\frac{8}{15} > \frac{4}{15} > \frac{3}{15}$.
Thus,the probability of drawing one white and one black ball is the highest.

(B) Total number of ways to select $5$ members from $10$ people ($6$ men + $4$ women) is given by ${}^{10}C_5 = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252$.
The number of ways to select a committee with no women (i.e.,all $5$ members are men) is given by ${}^{6}C_5 = 6$.
The number of ways to select at least one woman is (Total ways) - (Ways with no women) = $252 - 6 = 246$.
Therefore,the required probability is $P = \frac{246}{252} = \frac{41}{42}$.

(C) The total number of ways to form a three-digit number using the digits $1, 2, 3,$ and $4$ (without repetition) is $P(4, 3) = 4 \times 3 \times 2 = 24$.
$A$ number is divisible by $3$ if the sum of its digits is divisible by $3$.
Possible sets of $3$ digits from ${1, 2, 3, 4}$ whose sum is divisible by $3$ are:
$1) \{1, 2, 3\}$ (Sum $= 6$,which is divisible by $3$)
$2) \{2, 3, 4\}$ (Sum $= 9$,which is divisible by $3$)
For each set,the number of ways to arrange the $3$ digits is $3! = 6$.
Total favorable outcomes $= 6 + 6 = 12$.
Required probability $= \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{12}{24} = \frac{1}{2}$.

(B) Total number of ways to draw $3$ cards from $52$ cards is given by ${}^{52}C_3 = \frac{52 \times 51 \times 50}{3 \times 2 \times 1} = 22100$.
The number of ways to select $1$ king from $4$,$1$ queen from $4$,and $1$ jack from $4$ is ${}^4C_1 \times {}^4C_1 \times {}^4C_1 = 4 \times 4 \times 4 = 64$.
The required probability is $P = \frac{64}{22100}$.
Simplifying the fraction by dividing both numerator and denominator by $4$,we get $P = \frac{16}{5525}$.

(C) The word '$UNIVERSITY$' contains $10$ letters,where '$I$' appears $2$ times and all other letters are distinct.
Total number of arrangements $ = \frac{10!}{2!}$.
To find the probability that both '$I$'s come together,we treat the two '$I$'s as a single unit $(II)$. Now we have $9$ units to arrange: $(U, N, V, E, R, S, T, Y, (II))$.
The number of ways to arrange these $9$ units is $9!$.
Probability that both '$I$'s come together $ = \frac{9!}{\frac{10!}{2!}} = \frac{9! \times 2!}{10!} = \frac{2}{10} = \frac{1}{5}$.
Therefore,the probability that both '$I$'s do not come together $ = 1 - \frac{1}{5} = \frac{4}{5}$.

(A) Let $E_i$ be the event that the $i^{th}$ object occupies the $i^{th}$ place.
The total number of ways to arrange $n$ objects in $n$ places is $n!$.
The number of ways to choose $3$ objects out of $n$ that occupy their corresponding places is given by $\binom{n}{3}$.
The remaining $(n-3)$ objects can be arranged in $(n-3)!$ ways.
Thus,the number of ways where at least $3$ objects are in their correct places is $\binom{n}{3} \times (n-3)!$.
The probability is $\frac{\binom{n}{3} \times (n-3)!}{n!} = \frac{n!}{3!(n-3)!} \times \frac{(n-3)!}{n!} = \frac{1}{3!} = \frac{1}{6}$.

(C) The total number of outcomes in $5$ throws of a $6$-faced die is $6^5 = 7776$.
Let $X$ be the outcome of a single throw. The possible values are $0$ (blank),$2$,and $3$.
We want the sum of $5$ throws to be $12$.
Let $n_0, n_2, n_3$ be the number of times $0, 2, 3$ appear respectively,where $n_0 + n_2 + n_3 = 5$.
The sum is $0(n_0) + 2(n_2) + 3(n_3) = 12$.
Case $1$: If $n_0 = 1$,then $2n_2 + 3n_3 = 12$. If $n_3 = 4$,then $2n_2 = 0 \implies n_2 = 0$. This gives $(n_0, n_2, n_3) = (1, 0, 4)$.
The number of ways is $\frac{5!}{1!0!4!} = 5$.
Case $2$: If $n_0 = 0$,then $2n_2 + 3n_3 = 12$. If $n_3 = 2$,then $2n_2 = 6 \implies n_2 = 3$. This gives $(n_0, n_2, n_3) = (0, 3, 2)$.
The number of ways is $\frac{5!}{0!3!2!} = 10$.
Total favorable outcomes = $5 + 10 = 15$.
Probability = $\frac{15}{6^5} = \frac{15}{7776} = \frac{5}{2592}$.

(C) For ${p_1}$: Two persons throw a die. Total outcomes $= 6 \times 6 = 36$. Favourable outcomes (equal values) are $(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)$,which are $6$ outcomes. So,${p_1} = \frac{6}{36} = \frac{1}{6}$.
For ${p_2}$: Four persons throw a die. Total outcomes $= 6^4 = 1296$. We need the probability that exactly three persons get the same value.
Step $1$: Choose the value that appears three times: ${}^6C_1 = 6$ ways.
Step $2$: Choose the three persons out of four who get this value: ${}^4C_3 = 4$ ways.
Step $3$: Choose the value for the fourth person (must be different from the first value): ${}^5C_1 = 5$ ways.
Favourable outcomes $= 6 \times 4 \times 5 = 120$.
So,${p_2} = \frac{120}{1296} = \frac{5}{54}$.
Comparing ${p_1} = \frac{1}{6} = \frac{9}{54}$ and ${p_2} = \frac{5}{54}$,we get ${p_1} > {p_2}$.

(A) The total number of ways to arrange $n$ cadets in a row is $n!$.
To find the number of favorable cases where two particular cadets stand side by side,we treat these two cadets as a single unit.
Now,we have $(n - 1)$ units to arrange,which can be done in $(n - 1)!$ ways.
Within the single unit,the two particular cadets can be arranged in $2! = 2$ ways.
Thus,the total number of favorable arrangements is $2 \times (n - 1)!$.
The required probability is the ratio of favorable cases to the total number of cases:
$P = \frac{2 \times (n - 1)!}{n!} = \frac{2 \times (n - 1)!}{n \times (n - 1)!} = \frac{2}{n}$.

(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 $8$ prime numbers is given by $^8C_2 = \frac{8 \times 7}{2 \times 1} = 28$.
The probability that both numbers are prime is $P = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{28}{190}$.
Simplifying the fraction,we get $P = \frac{14}{95}$.

(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 out of $15$ is ${}^{15}C_2 = \frac{15 \times 14}{2 \times 1} = 105$.
The number of ways to choose $2$ non-red balls out of $9$ is ${}^{9}C_2 = \frac{9 \times 8}{2 \times 1} = 36$.
The probability that none of the balls is red is $P = \frac{{}^{9}C_2}{{}^{15}C_2} = \frac{36}{105}$.
Simplifying the fraction by dividing by $3$,we get $P = \frac{12}{35}$.

(B) Total number of white balls = $3$.
Total number of black balls = $5$.
Total number of balls in the bag = $3 + 5 = 8$.
The probability of drawing a black ball is given by the ratio of the number of black balls to the total number of balls.
Probability $P(\text{black}) = \frac{\text{Number of black balls}}{\text{Total number of balls}} = \frac{5}{8}$.

(C) regular hexagon has $6$ vertices. The total number of ways to choose $3$ vertices out of $6$ is given by the combination formula ${}^nC_r = \frac{n!}{r!(n-r)!}$.
Total triangles $= {}^6C_3 = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$.
In a regular hexagon,an equilateral triangle is formed by connecting every second vertex. There are only $2$ such triangles possible (by taking vertices ${1, 3, 5}$ and ${2, 4, 6}$).
Therefore,the required probability $= \frac{\text{Number of equilateral triangles}}{\text{Total number of triangles}} = \frac{2}{20} = \frac{1}{10}$.

(B) Total number of fruits in the box $= 3 \text{ (mangoes)} + 3 \text{ (apples)} = 6 \text{ fruits}.$
We need to choose $2$ fruits out of $6$. The total number of ways to choose $2$ fruits is given by ${}^6C_2 = \frac{6 \times 5}{2 \times 1} = 15$.
The number of ways to choose $1$ mango out of $3$ is ${}^3C_1 = 3$.
The number of ways to choose $1$ apple out of $3$ is ${}^3C_1 = 3$.
The number of favorable outcomes (choosing $1$ mango and $1$ apple) is ${}^3C_1 \times {}^3C_1 = 3 \times 3 = 9$.
Therefore,the required probability $P = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{9}{15} = \frac{3}{5}.$

(A) Total number of ways to arrange $25$ books is $25!$.
Consider the $5$ volumes of Mathematics as distinct items. In any random arrangement of $25$ books,there are $5!$ possible relative orderings of these $5$ volumes.
Out of these $5!$ possible relative orderings,only $1$ ordering corresponds to the volumes being in increasing order (from left to right).
Since all $5!$ relative orderings are equally likely,the probability that the volumes appear in the specific increasing order is $\frac{1}{5!}$.

(C) Total members = $15$,Bowlers = $5$,Non-bowlers = $10$. We need to select $11$ members such that at least $3$ are bowlers.
The total number of ways to select $11$ members from $15$ is ${}^{15}C_{11} = {}^{15}C_4 = \frac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2 \times 1} = 1365$.
The number of ways to select at least $3$ bowlers is:
Case $1$: $3$ bowlers and $8$ non-bowlers: ${}^5C_3 \times {}^{10}C_8 = 10 \times 45 = 450$.
Case $2$: $4$ bowlers and $7$ non-bowlers: ${}^5C_4 \times {}^{10}C_7 = 5 \times 120 = 600$.
Case $3$: $5$ bowlers and $6$ non-bowlers: ${}^5C_5 \times {}^{10}C_6 = 1 \times 210 = 210$.
Total favorable ways = $450 + 600 + 210 = 1260$.
Probability = $\frac{1260}{1365} = \frac{12}{13}$.

(B) For ${P_1}$: Total balls = $13 + 14 + 15 = 42$. We select $4$ balls. The probability of getting exactly $2$ black balls is given by the hypergeometric distribution:
${P_1} = \frac{{{}^{15}{C_2} \times {}^{27}{C_2}}}{{{}^{42}{C_4}}} = \frac{105 \times 351}{111930} = \frac{36855}{111930} \approx 0.3292$.
For ${P_2}$: The number of balls is doubled. Total balls = $26 + 28 + 30 = 84$. We select $8$ balls. The probability of getting exactly $4$ black balls is:
${P_2} = \frac{{{}^{30}{C_4} \times {}^{54}{C_4}}}{{{}^{84}{C_8}}}$.
Using the property of the hypergeometric distribution,as the population size $N$ and sample size $n$ are scaled by a factor $k=2$,the probability of getting a proportional number of successes $(k \times r)$ generally decreases as the sample size increases relative to the population. Calculating the values,we find ${P_1} \approx 0.329$ and ${P_2} \approx 0.273$.
Therefore,${P_1} > {P_2}$.

(B) Total number of ways to arrange $m$ rupee coins and $n$ ten paise coins in a line is given by the permutation of multiset formula: $\frac{(m+n)!}{m!n!}$.
If the extreme coins are both 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 $\frac{(m+n-2)!}{m!(n-2)!}$.
The required probability is the ratio of favorable arrangements to total arrangements:
$P = \frac{\frac{(m+n-2)!}{m!(n-2)!}}{\frac{(m+n)!}{m!n!}}$
$P = \frac{(m+n-2)!}{m!(n-2)!} \times \frac{m!n!}{(m+n)!}$
$P = \frac{n(n-1)}{(m+n)(m+n-1)}$.

(C) The total number of functions from a set $A$ with $n$ elements to itself is given by $n^n$.
An injection (one-to-one function) from a finite set $A$ to itself is necessarily a bijection (one-to-one and onto function).
The total number of bijections (permutations) of a set with $n$ elements is $n!$.
Therefore,the probability that a randomly selected mapping is an injection is the ratio of the number of bijections to the total number of functions:
$P = \frac{n!}{n^n} = \frac{n \times (n-1)!}{n \times n^{n-1}} = \frac{(n-1)!}{n^{n-1}}$.

(A) Total number of pencils = $12 + 6 + 2 = 20$.
Number of good (non-defective) pencils = $12$.
Probability of choosing a non-defective pencil = $\frac{\text{Number of good pencils}}{\text{Total number of pencils}}$.
Probability = $\frac{12}{20} = \frac{3}{5}$.

(C) Total mangoes $= 10$. Good mangoes $= 6$. Rotten mangoes $= 4$.
Let $G_1$ be the event that the first mango is good and $G_2$ be the event that the second mango is good.
We are given that at least one mango is good. Let $A$ be the event that at least one mango is good.
Total ways to select $2$ mangoes from $10$ is $^{10}C_2 = \frac{10 \times 9}{2} = 45$.
Ways to select $2$ rotten mangoes $= ^4C_2 = 6$.
Ways to select at least one good mango $= 45 - 6 = 39$.
Ways to select $2$ good mangoes $= ^6C_2 = 15$.
The probability that both are good given that at least one is good is $\frac{15}{39} = \frac{5}{13}$.
Wait,re-evaluating: If one is found to be good,the sample space is restricted to cases where at least one is good. The number of ways to pick at least one good mango is $39$. The number of ways to pick two good mangoes is $15$. Thus,the probability is $\frac{15}{39} = \frac{5}{13}$.

(A) Total number of ways to select $2$ persons out of $13$ is ${}^{13}C_2 = \frac{13 \times 12}{2 \times 1} = 78$.
We want the probability that at least one woman is selected.
This can happen in two ways: selecting $1$ woman and $1$ man,or selecting $2$ women.
Number of ways to select $1$ woman and $1$ man = ${}^5C_1 \times {}^8C_1 = 5 \times 8 = 40$.
Number of ways to select $2$ women = ${}^5C_2 = \frac{5 \times 4}{2 \times 1} = 10$.
Total favorable outcomes = $40 + 10 = 50$.
Probability = $\frac{50}{78} = \frac{25}{39}$.