Probability Questions in English

(D) Let $P(A), P(B), P(C),$ and $P(D)$ be the probabilities that horses $A, B, C,$ and $D$ win the race,respectively.
Given the odds in favour of $A, B, C,$ and $D$ are $1:3, 1:4, 1:5,$ and $1:6,$ the probabilities are calculated as follows:
$P(A) = \frac{1}{1+3} = \frac{1}{4}$
$P(B) = \frac{1}{1+4} = \frac{1}{5}$
$P(C) = \frac{1}{1+5} = \frac{1}{6}$
$P(D) = \frac{1}{1+6} = \frac{1}{7}$
Since these events are mutually exclusive (only one horse can win the race),the probability that one of them wins is the sum of their individual probabilities:
$P(\text{one of them wins}) = P(A) + P(B) + P(C) + P(D)$
$= \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7}$
To add these fractions,find the least common multiple of $4, 5, 6,$ and $7$,which is $420$.
$= \frac{105 + 84 + 70 + 60}{420} = \frac{319}{420}$
Thus,the correct option is $D$.

(B) Let $A$ and $B$ be the events that the chartered accountant is selected in firms $X$ and $Y$,respectively.
Given: $P(A) = 0.7$,$P(\bar{B}) = 0.5$,and $P(\bar{A} \cup \bar{B}) = 0.6$.
We know that $P(\bar{A}) = 1 - P(A) = 1 - 0.7 = 0.3$.
Also,$P(B) = 1 - P(\bar{B}) = 1 - 0.5 = 0.5$.
By De Morgan's Law,$\bar{A} \cup \bar{B} = \overline{A \cap B}$.
Therefore,$P(\overline{A \cap B}) = 0.6$.
This implies $P(A \cap B) = 1 - P(\overline{A \cap B}) = 1 - 0.6 = 0.4$.
The probability that he is selected in at least one firm is $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
$P(A \cup B) = 0.7 + 0.5 - 0.4 = 0.8$.
The probability that he is selected in exactly one firm is $P(A \cup B) - P(A \cap B) = 0.8 - 0.4 = 0.4$.
Wait,the question asks for the probability that he will be selected in $1$ of the firms (meaning at least one). Thus,the answer is $0.8$.

(NONE OF THE ABOVE (CALCULATED: 43:34)) Since the odds against event $A$ are $8:3$,the probability of the occurrence of event $A$ is $P(A) = \frac{3}{8+3} = \frac{3}{11}$.
Similarly,the odds against event $B$ are $5:2$,so the probability of the occurrence of event $B$ is $P(B) = \frac{2}{5+2} = \frac{2}{7}$.
Since events $A, B, C$ are mutually exclusive and exhaustive,the sum of their probabilities must be $1$.
Therefore,$P(A) + P(B) + P(C) = 1$.
Substituting the values: $\frac{3}{11} + \frac{2}{7} + P(C) = 1$.
Calculating the sum: $\frac{21 + 22}{77} + P(C) = 1 \Rightarrow \frac{43}{77} + P(C) = 1$.
Thus,$P(C) = 1 - \frac{43}{77} = \frac{34}{77}$.
The probability of the non-occurrence of $C$ is $P(\bar{C}) = 1 - P(C) = 1 - \frac{34}{77} = \frac{43}{77}$.
The odds against $C$ are given by $P(\bar{C}) : P(C) = \frac{43}{77} : \frac{34}{77} = 43:34$.

(B) Let $P(A), P(B), P(C),$ and $P(D)$ be the probabilities of students $A, B, C,$ and $D$ solving the problem respectively.
Given: $P(A) = \frac{1}{3}, P(B) = \frac{1}{4}, P(C) = \frac{1}{5}, P(D) = \frac{1}{6}$.
The probability that a student fails to solve the problem is $1 - P(\text{solving})$.
$P(A') = 1 - \frac{1}{3} = \frac{2}{3}$
$P(B') = 1 - \frac{1}{4} = \frac{3}{4}$
$P(C') = 1 - \frac{1}{5} = \frac{4}{5}$
$P(D') = 1 - \frac{1}{6} = \frac{5}{6}$
Since the events are independent,the probability that none of them solve the problem is $P(A') \times P(B') \times P(C') \times P(D') = \frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} \times \frac{5}{6} = \frac{2}{6} = \frac{1}{3}$.
The probability that the problem will be solved is $1 - P(\text{none solve}) = 1 - \frac{1}{3} = \frac{2}{3}$.

(C) Let $E_1$ be the event of drawing a white ball from the first bag.
Total balls in the first bag $= 4 + 2 = 6$.
Probability $P(E_1) = \frac{4}{6} = \frac{2}{3}$.
Let $E_2$ be the event of drawing a white ball from the second bag.
Total balls in the second bag $= 3 + 5 = 8$.
Probability $P(E_2) = \frac{3}{8}$.
Since the events are independent,the probability that both balls are white is $P(E_1 \cap E_2) = P(E_1) \times P(E_2)$.
$P(E_1 \cap E_2) = \frac{2}{3} \times \frac{3}{8} = \frac{6}{24} = \frac{1}{4}$.

(A) Let $E_1$ be the event of drawing a black ball from the first bag.
Total balls in the first bag $= 4 + 2 = 6$.
Number of black balls in the first bag $= 2$.
$P(E_1) = \frac{2}{6} = \frac{1}{3}$.
Let $E_2$ be the event of drawing a black ball from the second bag.
Total balls in the second bag $= 3 + 5 = 8$.
Number of black balls in the second bag $= 5$.
$P(E_2) = \frac{5}{8}$.
Since the events are independent,the probability that both balls are black is $P(E_1 \cap E_2) = P(E_1) \times P(E_2)$.
$P(E_1 \cap E_2) = \frac{1}{3} \times \frac{5}{8} = \frac{5}{24}$.

(C) Let $B_1$ be the first bag and $B_2$ be the second bag.
In $B_1$,total balls $= 4 + 2 = 6$. Probability of white $(W_1) = \frac{4}{6} = \frac{2}{3}$,probability of black $(Bl_1) = \frac{2}{6} = \frac{1}{3}$.
In $B_2$,total balls $= 3 + 5 = 8$. Probability of white $(W_2) = \frac{3}{8}$,probability of black $(Bl_2) = \frac{5}{8}$.
The event 'one is white and one is black' occurs in two mutually exclusive cases: $(W_1 \text{ and } Bl_2)$ or $(Bl_1 \text{ and } W_2)$.
Probability $= P(W_1) \times P(Bl_2) + P(Bl_1) \times P(W_2)$
$= (\frac{2}{3} \times \frac{5}{8}) + (\frac{1}{3} \times \frac{3}{8})$
$= \frac{10}{24} + \frac{3}{24} = \frac{13}{24}$.

(A) Total balls in the urn = $25$.
Odd numbers between $1$ and $25$ are ${1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25}$.
Number of odd balls = $13$.
Probability of success $(p)$ = $\frac{13}{25}$.
Since the balls are drawn with replacement,the trials are independent.
The probability of getting two successes in two draws is $P = p \times p$.
$P = \frac{13}{25} \times \frac{13}{25} = \frac{169}{625}$.

(B) Total balls = $25$.
Odd numbers between $1$ and $25$ are ${1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25}$,which are $13$ in total.
Probability of success $(p)$ = $\frac{13}{25}$.
Probability of failure $(q)$ = $1 - p = 1 - \frac{13}{25} = \frac{12}{25}$.
Since $2$ balls are drawn with replacement,this follows a binomial distribution with $n = 2$.
The probability of getting exactly one success is given by $P(X = 1) = ^nC_1 \cdot p^1 \cdot q^{n-1} = 2 \cdot p \cdot q$.
$P(X = 1) = 2 \times \frac{13}{25} \times \frac{12}{25} = \frac{312}{625}$.

(C) Total balls $= 25$.
Odd numbers are ${1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25}$,which are $13$ in total.
Probability of success $(p)$ $= \frac{13}{25}$.
Probability of failure $(q)$ $= 1 - p = 1 - \frac{13}{25} = \frac{12}{25}$.
Since $2$ balls are drawn with replacement,the probability of getting no success in $2$ trials is $q \times q = \left(\frac{12}{25}\right)^2 = \frac{144}{625}$.
The probability of getting at least one success $= 1 - P(\text{no success}) = 1 - \frac{144}{625} = \frac{625 - 144}{625} = \frac{481}{625}$.

(D) Total balls = $25$.
Odd numbers between $1$ and $25$ are ${1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25}$,which are $13$ in total.
Probability of success $(p)$ = $\frac{13}{25}$.
Probability of no success $(q)$ = $1 - p = 1 - \frac{13}{25} = \frac{12}{25}$.
Since $2$ balls are drawn with replacement,the trials are independent.
Probability of getting no success in $2$ draws = $q \times q = \frac{12}{25} \times \frac{12}{25} = \frac{144}{625}$.

(D) The urn contains $25$ balls numbered $1$ to $25$.
An odd number is considered a 'success'. The odd numbers between $1$ and $25$ are $1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25$.
There are $13$ odd numbers.
Probability of success $(p)$ = $\frac{13}{25}$.
Probability of failure $(q)$ = $1 - p = 1 - \frac{13}{25} = \frac{12}{25}$.
We are drawing $2$ balls with replacement.
Since we are only performing $2$ trials,it is impossible to get $3$ successes.
Therefore,the probability of getting $3$ successes is $0$.

(C) Total balls $= 25$. The odd numbers between $1$ and $25$ are ${1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25}$,which are $13$ in total.
Probability of success $(p)$ $= 13/25$.
Probability of failure $(q)$ $= 1 - p = 1 - 13/25 = 12/25$.
Since $2$ balls are drawn with replacement,this follows a binomial distribution $B(n, p)$ with $n = 2$.
The probability of getting exactly $2$ successes is given by $P(X = 2) = ^nC_r \cdot p^r \cdot q^{n-r}$.
Here,$n = 2$ and $r = 2$.
$P(X = 2) = ^2C_2 \cdot (13/25)^2 \cdot (12/25)^0 = 1 \cdot (169/625) \cdot 1 = 169/625$.

(C) The total number of balls is $25$. The odd numbers between $1$ and $25$ are ${1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25}$,which are $13$ in total.
Probability of success $(p)$ $= \frac{13}{25}$.
Probability of failure $(q)$ $= 1 - p = 1 - \frac{13}{25} = \frac{12}{25}$.
Since $2$ balls are drawn with replacement,this follows a binomial distribution $B(n, p)$ where $n = 2$.
We need to find the probability of getting at most $2$ successes,i.e.,$P(X \le 2) = P(X=0) + P(X=1) + P(X=2)$.
Since the maximum number of successes possible in $2$ trials is $2$,$P(X \le 2)$ covers all possible outcomes.
The sum of probabilities of all possible outcomes in a probability distribution is always $1$.
Therefore,$P(X \le 2) = 1$.

(NONE) The total number of balls is $25$. The odd numbers between $1$ and $25$ are ${1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25}$,which are $13$ in total.
Probability of success $(p)$ $= \frac{13}{25}$.
Probability of failure $(q)$ $= 1 - p = 1 - \frac{13}{25} = \frac{12}{25}$.
Since $2$ balls are drawn with replacement,this follows a Binomial distribution $B(n, p)$ where $n = 2$.
We need the probability of at least $2$ successes,which is $P(X \ge 2) = P(X = 2)$.
$P(X = 2) = \binom{2}{2} p^2 q^0 = 1 \times (\frac{13}{25})^2 \times 1 = \frac{169}{625}$.

(C) Let $A$ be the event that the first card is a diamond.
Let $B$ be the event that the second card is a king.
Since the first card is replaced before the second is drawn,the events $A$ and $B$ are independent.
The probability of drawing a diamond is $P(A) = \frac{13}{52} = \frac{1}{4}$.
The probability of drawing a king is $P(B) = \frac{4}{52} = \frac{1}{13}$.
Since the events are independent,the required probability is $P(A \cap B) = P(A) \times P(B)$.
Therefore,the required probability $= \frac{1}{4} \times \frac{1}{13} = \frac{1}{52}$.

(A) Let $A$ be the event that the husband is selected and $B$ be the event that the wife is selected.
Given: $P(A) = \frac{1}{7}$ and $P(B) = \frac{1}{5}$.
Then,the probability of not being selected is:
$P(\bar{A}) = 1 - P(A) = 1 - \frac{1}{7} = \frac{6}{7}$
$P(\bar{B}) = 1 - P(B) = 1 - \frac{1}{5} = \frac{4}{5}$
The probability that only $1$ of them is selected is given by the case where the husband is selected and the wife is not,$OR$ the wife is selected and the husband is not:
$P(\text{only one}) = P(A) \cdot P(\bar{B}) + P(B) \cdot P(\bar{A})$
$= \left(\frac{1}{7} \times \frac{4}{5}\right) + \left(\frac{1}{5} \times \frac{6}{7}\right)$
$= \frac{4}{35} + \frac{6}{35} = \frac{10}{35} = \frac{2}{7}$

(B) Let $P(H)$ be the probability of the husband's selection and $P(W)$ be the probability of the wife's selection.
Given,$P(H) = \frac{1}{7}$ and $P(W) = \frac{1}{5}$.
Since the selection of the husband and the wife are independent events,the probability that both are selected is given by the product of their individual probabilities.
$P(H \cap W) = P(H) \times P(W)$
$P(H \cap W) = \frac{1}{7} \times \frac{1}{5} = \frac{1}{35}$.

(C) Let $A$ be the event that the husband is selected and $B$ be the event that the wife is selected.
Given: $P(A) = \frac{1}{7}$ and $P(B) = \frac{1}{5}$.
The probability that the husband is not selected is $P(\bar{A}) = 1 - P(A) = 1 - \frac{1}{7} = \frac{6}{7}$.
The probability that the wife is not selected is $P(\bar{B}) = 1 - P(B) = 1 - \frac{1}{5} = \frac{4}{5}$.
Since the events are independent,the probability that none of them will be selected is $P(\bar{A} \cap \bar{B}) = P(\bar{A}) \times P(\bar{B})$.
$P(\bar{A} \cap \bar{B}) = \frac{6}{7} \times \frac{4}{5} = \frac{24}{35}$.

(D) Let $A$ be the event that the husband is selected and $B$ be the event that the wife is selected.
Given: $P(A) = 1/7$ and $P(B) = 1/5$.
The probability that the husband is not selected is $P(\bar{A}) = 1 - 1/7 = 6/7$.
The probability that the wife is not selected is $P(\bar{B}) = 1 - 1/5 = 4/5$.
The probability that at least one of them is selected is given by $1 - P(\text{neither is selected})$.
$P(\text{at least one}) = 1 - P(\bar{A}) \times P(\bar{B})$.
$P(\text{at least one}) = 1 - (6/7 \times 4/5) = 1 - 24/35$.
$P(\text{at least one}) = (35 - 24) / 35 = 11/35$.

(A) Let $A$ be the event that the husband will be alive $25$ years hence.
Let $B$ be the event that the wife will be alive $25$ years hence.
Given that the events $A$ and $B$ are independent,the probability that both will be alive is given by the product of their individual probabilities.
$P(A) = 0.3$
$P(B) = 0.4$
Since the events are independent,the probability that both will be alive is $P(A \cap B) = P(A) \times P(B)$.
$P(A \cap B) = 0.3 \times 0.4 = 0.12$.

(B) Let $A$ be the event that the man is alive after $25$ years and $B$ be the event that his wife is alive after $25$ years.
Given: $P(A) = 0.3$ and $P(B) = 0.4$.
The probability that the wife will not be alive after $25$ years is $P(\bar{B}) = 1 - P(B) = 1 - 0.4 = 0.6$.
We need to find the probability that only the man will be alive,which means the man is alive $AND$ the wife is not alive.
Required probability $= P(A \cap \bar{B}) = P(A) \times P(\bar{B}) = 0.3 \times 0.6 = 0.18$.

(C) Let $A$ be the event that the man is alive after $25$ years and $B$ be the event that the wife is alive after $25$ years.
Given: $P(A) = 0.3$ and $P(B) = 0.4$.
The probability that the man is not alive is $P(A') = 1 - P(A) = 1 - 0.3 = 0.7$.
We need to find the probability that only the woman will be alive,which means the woman is alive $(B)$ and the man is not alive $(A')$.
Since these are independent events,the required probability is $P(A' \cap B) = P(A') \times P(B)$.
Required probability $= 0.7 \times 0.4 = 0.28$.

(D) Let $A$ be the event that the man is alive and $B$ be the event that the wife is alive after $25$ years.
Given: $P(A) = 0.3$ and $P(B) = 0.4$.
The probability that the man is not alive is
$P(\bar{A}) = 1 - 0.3 = 0.7$.
The probability that the wife is not alive is
$P(\bar{B}) = 1 - 0.4 = 0.6$.
The probability that at least $1$ of them will be alive is given by
$1 - P(\text{none of them is alive})$.
Since the events are independent,
$P(\bar{A} \cap \bar{B}) = P(\bar{A}) \times P(\bar{B}) = 0.7 \times 0.6 = 0.42$.
Therefore, the required probability
$= 1 - 0.42 = 0.58$.

(B) Let the two men be $A$ and $B$. Let $P(A)$ be the probability that $A$ speaks the truth and $P(B)$ be the probability that $B$ speaks the truth.
Given: $P(A) = \frac{80}{100} = 0.8$ and $P(B) = \frac{90}{100} = 0.9$.
The probability that $A$ lies is $P(\bar{A}) = 1 - 0.8 = 0.2$.
The probability that $B$ lies is $P(\bar{B}) = 1 - 0.9 = 0.1$.
They contradict each other if one speaks the truth and the other lies.
This can happen in two mutually exclusive ways: ($A$ speaks truth $AND$ $B$ lies) $OR$ ($B$ speaks truth $AND$ $A$ lies).
Required Probability $= P(A) \times P(\bar{B}) + P(B) \times P(\bar{A})$.
$= (0.8 \times 0.1) + (0.9 \times 0.2)$.
$= 0.08 + 0.18 = 0.26$.
$= \frac{26}{100} = \frac{13}{50}$.

(A) Let $R_A, R_B, R_C$ be the events of drawing a red ball from urns $A, B, C$ respectively,and $B_A, B_B, B_C$ be the events of drawing a black ball from urns $A, B, C$ respectively.
Probabilities are:
$P(R_A) = \frac{4}{7}, P(B_A) = \frac{3}{7}$
$P(R_B) = \frac{5}{9}, P(B_B) = \frac{4}{9}$
$P(R_C) = \frac{4}{8} = \frac{1}{2}, P(B_C) = \frac{4}{8} = \frac{1}{2}$
To get $2$ red balls and $1$ black ball,the possible cases are:
$1$. $(R_A, R_B, B_C) = \frac{4}{7} \times \frac{5}{9} \times \frac{1}{2} = \frac{20}{126}$
$2$. $(R_A, B_B, R_C) = \frac{4}{7} \times \frac{4}{9} \times \frac{1}{2} = \frac{16}{126}$
$3$. $(B_A, R_B, R_C) = \frac{3}{7} \times \frac{5}{9} \times \frac{1}{2} = \frac{15}{126}$
Total probability $= \frac{20+16+15}{126} = \frac{51}{126} = \frac{17}{42}$.

(C) Let $A, B, C,$ and $D$ be the events that the plane is hit by the first,second,third,and fourth shots,respectively.
The probabilities of hitting the plane are $P(A) = 0.4, P(B) = 0.3, P(C) = 0.2,$ and $P(D) = 0.1$.
The probabilities of missing the plane in each shot are:
$P(\bar{A}) = 1 - 0.4 = 0.6$
$P(\bar{B}) = 1 - 0.3 = 0.7$
$P(\bar{C}) = 1 - 0.2 = 0.8$
$P(\bar{D}) = 1 - 0.1 = 0.9$
The gun hits the plane if at least one shot hits the target. This is the complement of the event that all shots miss the target.
Required probability $= 1 - P(\text{all shots miss})$
$= 1 - [P(\bar{A}) \times P(\bar{B}) \times P(\bar{C}) \times P(\bar{D})]$
$= 1 - [0.6 \times 0.7 \times 0.8 \times 0.9]$
$= 1 - 0.3024$
$= 0.6976$

(B) Let $A$ be the event that $A$ solves the problem and $B$ be the event that $B$ solves the problem.
Given $P(A) = \frac{90}{100} = 0.9$ and $P(B) = \frac{70}{100} = 0.7$.
The probability that $A$ does not solve the problem is $P(\bar{A}) = 1 - P(A) = 1 - 0.9 = 0.1$.
The probability that $B$ does not solve the problem is $P(\bar{B}) = 1 - P(B) = 1 - 0.7 = 0.3$.
The probability that at least one of them solves the problem is $1 - P(\text{none solves the problem})$.
$P(\text{none solves}) = P(\bar{A}) \times P(\bar{B}) = 0.1 \times 0.3 = 0.03$.
Therefore,the required probability is $1 - 0.03 = 0.97 = \frac{97}{100}$.

(B) Let $p$ be the probability of getting a head,so $p = \frac{1}{2}$.
Let $q$ be the probability of getting a tail,so $q = 1 - p = \frac{1}{2}$.
$A$ starts the game. $A$ wins if $A$ gets a head in the $1^{st}$ throw,or $A$ gets a tail,$B$ gets a tail,and $A$ gets a head in the $3^{rd}$ throw,and so on.
The probability of $A$ winning is given by the sum of an infinite geometric series:
$P(A) = p + qqp + qqqqp + \dots$
$P(A) = p(1 + q^2 + q^4 + \dots)$
This is an infinite geometric series with first term $a = p$ and common ratio $r = q^2$.
The sum is $S = \frac{a}{1 - r} = \frac{p}{1 - q^2}$.
Substituting the values $p = \frac{1}{2}$ and $q = \frac{1}{2}$:
$P(A) = \frac{1/2}{1 - (1/2)^2} = \frac{1/2}{1 - 1/4} = \frac{1/2}{3/4} = \frac{1}{2} \times \frac{4}{3} = \frac{2}{3}$.

(A) Let $p$ be the probability of getting a '$6$' in a single throw,so $p = \frac{1}{6}$.
Let $q$ be the probability of not getting a '$6$',so $q = 1 - p = 1 - \frac{1}{6} = \frac{5}{6}$.
$A$ throws first. $B$ wins if $A$ fails,then $B$ succeeds,or $A$ fails,$B$ fails,$A$ fails,$B$ succeeds,and so on.
$P(B \text{ winning}) = qp + qqqp + qqqqqp + \dots$
This is an infinite geometric series with first term $a = qp = \frac{5}{6} \times \frac{1}{6} = \frac{5}{36}$ and common ratio $r = q^2 = (\frac{5}{6})^2 = \frac{25}{36}$.
The sum of an infinite geometric series is $S = \frac{a}{1 - r}$.
$P(B \text{ winning}) = \frac{5/36}{1 - 25/36} = \frac{5/36}{11/36} = \frac{5}{11}$.

(D) The word $SOCIETY$ contains $7$ distinct letters: $S, O, C, I, E, T, Y$.
The total number of ways to arrange these $7$ letters is $n(S) = 7!$.
The vowels in the word are $O, I, E$ (total $3$ vowels).
To find the number of arrangements where the $3$ vowels come together,we treat the group $(O, I, E)$ as a single unit.
Now,we have $5$ units to arrange: $(OIE), S, C, T, Y$.
These $5$ units can be arranged in $5!$ ways.
Within the group $(OIE)$,the $3$ vowels can be arranged among themselves in $3!$ ways.
Thus,the number of favorable outcomes is $n(A) = 5! \times 3!$.
The probability $P(A) = \frac{n(A)}{n(S)} = \frac{5! \times 3!}{7!} = \frac{5! \times 6}{7 \times 6 \times 5!} = \frac{6}{42} = \frac{1}{7}$.

(A) The word $DAUGHTER$ contains $8$ distinct letters $(D, A, U, G, H, T, E, R)$.
The total number of ways to arrange these $8$ letters is $n(S) = 8!$.
If the letter $D$ is fixed at the first position,we need to arrange the remaining $7$ letters in the remaining $7$ positions.
The number of favorable outcomes is $n(A) = 7!$.
Therefore,the required probability is $P(A) = \frac{n(A)}{n(S)} = \frac{7!}{8!} = \frac{7!}{8 \times 7!} = \frac{1}{8}$.

(B) Total number of balls = $3$ (red) $+ 5$ (yellow) $+ 7$ (pink) $= 15$ balls.
Number of favorable outcomes (pink or red) = $7 + 3 = 10$.
Probability $P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$.
$P(E) = \frac{10}{15} = \frac{2}{3}$.

(D) Total number of balls $= 13 + 7 = 20$.
The total number of ways to draw $2$ balls from $20$ is given by the sample space $n(S) = {}^{20}C_{2} = \frac{20 \times 19}{2 \times 1} = 190$.
For the balls to be of the same colour,they must either both be white or both be black.
Number of ways to choose $2$ white balls from $13$ is ${}^{13}C_{2} = \frac{13 \times 12}{2 \times 1} = 78$.
Number of ways to choose $2$ black balls from $7$ is ${}^{7}C_{2} = \frac{7 \times 6}{2 \times 1} = 21$.
Total number of favorable outcomes $n(E) = 78 + 21 = 99$.
Therefore,the probability $P(E) = \frac{n(E)}{n(S)} = \frac{99}{190}$.

(D) Total number of marbles in the urn $= 4 + 5 + 2 + 3 = 14$.
Total number of ways to select $2$ marbles out of $14$ is given by ${}^{14}C_2 = \frac{14 \times 13}{2 \times 1} = 91$.
The event 'both are red or at least $1$ is red' is equivalent to the event 'at least $1$ is red'.
Number of ways to select $0$ red marbles $= {}^{12}C_2 = \frac{12 \times 11}{2 \times 1} = 66$.
Number of ways to select at least $1$ red marble $= \text{Total ways} - \text{Ways to select } 0 \text{ red marbles} = 91 - 66 = 25$.
Therefore,the required probability $= 25/91$.

(B) Total number of marbles $= 4 + 5 + 2 + 3 = 14$.
Total number of ways to draw $3$ marbles from $14$ is ${}^{14}C_{3} = \frac{14 \times 13 \times 12}{3 \times 2 \times 1} = 364$.
Number of non-yellow marbles $= 14 - 3 = 11$.
Number of ways to draw $3$ marbles such that none of them is yellow is ${}^{11}C_{3} = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165$.
Probability that no marble is yellow $= \frac{165}{364}$.
Probability that at least $1$ marble is yellow $= 1 - P(\text{no yellow marble}) = 1 - \frac{165}{364} = \frac{364 - 165}{364} = \frac{199}{364}$.

(C) Total number of marbles $= 4 + 5 + 2 + 3 = 14$.
Total ways to draw $8$ marbles from $14$ is given by ${}^{14}C_{8}$.
Since ${}^{n}C_{r} = {}^{n}C_{n-r}$,${}^{14}C_{8} = {}^{14}C_{6} = \frac{14 \times 13 \times 12 \times 11 \times 10 \times 9}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 3003$.
To have an equal number of marbles of each colour,we must draw $2$ marbles of each of the $4$ colours ($2$ green,$2$ blue,$2$ red,$2$ yellow).
Number of favourable outcomes $= {}^{4}C_{2} \times {}^{5}C_{2} \times {}^{2}C_{2} \times {}^{3}C_{2}$.
$= 6 \times 10 \times 1 \times 3 = 180$.
Required probability $= \frac{180}{3003} = \frac{60}{1001}$.

(D) Total number of marbles = $4 + 5 + 2 + 3 = 14$.
Total number of ways to draw $3$ marbles from $14$ is given by ${}^{14}C_{3} = \frac{14 \times 13 \times 12}{3 \times 2 \times 1} = 364$.
We want to find the probability that none of the $3$ marbles is green.
This means all $3$ marbles must be chosen from the non-green marbles (blue,red,and yellow).
Number of non-green marbles = $5 + 2 + 3 = 10$.
Number of ways to choose $3$ non-green marbles = ${}^{10}C_{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$.
Therefore,the required probability = $\frac{\text{Favourable outcomes}}{\text{Total outcomes}} = \frac{120}{364}$.
Simplifying the fraction by dividing both numerator and denominator by $4$,we get $\frac{30}{91}$.

(A) Total number of marbles $= 4 + 5 + 2 + 3 = 14$.
Total number of ways to draw $4$ marbles from $14$ is given by ${}^{14}C_{4}$.
${}^{14}C_{4} = \frac{14 \times 13 \times 12 \times 11}{4 \times 3 \times 2 \times 1} = 1001$.
Number of ways to choose $2$ blue marbles from $5$ blue marbles is ${}^{5}C_{2} = \frac{5 \times 4}{2 \times 1} = 10$.
Number of ways to choose $2$ red marbles from $2$ red marbles is ${}^{2}C_{2} = 1$.
Total number of favourable outcomes $= {}^{5}C_{2} \times {}^{2}C_{2} = 10 \times 1 = 10$.
Required probability $= \frac{\text{Favourable outcomes}}{\text{Total outcomes}} = \frac{10}{1001}$.

(A) Total number of children $= 5 + 3 = 8$.
Number of ways to select $4$ children out of $8$ is given by the combination formula ${}^{n}C_{r} = \frac{n!}{r!(n-r)!}$.
Total ways $= {}^{8}C_{4} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70$.
Number of ways to select $4$ girls out of $5$ is ${}^{5}C_{4} = {}^{5}C_{1} = 5$.
Probability of selecting $4$ girls $= \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{5}{70} = \frac{1}{14}$.

(B) Total number of balls $= 3 + 4 = 7$.
Total ways to draw $3$ balls from $7$ is given by ${}^{7}C_{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$.
Case $1$: All $3$ balls are blue. The number of ways to choose $3$ blue balls from $3$ is ${}^{3}C_{3} = 1$.
Case $2$: All $3$ balls are red. The number of ways to choose $3$ red balls from $4$ is ${}^{4}C_{3} = 4$.
Since these are mutually exclusive events,the total number of favorable outcomes is $1 + 4 = 5$.
Required probability $= \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{5}{35} = \frac{1}{7}$.

(D) Total number of marbles $= 4 + 5 + 3 = 12$.
$n(S) =$ Total ways to pick $3$ marbles out of $12 = {}^{12}C_{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220$.
$n(E) =$ Favourable outcomes (either all green or all red).
Ways to pick $3$ green marbles from $3 = {}^{3}C_{3} = 1$.
Ways to pick $3$ red marbles from $4 = {}^{4}C_{3} = 4$.
Total favourable outcomes $n(E) = 1 + 4 = 5$.
Required probability $= \frac{n(E)}{n(S)} = \frac{5}{220} = \frac{1}{44}$.

(D) Total number of marbles $= 4 + 5 + 3 = 12$.
Number of ways to draw $2$ marbles out of $12$ is given by $n(S) = {}^{12}C_2 = \frac{12 \times 11}{2 \times 1} = 66$.
Number of ways to draw $2$ red marbles out of $4$ red marbles is given by $n(E) = {}^{4}C_2 = \frac{4 \times 3}{2 \times 1} = 6$.
The probability of drawing $2$ red marbles is $P(E) = \frac{n(E)}{n(S)} = \frac{6}{66} = \frac{1}{11}$.
Since $\frac{1}{11}$ is not among the given options,the correct answer is 'None of these'.

(B) Total number of marbles $= 4 + 5 + 3 = 12$.
Total ways to pick $3$ marbles from $12$ is $n(S) = {}^{12}C_3 = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220$.
To find the probability that at least $1$ marble is blue,we use the complement rule: $P(\text{at least } 1 \text{ blue}) = 1 - P(\text{no blue marble})$.
Number of non-blue marbles $= 4 \text{ (red)} + 3 \text{ (green)} = 7$.
Ways to pick $3$ marbles such that none are blue is $n(E') = {}^{7}C_3 = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$.
Probability of picking no blue marbles $P(E') = \frac{35}{220} = \frac{7}{44}$.
Therefore,the probability of picking at least $1$ blue marble is $1 - \frac{7}{44} = \frac{37}{44}$.