Conditional probability Questions in English

(C) Given that $A$ and $B$ are independent events,$A$ and $B^c$ are also independent events.
$P(B) = \frac{2}{7} \implies P(B^c) = 1 - \frac{2}{7} = \frac{5}{7}$.
We know that $P(A \cup B^c) = P(A) + P(B^c) - P(A \cap B^c)$.
Since $A$ and $B^c$ are independent,$P(A \cap B^c) = P(A) \cdot P(B^c)$.
Substituting the values:
$0.8 = P(A) + \frac{5}{7} - P(A) \cdot \frac{5}{7}$.
$0.8 - \frac{5}{7} = P(A) \cdot (1 - \frac{5}{7})$.
$\frac{5.6 - 5}{7} = P(A) \cdot \frac{2}{7}$.
$0.6 = 2 \cdot P(A)$.
$P(A) = 0.3$.

(A) We know that the conditional probability formula is given by $P(B \mid A) = \frac{P(A \cap B)}{P(A)}$.
Substituting the given values,we get $P(A \cap B) = P(B \mid A) \times P(A) = \frac{2}{3} \times \frac{1}{4} = \frac{2}{12} = \frac{1}{6}$.
Now,we use the conditional probability formula for $P(A \mid B)$:
$P(A \mid B) = \frac{P(A \cap B)}{P(B)}$.
Rearranging for $P(B)$,we get $P(B) = \frac{P(A \cap B)}{P(A \mid B)}$.
Substituting the values,$P(B) = \frac{1/6}{1/2} = \frac{1}{6} \times 2 = \frac{1}{3}$.

(A) Let $H$ denote the event of getting a head and $T$ denote the event of getting a tail. The experiment stops when a head appears or after three tosses.
The sample space of the experiment is $S = \{H, TH, TTH, TTT\}$.
Let $A$ be the event that a head does not appear on the first toss. This means the first toss is a tail $(T)$.
The outcomes corresponding to event $A$ are $\{TH, TTH, TTT\}$.
The probability $P(A) = P(T) = \frac{1}{2}$.
Let $B$ be the event that the coin is tossed thrice. The outcomes corresponding to event $B$ are $\{TTH, TTT\}$.
The intersection $A \cap B$ represents the event that the first toss is a tail $AND$ the coin is tossed thrice. This corresponds to the outcomes $\{TTH, TTT\}$.
$P(A \cap B) = P(TTH) + P(TTT) = \frac{1}{8} + \frac{1}{8} = \frac{2}{8} = \frac{1}{4}$.
We need to find the conditional probability $P(B|A)$:
$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{2}$.

(C) The sum of probabilities for a random variable is $1$.
Given $P(X=r) = K r^3$ for $r \in \{1, 2, 3, 4\}$.
Summing these: $K(1^3) + K(2^3) + K(3^3) + K(4^3) = 1$.
$K(1 + 8 + 27 + 64) = 1 \Rightarrow 100K = 1 \Rightarrow K = \frac{1}{100}$.
We need to find $P\left(\left.\frac{1}{2} < X < \frac{5}{2} \right\rvert X > 1\right)$.
This is equivalent to $P(X=2 \mid X \in \{2, 3, 4\})$.
By the definition of conditional probability: $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$.
Here $A = \{X=2\}$ and $B = \{X=2, 3, 4\}$.
$P(A \cap B) = P(X=2) = 8K$.
$P(B) = P(X=2) + P(X=3) + P(X=4) = 8K + 27K + 64K = 99K$.
Thus,$P(A \mid B) = \frac{8K}{99K} = \frac{8}{99}$.
Therefore,$K = \frac{1}{100}$ and the conditional probability is $\frac{8}{99}$.

(A) The possible sums of two dice that are prime numbers are $2, 3, 5, 7, 11$.
The sample space $S$ of pairs $(x, y)$ where $x+y$ is prime is:
Sum $= 2: (1, 1)$
Sum $= 3: (1, 2), (2, 1)$
Sum $= 5: (1, 4), (4, 1), (2, 3), (3, 2)$
Sum $= 7: (1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3)$
Sum $= 11: (5, 6), (6, 5)$
Total number of outcomes $n(S) = 1 + 2 + 4 + 6 + 2 = 15$.
We need the probability that at least one number is a multiple of $3$ (i.e.,$3$ or $6$).
The favorable outcomes are: $(2, 3), (3, 2), (1, 6), (6, 1), (3, 4), (4, 3), (5, 6), (6, 5)$.
Number of favorable outcomes $n(E) = 8$.
The required probability $P(E) = \frac{n(E)}{n(S)} = \frac{8}{15}$.

(B) Given $P(A / B) = \frac{P(A \cap B)}{P(B)} = \frac{3}{10} \implies P(A \cap B) = \frac{3}{10} P(B)$.
$P(B / A) = \frac{P(A \cap B)}{P(A)} = \frac{4}{5} \implies P(A) = \frac{5}{4} P(A \cap B) = \frac{5}{4} \times \frac{3}{10} P(B) = \frac{3}{8} P(B)$.
We know $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Given $P(A \cup B) = K P(B)$,so $K P(B) = P(A) + P(B) - P(A \cap B)$.
Dividing by $P(B)$,we get $K = \frac{P(A)}{P(B)} + 1 - \frac{P(A \cap B)}{P(B)}$.
Substituting the values: $K = \frac{3}{8} + 1 - \frac{3}{10}$.
$K = \frac{15 + 40 - 12}{40} = \frac{43}{40}$.
Therefore,$\frac{1}{K} = \frac{40}{43}$.

(C) Let $E$ be the event of selecting a science book.
Let $A$ be the event of selecting a book from group $A$,and $B$ be the event of selecting a book from group $B$.
The probability of getting $2$ or $5$ on a die is $P(A) = \frac{2}{6} = \frac{1}{3}$.
The probability of not getting $2$ or $5$ is $P(B) = 1 - \frac{1}{3} = \frac{2}{3}$.
In group $A$,there are $8$ science and $5$ engineering books,total $13$ books. So,$P(E|A) = \frac{8}{13}$.
In group $B$,there are $6$ science and $7$ engineering books,total $13$ books. So,$P(E|B) = \frac{6}{13}$.
Using the law of total probability:
$P(E) = P(A) \times P(E|A) + P(B) \times P(E|B)$
$P(E) = \frac{1}{3} \times \frac{8}{13} + \frac{2}{3} \times \frac{6}{13}$
$P(E) = \frac{8}{39} + \frac{12}{39} = \frac{20}{39}$

(B) Let $S = \{1, 2, \ldots, 100\}$. The total number of elements is $100$.
Let $A$ be the event that the number is divisible by $2$. The elements of $A$ are $\{2, 4, 6, \ldots, 100\}$. The number of elements in $A$ is $n(A) = 50$.
Let $B$ be the event that the number is divisible by $3$ or $5$. We want to find the conditional probability $P(B|A) = \frac{n(A \cap B)}{n(A)}$.
$A \cap B$ is the set of numbers in $\{1, 2, \ldots, 100\}$ that are divisible by $2$ $AND$ (divisible by $3$ $OR$ divisible by $5$).
This is equivalent to numbers divisible by $6$ or $10$.
Numbers divisible by $6$ up to $100$: $\lfloor \frac{100}{6} \rfloor = 16$.
Numbers divisible by $10$ up to $100$: $\lfloor \frac{100}{10} \rfloor = 10$.
Numbers divisible by both $6$ and $10$ (i.e.,divisible by $30$): $\lfloor \frac{100}{30} \rfloor = 3$.
By the Principle of Inclusion-Exclusion,$n(A \cap B) = 16 + 10 - 3 = 23$.
The required probability is $\frac{n(A \cap B)}{n(A)} = \frac{23}{50}$.

(B) Let $A$ be the event that the sum of the numbers appearing on the two dice is $6$. The possible outcomes for $A$ are: $(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)$.
Thus,the number of outcomes in $A$ is $n(A) = 5$.
Let $B$ be the event that the number $1$ appears at least once.
We are looking for the conditional probability $P(B|A)$,which is defined as $P(B|A) = \frac{n(A \cap B)}{n(A)}$.
The intersection $A \cap B$ represents the outcomes where the sum is $6$ $AND$ the number $1$ appears at least once. These outcomes are: $(1, 5)$ and $(5, 1)$.
Thus,$n(A \cap B) = 2$.
Therefore,the required probability is $P(B|A) = \frac{2}{5}$.

(A) Given that,$P(A)=\frac{1}{3}$,$P(A \cap B)=\frac{1}{5}$,$P(A \cup B)=\frac{3}{5}$.
We know that,$P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Substituting the values: $\frac{3}{5} = \frac{1}{3} + P(B) - \frac{1}{5}$.
$P(B) = \frac{3}{5} + \frac{1}{5} - \frac{1}{3} = \frac{9+3-5}{15} = \frac{7}{15}$.
Now,matching the items:
$A$. $P(\frac{A}{B}) = \frac{P(A \cap B)}{P(B)} = \frac{1/5}{7/15} = \frac{3}{7}$ (Matches $(v)$).
$B$. $P(\bar{B}) = 1 - P(B) = 1 - \frac{7}{15} = \frac{8}{15}$ (Matches $(iii)$).
$C$. $P(A \cap \bar{B}) = P(A) - P(A \cap B) = \frac{1}{3} - \frac{1}{5} = \frac{2}{15}$ (Matches $(i)$).
$D$. $P(B \cap \bar{A}) = P(B) - P(A \cap B) = \frac{7}{15} - \frac{1}{5} = \frac{4}{15}$ (Matches $(ii)$).
Thus,the correct matching is: $A-(v), B-(iii), C-(i), D-(ii)$.

(A) Given: $P(A | B) = 0.6$,$P(B | A) = 0.3$,and $P(A) = 0.1$.
Using the definition of conditional probability,$P(B | A) = \frac{P(A \cap B)}{P(A)}$.
Substituting the values: $0.3 = \frac{P(A \cap B)}{0.1} \implies P(A \cap B) = 0.03$.
Next,$P(A | B) = \frac{P(A \cap B)}{P(B)}$.
Substituting the values: $0.6 = \frac{0.03}{P(B)} \implies P(B) = \frac{0.03}{0.6} = 0.05$.
By De Morgan's Law,$P(\bar{A} \cap \bar{B}) = P(\overline{A \cup B}) = 1 - P(A \cup B)$.
Using the addition rule,$P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
$P(A \cup B) = 0.1 + 0.05 - 0.03 = 0.12$.
Therefore,$P(\bar{A} \cap \bar{B}) = 1 - 0.12 = 0.88$.

(A) Let $S$ denote success (passing) and $F$ denote failure. The probability of passing the first test is $P(S_1) = p$,so $P(F_1) = 1-p$.
For subsequent tests,$P(S_{n+1} | S_n) = p$ and $P(S_{n+1} | F_n) = \frac{p}{2}$.
The candidate is selected if they pass at least two tests. The possible favorable outcomes are $(S, S, S), (S, S, F), (S, F, S), (F, S, S)$.
$P(S, S, S) = P(S_1) \times P(S_2|S_1) \times P(S_3|S_2) = p \times p \times p = p^3$.
$P(S, S, F) = P(S_1) \times P(S_2|S_1) \times P(F_3|S_2) = p \times p \times (1-p) = p^2(1-p)$.
$P(S, F, S) = P(S_1) \times P(F_2|S_1) \times P(S_3|F_2) = p \times (1-p) \times \frac{p}{2} = \frac{p^2(1-p)}{2}$.
$P(F, S, S) = P(F_1) \times P(S_2|F_1) \times P(S_3|S_2) = (1-p) \times \frac{p}{2} \times p = \frac{p^2(1-p)}{2}$.
Summing these probabilities: $p^3 + p^2(1-p) + \frac{p^2(1-p)}{2} + \frac{p^2(1-p)}{2} = p^3 + p^2(1-p) + p^2(1-p) = p^3 + 2p^2(1-p) = p^3 + 2p^2 - 2p^3 = 2p^2 - p^3 = p^2(2-p)$.

(D) Given that $A$ and $B$ are mutually exclusive events,we have $P(A \cap B) = 0$.
By the definition of conditional probability,$P(A \mid \bar{B}) = \frac{P(A \cap \bar{B})}{P(\bar{B})}$.
Since $A = (A \cap B) \cup (A \cap \bar{B})$ and these are disjoint,we have $P(A) = P(A \cap B) + P(A \cap \bar{B})$.
Thus,$P(A \cap \bar{B}) = P(A) - P(A \cap B) = P(A) - 0 = P(A)$.
Also,$P(\bar{B}) = 1 - P(B)$.
Therefore,$P(A \mid \bar{B}) = \frac{P(A)}{1 - P(B)}$.

(C) Given: $P(A) = \frac{1}{2}, P(B) = \frac{1}{3}, P(A \cap B) = \frac{1}{4}$.
First,we find $P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{1/4}{1/3} = \frac{3}{4}$.
Next,we find $P(A^c \cap B^c) = P((A \cup B)^c) = 1 - P(A \cup B)$.
$P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{1}{2} + \frac{1}{3} - \frac{1}{4} = \frac{6+4-3}{12} = \frac{7}{12}$.
So,$P(A^c \cap B^c) = 1 - \frac{7}{12} = \frac{5}{12}$.
Also,$P(B^c) = 1 - P(B) = 1 - \frac{1}{3} = \frac{2}{3}$.
Then,$P(A^c|B^c) = \frac{P(A^c \cap B^c)}{P(B^c)} = \frac{5/12}{2/3} = \frac{5}{12} \times \frac{3}{2} = \frac{5}{8}$.
Finally,$P(A^c|B^c) + P(A|B) = \frac{5}{8} + \frac{3}{4} = \frac{5+6}{8} = \frac{11}{8}$.

(C) The total number of cards is $52$.
Since the cards are drawn with replacement,the events $A$ and $B$ are independent.
Event $A$ is drawing a face card in the first draw. There are $12$ face cards in a deck ($4$ Jacks,$4$ Queens,$4$ Kings).
Thus,$P(A) = \frac{12}{52} = \frac{3}{13}$.
Event $B$ is drawing a club card in the second draw. There are $13$ club cards in a deck.
Thus,$P(B) = \frac{13}{52} = \frac{1}{4}$.
Since the events are independent,$P(B|A) = P(B) = \frac{1}{4}$.
We need to find $P(\overline{B}|A)$.
Using the property of complementary events,$P(\overline{B}|A) = 1 - P(B|A)$.
$P(\overline{B}|A) = 1 - \frac{1}{4} = \frac{3}{4}$.
Therefore,the correct option is $C$.

(C) Let $W_P, R_P, B_P$ be the events of drawing a white,red,or blue ball from bag $P$ respectively.
Bag $P$ has $3+2+5 = 10$ balls. So,$P(W_P) = \frac{3}{10}, P(R_P) = \frac{2}{10}, P(B_P) = \frac{5}{10}$.
After transferring a ball to bag $Q$,bag $Q$ will have $10+1 = 11$ balls.
If a white ball is transferred,bag $Q$ has $3$ white,$3$ red,$5$ blue balls. $P(R|W_P) = \frac{3}{11}$.
If a red ball is transferred,bag $Q$ has $2$ white,$4$ red,$5$ blue balls. $P(R|R_P) = \frac{4}{11}$.
If a blue ball is transferred,bag $Q$ has $2$ white,$3$ red,$6$ blue balls. $P(R|B_P) = \frac{3}{11}$.
Using the law of total probability:
$P(R) = P(W_P) \times P(R|W_P) + P(R_P) \times P(R|R_P) + P(B_P) \times P(R|B_P)$
$P(R) = \frac{3}{10} \times \frac{3}{11} + \frac{2}{10} \times \frac{4}{11} + \frac{5}{10} \times \frac{3}{11}$
$P(R) = \frac{9}{110} + \frac{8}{110} + \frac{15}{110} = \frac{32}{110} = \frac{16}{55}$.

(C) Let $A$ be the event that the sum is a multiple of $3$,so $A = \{3, 6, 9, 12\}$.
Let $B$ be the event that the sum is an odd number,so $B = \{3, 5, 7, 9, 11\}$.
We need to find the conditional probability $P(A|B) = \frac{n(A \cap B)}{n(B)}$.
For the sum to be both a multiple of $3$ and odd,the sum must be $3$ or $9$.
The pairs $(x, y)$ such that $x+y = 3$ are $(1, 2)$ and $(2, 1)$.
The pairs $(x, y)$ such that $x+y = 9$ are $(3, 6), (6, 3), (4, 5), (5, 4)$.
Thus,$A \cap B = \{(1, 2), (2, 1), (3, 6), (6, 3), (4, 5), (5, 4)\}$ and $n(A \cap B) = 6$.
The total number of outcomes where the sum is odd $(B)$ is $18$ (since exactly half of the $36$ outcomes result in an odd sum).
Therefore,$P(A|B) = \frac{6}{18} = \frac{1}{3}$.

(A) Case $1$: With replacement.
$p_1 = P(\text{First is Queen}) \times P(\text{Second is Diamond}) = \frac{4}{52} \times \frac{13}{52} = \frac{1}{13} \times \frac{1}{4} = \frac{1}{52}$.
Case $2$: Without replacement.
Let $Q_1$ be the event that the first card is a queen and $D_2$ be the event that the second card is a diamond.
If the first card is the Queen of Diamonds,$P(Q_1 \cap D_2) = P(Q_1 \cap D_1) \times P(D_2 | Q_1 \cap D_1) = \frac{1}{52} \times \frac{12}{51}$.
If the first card is a Queen other than the Queen of Diamonds,$P(Q_1 \cap D_2) = P(Q_1 \cap D_1^c) \times P(D_2 | Q_1 \cap D_1^c) = \frac{3}{52} \times \frac{13}{51}$.
$p_2 = \frac{1 \times 12 + 3 \times 13}{52 \times 51} = \frac{12 + 39}{52 \times 51} = \frac{51}{52 \times 51} = \frac{1}{52}$.
Therefore,$\frac{p_1}{p_2} = \frac{1/52}{1/52} = 1$.

(C) The set is $S = \{1, 2, 3, \ldots, 13\}$. There are $7$ odd numbers $\{1, 3, 5, 7, 9, 11, 13\}$ and $6$ even numbers $\{2, 4, 6, 8, 10, 12\}$.
Let $E$ be the event that the sum of the two selected numbers is even. This happens if both numbers are odd or both numbers are even.
Number of ways to select two odd numbers = $^7C_2 = \frac{7 \times 6}{2} = 21$.
Number of ways to select two even numbers = $^6C_2 = \frac{6 \times 5}{2} = 15$.
Total ways for the sum to be even = $21 + 15 = 36$.
Let $A$ be the event that both numbers are odd. We want to find $P(A|E) = \frac{n(A \cap E)}{n(E)}$.
Since $A \cap E$ is the event that both numbers are odd,$n(A \cap E) = 21$.
Therefore,$P(A|E) = \frac{21}{36} = \frac{7}{12}$.

(C) We have,$P(E_1 \cap E_2) = P(E_1) \cdot P(E_2 | E_1)$.
$\therefore P(E_1 \cap E_2) = \frac{1}{4} \cdot \frac{1}{3} = \frac{1}{12}$.
Now,$P(E_1 | E_2) = \frac{P(E_1 \cap E_2)}{P(E_2)}$.
$\Rightarrow \frac{1}{2} = \frac{1/12}{P(E_2)}$.
$\Rightarrow P(E_2) = \frac{1}{12} \times 2 = \frac{1}{6}$.
$\therefore P(\bar{E}_2) = 1 - P(E_2) = 1 - \frac{1}{6} = \frac{5}{6}$.
Now,$P(E_1 | \bar{E}_2) = \frac{P(E_1 \cap \bar{E}_2)}{P(\bar{E}_2)} = \frac{P(E_1) - P(E_1 \cap E_2)}{P(\bar{E}_2)}$.
$= \frac{\frac{1}{4} - \frac{1}{12}}{\frac{5}{6}} = \frac{\frac{3-1}{12}}{\frac{5}{6}} = \frac{\frac{2}{12}}{\frac{5}{6}} = \frac{1}{6} \times \frac{6}{5} = \frac{1}{5}$.

(B) Given that $P(A) = \frac{4}{9}$ and $P(A \cap \bar{B}) = \frac{3}{7}$.
We know that $P(A) = P(A \cap B) + P(A \cap \bar{B})$.
Therefore,$P(A \cap B) = P(A) - P(A \cap \bar{B})$.
Substituting the values,we get $P(A \cap B) = \frac{4}{9} - \frac{3}{7} = \frac{28 - 27}{63} = \frac{1}{63}$.
Now,the conditional probability $P\left(\frac{B}{A}\right)$ is defined as $P\left(\frac{B}{A}\right) = \frac{P(A \cap B)}{P(A)}$.
Substituting the values,$P\left(\frac{B}{A}\right) = \frac{\frac{1}{63}}{\frac{4}{9}} = \frac{1}{63} \times \frac{9}{4} = \frac{1}{7 \times 4} = \frac{1}{28}$.

(C) Given: $P(X)=\frac{1}{3}$,$P(X|Y)=\frac{1}{2}$,and $P(Y|X)=\frac{2}{5}$.
Using the definition of conditional probability,$P(Y|X) = \frac{P(X \cap Y)}{P(X)}$.
Therefore,$P(X \cap Y) = P(Y|X) \times P(X) = \frac{2}{5} \times \frac{1}{3} = \frac{2}{15}$.
Now,using $P(X|Y) = \frac{P(X \cap Y)}{P(Y)}$,we have $\frac{1}{2} = \frac{2/15}{P(Y)}$.
This implies $P(Y) = 2 \times \frac{2}{15} = \frac{4}{15}$.
Thus,option $C$ is correct.

(D) Total mangoes $= 10$. Good mangoes $= 6$. Spoiled mangoes $= 4$.
Let $E$ be the event that at least one mango is good,and $F$ be the event that both mangoes are good.
The number of ways to select $2$ mangoes from $10$ is $^{10}C_2 = \frac{10 \times 9}{2} = 45$.
The number of ways to select $2$ spoiled mangoes is $^4C_2 = \frac{4 \times 3}{2} = 6$.
The number of ways to select at least one good mango is $45 - 6 = 39$. So,$P(E) = \frac{39}{45}$.
The number of ways to select $2$ good mangoes is $^6C_2 = \frac{6 \times 5}{2} = 15$. So,$P(F) = \frac{15}{45}$.
Since $F \subset E$,$P(E \cap F) = P(F) = \frac{15}{45}$.
The conditional probability that the other is good given that one is good is $P(F|E) = \frac{P(F \cap E)}{P(E)} = \frac{15/45}{39/45} = \frac{15}{39} = \frac{5}{13}$.

(A) Let $X$ be the event that the sum of the numbers on the two dice is greater than $9$.
Let $Y$ be the event that the number on die $B$ is $5$.
The sample space for event $Y$ is $\{(1,5), (2,5), (3,5), (4,5), (5,5), (6,5)\}$. Thus,$n(Y) = 6$.
The event $X \cap Y$ represents the outcomes where the number on die $B$ is $5$ $AND$ the sum is greater than $9$.
The possible outcomes for $X \cap Y$ are $\{(5,5), (6,5)\}$. Thus,$n(X \cap Y) = 2$.
The conditional probability is given by $P(X|Y) = \frac{n(X \cap Y)}{n(Y)}$.
$P(X|Y) = \frac{2}{6} = \frac{1}{3}$.

(C) Given $P(\bar{A})=0.3$,so $P(A)=1-0.3=0.7$.
Given $P(B)=0.4$,so $P(\bar{B})=1-0.4=0.6$.
We know that $P(A \cap \bar{B}) = P(A) - P(A \cap B) = 0.5$.
Substituting $P(A)=0.7$,we get $0.7 - P(A \cap B) = 0.5$,which implies $P(A \cap B) = 0.2$.
Now,we need to find $P(B \mid A \cup \bar{B}) = \frac{P(B \cap (A \cup \bar{B}))}{P(A \cup \bar{B})}$.
First,calculate the denominator: $P(A \cup \bar{B}) = P(A) + P(\bar{B}) - P(A \cap \bar{B}) = 0.7 + 0.6 - 0.5 = 0.8$.
Next,calculate the numerator: $P(B \cap (A \cup \bar{B})) = P((B \cap A) \cup (B \cap \bar{B})) = P((B \cap A) \cup \emptyset) = P(A \cap B) = 0.2$.
Therefore,$P(B \mid A \cup \bar{B}) = \frac{0.2}{0.8} = 0.25$.

(A) Let $A$ be the event that a flight departs on time and $B$ be the event that a flight arrives on time.
Given probabilities are:
$P(A) = 80 \% = 0.80$
$P(B) = 70 \% = 0.70$
$P(A \cap B) = 65 \% = 0.65$
We need to find the conditional probability that a flight arrives on time given that it has departed on time,which is $P(B|A)$.
Using the formula for conditional probability:
$P(B|A) = \frac{P(A \cap B)}{P(A)}$
Substituting the values:
$P(B|A) = \frac{0.65}{0.80} = \frac{65}{80} = \frac{13}{16}$
Thus,the probability is $\frac{13}{16}$.

(A) Let $G$ denote a good article and $B$ denote a bad article. Total articles $= n + m$.
We are picking $2$ articles in succession without replacement.
The second article is bad in two mutually exclusive cases:
$1$. The first article is bad and the second article is bad $(B_1 \cap B_2)$.
$2$. The first article is good and the second article is bad $(G_1 \cap B_2)$.
The probability is given by:
$P(B_2) = P(B_1 \cap B_2) + P(G_1 \cap B_2)$
$P(B_2) = P(B_1) \cdot P(B_2|B_1) + P(G_1) \cdot P(B_2|G_1)$
$P(B_2) = \left( \frac{m}{n+m} \right) \cdot \left( \frac{m-1}{n+m-1} \right) + \left( \frac{n}{n+m} \right) \cdot \left( \frac{m}{n+m-1} \right)$
$P(B_2) = \frac{m(m-1) + nm}{(n+m)(n+m-1)}$
$P(B_2) = \frac{m^2 - m + nm}{(n+m)(n+m-1)}$
$P(B_2) = \frac{m(m + n - 1)}{(n+m)(n+m-1)}$
$P(B_2) = \frac{m}{n+m}$

(C) Given that,$P(E) = 1/5$ and $P(F|E) = 1/10$.
We know that the probability of both events occurring is given by $P(E \cap F) = P(E) \cdot P(F|E)$.
Substituting the values,we get $P(E \cap F) = (1/5) \cdot (1/10) = 1/50$.
The probability of non-occurrence of at least one of the events $E$ and $F$ is equivalent to the complement of the event that both $E$ and $F$ occur.
Thus,the required probability is $1 - P(E \cap F)$.
Calculating this,we get $1 - 1/50 = 49/50$.

(B) The total number of ways to choose $2$ cards from the same suit is given by choosing one suit out of $4$ and then choosing $2$ cards out of $13$ from that suit. Total ways $= 4 \times \binom{13}{2} = 4 \times \frac{13 \times 12}{2} = 312$.
Alternatively,since the condition is that the cards are from the same suit,we consider the sample space where both cards belong to the same suit. There are $4$ suits,and for each suit,there are $\binom{13}{2} = 78$ ways to pick $2$ cards. Total outcomes $= 4 \times 78 = 312$.
In each suit,the prime-numbered cards are ${2, 3, 5, 7}$ (total $4$ cards) and the face cards are ${J, Q, K}$ (total $3$ cards).
We need one face card and one prime-numbered card from the same suit.
For one specific suit,the number of ways to choose one face card and one prime card is $3 \times 4 = 12$.
Since there are $4$ suits,the total number of favourable outcomes is $4 \times (3 \times 4) = 48$.
However,the question implies the probability given that they are from the same suit.
Given the cards are from the same suit,the total ways to pick $2$ cards is $\binom{13}{2} = 78$.
The number of ways to pick one face card ($3$ options) and one prime card ($4$ options) is $3 \times 4 = 12$.
Since the order does not matter,we have $12$ ways.
Probability $= \frac{12}{78} = \frac{2}{13}$.

(D) Case $I$: $A$ white ball is transferred from urn $A$ to urn $B$.
Probability of selecting a white ball from $A$ is $P(W_1) = \frac{3}{3+5} = \frac{3}{8}$.
After transferring,urn $B$ contains $7$ white and $8$ black balls.
Probability of selecting a white ball from $B$ is $P(W_2|W_1) = \frac{7}{7+8} = \frac{7}{15}$.
Probability of this case $= P(W_1) \times P(W_2|W_1) = \frac{3}{8} \times \frac{7}{15} = \frac{21}{120} = \frac{7}{40}$.
Case $II$: $A$ black ball is transferred from urn $A$ to urn $B$.
Probability of selecting a black ball from $A$ is $P(B_1) = \frac{5}{3+5} = \frac{5}{8}$.
After transferring,urn $B$ contains $6$ white and $9$ black balls.
Probability of selecting a white ball from $B$ is $P(W_2|B_1) = \frac{6}{6+9} = \frac{6}{15}$.
Probability of this case $= P(B_1) \times P(W_2|B_1) = \frac{5}{8} \times \frac{6}{15} = \frac{30}{120} = \frac{10}{40}$.
Total probability $= \frac{7}{40} + \frac{10}{40} = \frac{17}{40}$.

(B) Let $E_1, E_2, E_3$ be the events of choosing bags $B_1, B_2, B_3$ respectively.
The probabilities of choosing the bags are:
$P(E_1) = \frac{3}{6} = \frac{1}{2}$ (for outcomes $2, 3, 5$)
$P(E_2) = \frac{2}{6} = \frac{1}{3}$ (for outcomes $4, 6$)
$P(E_3) = \frac{1}{6}$ (for outcome $1$)
Let $G$ be the event of drawing a green ball. The conditional probabilities are:
$P(G|E_1) = \frac{3}{2+3+2} = \frac{3}{7}$
$P(G|E_2) = \frac{2}{3+2+2} = \frac{2}{7}$
$P(G|E_3) = \frac{2}{2+2+3} = \frac{2}{7}$
Using the Law of Total Probability:
$P(G) = P(E_1)P(G|E_1) + P(E_2)P(G|E_2) + P(E_3)P(G|E_3)$
$P(G) = \left(\frac{3}{6} \times \frac{3}{7}\right) + \left(\frac{2}{6} \times \frac{2}{7}\right) + \left(\frac{1}{6} \times \frac{2}{7}\right)$
$P(G) = \frac{9}{42} + \frac{4}{42} + \frac{2}{42} = \frac{15}{42} = \frac{5}{14}$

(A) Given: $P(A)=\frac{1}{4}$,$P(A|B)=\frac{1}{4}$,$P(B|A)=\frac{1}{2}$.
We know $P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{1}{4} \implies P(A \cap B) = \frac{1}{4}P(B)$.
Also $P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{1}{2} \implies P(A \cap B) = \frac{1}{2}P(A) = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8}$.
Equating the two,$\frac{1}{4}P(B) = \frac{1}{8} \implies P(B) = \frac{1}{2}$.
$(I)$ $P(\bar{A}|\bar{B}) = \frac{P(\bar{A} \cap \bar{B})}{P(\bar{B})} = \frac{1-P(A \cup B)}{1-P(B)} = \frac{1-[P(A)+P(B)-P(A \cap B)]}{1-P(B)} = \frac{1-[\frac{1}{4}+\frac{1}{2}-\frac{1}{8}]}{1-\frac{1}{2}} = \frac{1-\frac{5}{8}}{\frac{1}{2}} = \frac{3/8}{1/2} = \frac{3}{4}$. Statement $(I)$ is correct.
$(II)$ Since $P(A \cap B) = \frac{1}{8} \neq 0$,$A$ and $B$ are not mutually exclusive. Statement $(II)$ is incorrect.
$(III)$ $P(A|B)+P(A|\bar{B}) = \frac{P(A \cap B)}{P(B)} + \frac{P(A \cap \bar{B})}{P(\bar{B})} = \frac{1/8}{1/2} + \frac{P(A)-P(A \cap B)}{1-P(B)} = \frac{1}{4} + \frac{1/4-1/8}{1/2} = \frac{1}{4} + \frac{1/8}{1/2} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \neq 1$. Statement $(III)$ is incorrect.
Thus,only $(I)$ is correct.

(B) Let $E_1$ be the event that the sum is $7$ and $E_2$ be the event that the sum is $11$.
The total number of outcomes when two dice are rolled is $6 \times 6 = 36$.
The outcomes for sum $7$ are $(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$,so $P(E_1) = \frac{6}{36} = \frac{1}{6}$.
The outcomes for sum $11$ are $(5,6), (6,5)$,so $P(E_2) = \frac{2}{36} = \frac{1}{18}$.
We are interested in the event that $E_1$ occurs before $E_2$. This means in any trial,we ignore outcomes where the sum is neither $7$ nor $11$.
The probability of getting either $7$ or $11$ is $P(E_1 \cup E_2) = P(E_1) + P(E_2) = \frac{6}{36} + \frac{2}{36} = \frac{8}{36} = \frac{2}{9}$.
Given that the sum is either $7$ or $11$,the conditional probability that the sum is $7$ is $P(E_1 | E_1 \cup E_2) = \frac{P(E_1)}{P(E_1) + P(E_2)} = \frac{6/36}{8/36} = \frac{6}{8} = \frac{3}{4}$.
Thus,the probability that $7$ appears before $11$ is $\frac{3}{4}$.

(C) Let $A$ be the event that the sum of the numbers on the two dice is a multiple of $4$. The possible sums are $4, 8, 12$.
The outcomes for event $A$ are: $(1, 3), (3, 1), (2, 2), (2, 6), (6, 2), (3, 5), (5, 3), (4, 4), (6, 6)$.
Total number of outcomes in $A$ is $n(A) = 9$.
Let $B$ be the event that the number $4$ appears at least once on the dice.
We need to find the conditional probability $p = P(B|A) = \frac{n(A \cap B)}{n(A)}$.
The intersection $A \cap B$ consists of outcomes where the sum is a multiple of $4$ $AND$ the number $4$ appears at least once.
From the set $A$,the outcomes containing at least one $4$ are: $(4, 4)$.
Thus,$n(A \cap B) = 1$.
Therefore,$p = P(B|A) = \frac{1}{9}$.
Finally,we calculate $3p + 2 = 3 \times \frac{1}{9} + 2 = \frac{1}{3} + 2 = \frac{7}{3}$.

(A) Given: $P(A^{C}) = 0.3 \implies P(A) = 1 - 0.3 = 0.7$.
$P(B) = 0.4 \implies P(B^{C}) = 1 - 0.4 = 0.6$.
$P(A \cap B^{C}) = 0.5$.
We know $P(A) = P(A \cap B) + P(A \cap B^{C})$,so $P(A \cap B) = P(A) - P(A \cap B^{C}) = 0.7 - 0.5 = 0.2$.
We need to find $P(B \mid A \cup B^{C}) = \frac{P(B \cap (A \cup B^{C}))}{P(A \cup B^{C})}$.
Numerator: $P(B \cap (A \cup B^{C})) = P((B \cap A) \cup (B \cap B^{C})) = P(A \cap B) = 0.2$.
Denominator: $P(A \cup B^{C}) = P(A) + P(B^{C}) - P(A \cap B^{C}) = 0.7 + 0.6 - 0.5 = 0.8$.
Thus,$P(B \mid A \cup B^{C}) = \frac{0.2}{0.8} = \frac{1}{4}$.

(D) We are given that two integers $r$ and $s$ are drawn without replacement from the set $\{1, 2, \ldots, n\}$.
We need to find the conditional probability $P(r \leq k \mid s \leq k)$.
By the definition of conditional probability,$P(r \leq k \mid s \leq k) = \frac{P(r \leq k \cap s \leq k)}{P(s \leq k)}$.
First,the probability that $s \leq k$ is $P(s \leq k) = \frac{k}{n}$.
Next,the probability that both $r \leq k$ and $s \leq k$ is the probability of choosing two distinct integers from the set $\{1, 2, \ldots, k\}$ out of the total ways to choose two distinct integers from the set $\{1, 2, \ldots, n\}$.
This is given by $P(r \leq k \cap s \leq k) = \frac{k(k-1)}{n(n-1)}$.
Therefore,the conditional probability is $P(r \leq k \mid s \leq k) = \frac{\frac{k(k-1)}{n(n-1)}}{\frac{k}{n}} = \frac{k(k-1)}{n(n-1)} \times \frac{n}{k} = \frac{k-1}{n-1}$.

(B) Let $F$ be the event that the selected person is a female.
Let $A$ be the event that the selected person is aged above $40 \text{ yr}$.
We are given:
Total number of females $n(F) = 6$.
Number of females aged above $40 \text{ yr}$ is $n(A \cap F) = 3$.
We need to find the conditional probability $P(A|F)$,which is the probability that the person is aged above $40 \text{ yr}$ given that the person is a female.
The formula for conditional probability is $P(A|F) = \frac{n(A \cap F)}{n(F)}$.
Substituting the values:
$P(A|F) = \frac{3}{6} = \frac{1}{2}$.

(C) Given that $E$ and $F$ are independent events,we have $P(E \cap F) = P(E) \cdot P(F)$.
Using the formula $P(E \cup F) = P(E) + P(F) - P(E \cap F)$,we substitute the given values:
$0.5 = 0.3 + P(F) - 0.3 \cdot P(F)$
$0.2 = 0.7 \cdot P(F)$
$P(F) = \frac{0.2}{0.7} = \frac{2}{7}$.
For independent events,$P(E|F) = P(E)$ and $P(F|E) = P(F)$.
Therefore,$P(E|F) - P(F|E) = P(E) - P(F) = 0.3 - \frac{2}{7} = \frac{3}{10} - \frac{2}{7} = \frac{21 - 20}{70} = \frac{1}{70}$.

(B) Let $W_B$ be the event of picking a white ball from bag $B$,and $B_B$ be the event of picking a black ball from bag $B$.
$P(W_B) = \frac{6}{6+4} = \frac{6}{10} = \frac{3}{5}$
$P(B_B) = \frac{4}{6+4} = \frac{4}{10} = \frac{2}{5}$
If a white ball is transferred to bag $A$,bag $A$ now contains $10$ white and $8$ black balls (total $18$). The probability of drawing a white ball from bag $A$ is $P(W_A | W_B) = \frac{10}{18}$.
If a black ball is transferred to bag $A$,bag $A$ now contains $9$ white and $9$ black balls (total $18$). The probability of drawing a white ball from bag $A$ is $P(W_A | B_B) = \frac{9}{18}$.
Using the law of total probability:
$P(W_A) = P(W_B) \times P(W_A | W_B) + P(B_B) \times P(W_A | B_B)$
$P(W_A) = \frac{3}{5} \times \frac{10}{18} + \frac{2}{5} \times \frac{9}{18}$
$P(W_A) = \frac{30}{90} + \frac{18}{90} = \frac{48}{90} = \frac{8}{15}$
Thus,$p=8$ and $q=15$. Since $gcd(8,15)=1$,$p+q = 8+15 = 23$.

(B) Since $E$ and $F$ are independent events,the occurrence of one does not affect the probability of the other. Therefore,$E'$ and $F$ are independent,and $F'$ and $E$ are independent.
$P(E'/F) = P(E') = 1 - P(E) = 1 - \frac{3}{5} = \frac{2}{5}$.
Similarly,$P(F'/E) = P(F') = 1 - P(F) = 1 - \frac{3}{10} = \frac{7}{10}$.
Thus,$P(E'/F) + P(F'/E) = \frac{2}{5} + \frac{7}{10} = \frac{4}{10} + \frac{7}{10} = \frac{11}{10}$.

(A) We know that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Substituting the given values: $\frac{3}{4} = \frac{3}{8} + \frac{5}{8} - P(A \cap B)$.
$\frac{3}{4} = 1 - P(A \cap B) \Rightarrow P(A \cap B) = 1 - \frac{3}{4} = \frac{1}{4}$.
Now,calculate $P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{1/4}{5/8} = \frac{1}{4} \times \frac{8}{5} = \frac{2}{5}$.
Since $P(A'|B) = 1 - P(A|B)$,we have $P(A'|B) = 1 - \frac{2}{5} = \frac{3}{5}$.
Finally,$P(A'|B) - P(A|B) = \frac{3}{5} - \frac{2}{5} = \frac{1}{5}$.

(A) Given the equation $P(A) + P(B) - P(A \cap B) = P(A)$.
Subtracting $P(A)$ from both sides,we get $P(B) - P(A \cap B) = 0$,which implies $P(B) = P(A \cap B)$.
This equality indicates that $B \subseteq A$,meaning all outcomes of event $B$ are contained within event $A$.
By the definition of conditional probability,$P(A|B) = \frac{P(A \cap B)}{P(B)}$.
Since $P(A \cap B) = P(B)$,we substitute this into the formula:
$P(A|B) = \frac{P(B)}{P(B)} = 1$ (assuming $P(B) \neq 0$).

(B) We use the definition of conditional probability: $P(A'|B') = \frac{P(A' \cap B')}{P(B')}$.
By De Morgan's Law,$A' \cap B' = (A \cup B)'$.
Therefore,$P(A' \cap B') = P((A \cup B)') = 1 - P(A \cup B)$.
Given $P(A \cup B) = \frac{3}{11}$,we have $P(A' \cap B') = 1 - \frac{3}{11} = \frac{8}{11}$.
Also,$P(B') = 1 - P(B) = 1 - \frac{2}{11} = \frac{9}{11}$.
Substituting these values into the formula:
$P(A'|B') = \frac{8/11}{9/11} = \frac{8}{9}$.