Conditional probability Questions in English

(B) Given that $P(A \mid B) = 1$.
By the definition of conditional probability,$P(A \mid B) = \frac{P(A \cap B)}{P(B)}$.
Since $P(A \mid B) = 1$,we have $\frac{P(A \cap B)}{P(B)} = 1$,which implies $P(A \cap B) = P(B)$.
This equality holds if and only if $B \subset A$.
Therefore,the correct option is $B$.

(D) 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 condition $P(A \cap B) = P(B)$ indicates that the event $B$ is a subset of event $A$ (i.e.,$B \subseteq A$).
By the definition of conditional probability,$P(A \mid B) = \frac{P(A \cap B)}{P(B)}$.
Substituting $P(A \cap B) = P(B)$ into the formula,we get $P(A \mid B) = \frac{P(B)}{P(B)} = 1$ (assuming $P(B) \neq 0$).
Therefore,the correct option is $D$.

(D) Given that $P(B \mid A) = 1$.
By the definition of conditional probability,$P(B \mid A) = \frac{P(A \cap B)}{P(A)}$.
Since $P(B \mid A) = 1$,we have $\frac{P(A \cap B)}{P(A)} = 1$,which implies $P(A \cap B) = P(A)$.
This equality holds if and only if $A \subseteq B$.
Therefore,the correct option is $D$.

(A) We are given that $P(A) = \frac{6}{11}$,$P(B) = \frac{5}{11}$,and $P(A \cup B) = \frac{7}{11}$.
Using the addition theorem of probability,$P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Substituting the given values: $\frac{7}{11} = \frac{6}{11} + \frac{5}{11} - P(A \cap B)$.
$\frac{7}{11} = \frac{11}{11} - P(A \cap B)$.
$P(A \cap B) = 1 - \frac{7}{11} = \frac{4}{11}$.
Now,the conditional probability $P(A \mid B)$ is defined as $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$.
$P(A \mid B) = \frac{4/11}{5/11} = \frac{4}{5}$.
Therefore,the correct option is $A$.

(A) Given that $6 P(A) = 1 \implies P(A) = \frac{1}{6}$.
Given that $8 P(B) = 1 \implies P(B) = \frac{1}{8}$.
Given that $14 P(A \cap B) = 1 \implies P(A \cap B) = \frac{1}{14}$.
We need to find $P(A' \mid B)$.
Using the conditional probability formula,$P(A' \mid B) = \frac{P(A' \cap B)}{P(B)}$.
We know that $P(A' \cap B) = P(B) - P(A \cap B)$.
Substituting the values: $P(A' \cap B) = \frac{1}{8} - \frac{1}{14} = \frac{7 - 4}{56} = \frac{3}{56}$.
Now,$P(A' \mid B) = \frac{3/56}{1/8} = \frac{3}{56} \times 8 = \frac{3}{7}$.
Thus,the correct option is $A$.

(B) Given the equation: $P(A) + P(B) - P(A \cap B) = P(A)$.
Subtracting $P(A)$ from both sides,we get: $P(B) - P(A \cap B) = 0$.
This implies $P(B) = P(A \cap B)$.
By the definition of conditional probability,$P(A|B) = \frac{P(A \cap B)}{P(B)}$.
Substituting $P(A \cap B) = P(B)$ into the formula,we get $P(A|B) = \frac{P(B)}{P(B)} = 1$ (assuming $P(B) \neq 0$).

(C) Given that $2 P(A) = \frac{5}{13}$,we have $P(A) = \frac{5}{26}$.
Given that $P(B) = \frac{5}{13}$.
We know that $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$.
Substituting the values,$\frac{2}{5} = \frac{P(A \cap B)}{5/13}$.
Therefore,$P(A \cap B) = \frac{2}{5} \times \frac{5}{13} = \frac{2}{13}$.
Using the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$:
$P(A \cup B) = \frac{5}{26} + \frac{5}{13} - \frac{2}{13}$.
$P(A \cup B) = \frac{5}{26} + \frac{10}{26} - \frac{4}{26}$.
$P(A \cup B) = \frac{5 + 10 - 4}{26} = \frac{11}{26}$.
Thus,the correct option is $C$.

(A) Given that $P(B \mid A) = 1$.
By the definition of conditional probability,$P(B \mid A) = \frac{P(A \cap B)}{P(A)}$.
Since $P(B \mid A) = 1$,we have $\frac{P(A \cap B)}{P(A)} = 1$,which implies $P(A \cap B) = P(A)$.
This equality $P(A \cap B) = P(A)$ holds if and only if $A \subseteq B$.
Therefore,the correct option is $A$.

(C) Given that $A \subset B$,it implies that $A \cap B = A$.
By the definition of conditional probability,$P(A \mid B) = \frac{P(A \cap B)}{P(B)}$.
Substituting $A \cap B = A$,we get $P(A \mid B) = \frac{P(A)}{P(B)}$.
Since $A \subset B$,we have $P(A) \leq P(B)$,which implies $\frac{1}{P(B)} \leq \frac{1}{P(A)}$.
Multiplying both sides by $P(A)$ (assuming $P(A) > 0$),we get $\frac{P(A)}{P(B)} \leq 1$.
Also,since $P(B) \leq 1$,we have $\frac{P(A)}{P(B)} \geq P(A)$.
Thus,$P(A \mid B) \geq P(A)$.

(A) Given that $P(E)=0.8$,$P(F)=0.5$,and $P(F \mid E)=0.4$.
Using the definition of conditional probability,$P(F \mid E) = \frac{P(E \cap F)}{P(E)}$.
Substituting the values,$0.4 = \frac{P(E \cap F)}{0.8}$.
Therefore,$P(E \cap F) = 0.4 \times 0.8 = 0.32$.
Now,we need to find $P(E \mid F)$.
Using the formula $P(E \mid F) = \frac{P(E \cap F)}{P(F)}$.
Substituting the values,$P(E \mid F) = \frac{0.32}{0.5} = 0.64$.
Thus,the correct option is $A$.

(D) We know that for any event $A$,$P(A) + P(A^{\prime}) = 1$,which implies $P(A) = 1 - P(A^{\prime})$.
Applying this to the conditional probability $P(A \mid B^{\prime})$,we have:
$P(A \mid B^{\prime}) = 1 - P(A^{\prime} \mid B^{\prime})$.
This follows from the property of conditional probability where the sum of probabilities of complementary events given the same condition is $1$.

(B) Let $H_1$ be the event of getting a head on the first coin and $H_2$ be the event of getting a head on the second coin.
Since the two coin tosses are independent events,the outcome of the first coin does not affect the outcome of the second coin.
Therefore,$P(H_2 | H_1) = P(H_2)$.
For a fair coin,the probability of getting a head is $P(H_2) = \frac{1}{2}$.
Thus,the probability of getting a head on the second coin given that a head is obtained on the first coin is $\frac{1}{2}$.

(C) Given that $P(A)=0.25, P(B)=0.55$ and $P(A \cup B)=0.65$.
First,we find $P(A \cap B)$ using the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
$0.65 = 0.25 + 0.55 - P(A \cap B)$
$0.65 = 0.80 - P(A \cap B)$
$P(A \cap B) = 0.80 - 0.65 = 0.15$.
Now,we need to find $P(B' \mid A)$.
Using the conditional probability formula,$P(B' \mid A) = \frac{P(B' \cap A)}{P(A)}$.
Since $P(B' \cap A) = P(A) - P(A \cap B)$,we have:
$P(B' \cap A) = 0.25 - 0.15 = 0.10$.
Therefore,$P(B' \mid A) = \frac{0.10}{0.25} = \frac{10}{25} = 0.4$.
Thus,the correct option is $C$.

(A) Given $P(A)=\frac{1}{3}$,$P(B)=\frac{1}{2}$ and $P(A \cap B)=\frac{1}{6}$.
We need to find $P(A^{\prime} | B)$.
Using the formula for conditional probability,$P(A^{\prime} | B) = \frac{P(A^{\prime} \cap B)}{P(B)}$.
Since $P(A^{\prime} \cap B) = P(B) - P(A \cap B)$,
$P(A^{\prime} | B) = \frac{P(B) - P(A \cap B)}{P(B)}$.
Substituting the values:
$P(A^{\prime} | B) = \frac{\frac{1}{2} - \frac{1}{6}}{\frac{1}{2}} = \frac{\frac{3-1}{6}}{\frac{1}{2}} = \frac{\frac{2}{6}}{\frac{1}{2}} = \frac{1}{3} \times 2 = \frac{2}{3}$.

(B) We are given that $P(A) \neq 0$. The conditional probability is defined as $P(B \mid A) = \frac{P(A \cap B)}{P(A)}$.
$(i)$ If $A \subset B$,then $A \cap B = A$. Therefore,$P(B \mid A) = \frac{P(A \cap B)}{P(A)} = \frac{P(A)}{P(A)} = 1$.
(ii) If $A \cap B = \phi$,then $P(A \cap B) = P(\phi) = 0$. Therefore,$P(B \mid A) = \frac{P(A \cap B)}{P(A)} = \frac{0}{P(A)} = 0$.
Thus,the values are $1$ and $0$ respectively.

(B) Given,$P(B) = \frac{3}{5}$,$P\left(\frac{A}{B}\right) = \frac{1}{2}$,and $P(A \cup B) = \frac{4}{5}$.
Since $P\left(\frac{A}{B}\right) = \frac{P(A \cap B)}{P(B)}$,we have:
$\frac{P(A \cap B)}{P(B)} = \frac{1}{2}$
$P(A \cap B) = \frac{1}{2} \times \frac{3}{5} = \frac{3}{10}$.
Using the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$:
$\frac{4}{5} = P(A) + \frac{3}{5} - \frac{3}{10}$
$P(A) = \frac{4}{5} - \frac{3}{5} + \frac{3}{10}$
$P(A) = \frac{1}{5} + \frac{3}{10} = \frac{2+3}{10} = \frac{5}{10} = \frac{1}{2}$.

(C) Let $E_{A}$ be the event that the sum of numbers on the dice is less than $6$.
The possible outcomes for $E_{A}$ are:
$E_{A} = \{(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2), (4,1)\}$
Thus,$n(E_{A}) = 10$.
Let $E_{B}$ be the event that the sum of numbers on the dice is $3$.
The possible outcomes for $E_{B}$ are:
$E_{B} = \{(1,2), (2,1)\}$
Thus,$n(E_{B}) = 2$.
The required conditional probability is $P(E_{B}|E_{A}) = \frac{n(E_{B} \cap E_{A})}{n(E_{A})} = \frac{2}{10} = \frac{1}{5}$.

(D) Given that $A$ and $B$ are non-mutually exclusive events,we have $P(A \cap B) \neq 0$.
Given the condition $P(A \mid B) = P(B \mid A)$.
Using the definition of conditional probability,we have:
$\frac{P(A \cap B)}{P(B)} = \frac{P(B \cap A)}{P(A)}$
Since $P(A \cap B) = P(B \cap A)$,we can divide both sides by $P(A \cap B)$ because $P(A \cap B) \neq 0$:
$\frac{1}{P(B)} = \frac{1}{P(A)}$
Therefore,$P(A) = P(B)$.

(A) Given: $P(A) = \frac{1}{4}$,$P(A|B) = \frac{1}{2}$,and $P(B|A) = \frac{2}{3}$.
We know that the conditional probability formula is $P(E_1|E_2) = \frac{P(E_1 \cap E_2)}{P(E_2)}$.
Using $P(B|A) = \frac{P(A \cap B)}{P(A)}$,we get:
$\frac{2}{3} = \frac{P(A \cap B)}{1/4}$
$P(A \cap B) = \frac{2}{3} \times \frac{1}{4} = \frac{1}{6}$.
Now,using $P(A|B) = \frac{P(A \cap B)}{P(B)}$,we have:
$\frac{1}{2} = \frac{1/6}{P(B)}$
$P(B) = 2 \times \frac{1}{6} = \frac{1}{3}$.

(D) Given,$P(A)=\frac{1}{2}$,$P(B)=\frac{1}{2}$ and $P(A \mid B)=\frac{1}{4}$.
Using the conditional probability formula,$P(A \mid B) = \frac{P(A \cap B)}{P(B)}$.
Substituting the values,$\frac{1}{4} = \frac{P(A \cap B)}{1/2}$.
Therefore,$P(A \cap B) = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8}$.
By De Morgan's Law,$P(A^{\prime} \cap B^{\prime}) = P((A \cup B)^{\prime}) = 1 - P(A \cup B)$.
Using the addition theorem,$P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{1}{2} + \frac{1}{2} - \frac{1}{8} = 1 - \frac{1}{8} = \frac{7}{8}$.
Thus,$P(A^{\prime} \cap B^{\prime}) = 1 - \frac{7}{8} = \frac{1}{8}$.

(D) Given that $P(A)=0.2$,$P(B)=0.6$ and $P(A \mid B)=0.5$.
We know that the conditional probability is defined as $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$.
Substituting the given values: $0.5 = \frac{P(A \cap B)}{0.6}$.
Therefore,$P(A \cap B) = 0.5 \times 0.6 = 0.3$.
We need to find $P(A^{\prime} \mid B)$.
Using the property of conditional probability,$P(A^{\prime} \mid B) = 1 - P(A \mid B)$.
Thus,$P(A^{\prime} \mid B) = 1 - 0.5 = 0.5 = \frac{1}{2}$.

(A) The total number of tickets is $ 17 $.
The even-numbered tickets are $ 2, 4, 6, 8, 10, 12, 14, 16 $.
Thus,the total number of even-numbered tickets is $ 8 $.
Let $ E_1 $ be the event that the first ticket drawn is even,and $ E_2 $ be the event that the second ticket drawn is even.
The probability of drawing an even ticket first is $ P(E_1) = \frac{8}{17} $.
Since the ticket is not replaced,the remaining number of tickets is $ 16 $,and the number of remaining even tickets is $ 7 $.
The conditional probability of drawing an even ticket second is $ P(E_2|E_1) = \frac{7}{16} $.
The probability that both tickets show even numbers is $ P(E_1 \cap E_2) = P(E_1) \times P(E_2|E_1) = \frac{8}{17} \times \frac{7}{16} = \frac{7}{34} $.

(C) Given that,$P(A \cap B) = \frac{7}{10}$ and $P(B) = \frac{17}{20}$.
We know the formula for conditional probability is $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$.
Substituting the given values into the formula:
$P(A \mid B) = \frac{\frac{7}{10}}{\frac{17}{20}}$
$P(A \mid B) = \frac{7}{10} \times \frac{20}{17}$
$P(A \mid B) = \frac{7 \times 2}{17} = \frac{14}{17}$.

(C) Statement $(I)$: If $E$ and $F$ are independent,then $P(E \cap F) = P(E)P(F)$. We know that $P(E^{\prime} \cap F^{\prime}) = P((E \cup F)^{\prime}) = 1 - P(E \cup F) = 1 - [P(E) + P(F) - P(E \cap F)] = 1 - P(E) - P(F) + P(E)P(F) = (1 - P(E))(1 - P(F)) = P(E^{\prime})P(F^{\prime})$. Thus,$E^{\prime}$ and $F^{\prime}$ are independent. Statement $(I)$ is true.
Statement $(II)$: If $E$ and $F$ are mutually exclusive,then $P(E \cap F) = 0$. For them to be independent,we require $P(E \cap F) = P(E)P(F)$. Since $P(E) > 0$ and $P(F) > 0$,$P(E)P(F) > 0$. Thus,$P(E \cap F) \neq P(E)P(F)$,meaning they cannot be independent. Statement $(II)$ is true.

(C) By the definition of conditional probability,we have:
$P(\bar{A} \mid \bar{B}) = \frac{P(\bar{A} \cap \bar{B})}{P(\bar{B})}$
Using De Morgan's Law,$\bar{A} \cap \bar{B} = \overline{A \cup B}$.
Therefore,$P(\bar{A} \cap \bar{B}) = P(\overline{A \cup B}) = 1 - P(A \cup B)$.
Substituting this into the formula,we get:
$P(\bar{A} \mid \bar{B}) = \frac{1 - P(A \cup B)}{P(\bar{B})}$

(B) Let $S$ be the set of outcomes where the sum of three dice is $8$. The possible combinations (unordered) are:
$(1, 1, 6) \rightarrow \frac{3!}{2!} = 3$ permutations
$(1, 2, 5) \rightarrow 3! = 6$ permutations
$(1, 3, 4) \rightarrow 3! = 6$ permutations
$(2, 2, 4) \rightarrow \frac{3!}{2!} = 3$ permutations
$(2, 3, 3) \rightarrow \frac{3!}{2!} = 3$ permutations
Total number of outcomes $n(S) = 3 + 6 + 6 + 3 + 3 = 21$.
Let $E$ be the event that at least one die shows a $4$. The favorable outcomes are from $(1, 3, 4)$ and $(2, 2, 4)$.
Number of favorable outcomes $n(E) = 6 + 3 = 9$.
The probability $P(E) = \frac{n(E)}{n(S)} = \frac{9}{21} = \frac{3}{7}$.

(C) Given $P(E_1) = \frac{1}{8}$,$P(E_1 \mid E_2) = \frac{1}{3}$,and $P(E_2 \mid E_1) = \frac{1}{4}$.
Using the definition of conditional probability,$P(E_1 \cap E_2) = P(E_2 \mid E_1) \times P(E_1) = \frac{1}{4} \times \frac{1}{8} = \frac{1}{32}$.
Then,$P(E_2) = \frac{P(E_1 \cap E_2)}{P(E_1 \mid E_2)} = \frac{1/32}{1/3} = \frac{3}{32}$. (Matches $IV$)
Now,$P(E_1 \cup E_2) = P(E_1) P(E_2) - P(E_1 \cap E_2) = \frac{1}{8} \frac{3}{32} - \frac{1}{32} = \frac{4 3-1}{32} = \frac{6}{32} = \frac{3}{16}$. (Matches $III$)
We know $P(\bar{E}_2) = 1 - P(E_2) = 1 - \frac{3}{32} = \frac{29}{32}$.
Also,$P(E_1 \cap \bar{E}_2) = P(E_1) - P(E_1 \cap E_2) = \frac{1}{8} - \frac{1}{32} = \frac{3}{32}$.
Thus,$P(E_1 \mid \bar{E}_2) = \frac{P(E_1 \cap \bar{E}_2)}{P(\bar{E}_2)} = \frac{3/32}{29/32} = \frac{3}{29}$. (Matches $I$)
Finally,$P(\bar{E}_1 \mid \bar{E}_2) = 1 - P(E_1 \mid \bar{E}_2) = 1 - \frac{3}{29} = \frac{26}{29}$. (Matches $II$)
The correct matching is $A-III, B-IV, C-I, D-II$.

(D) Let $P(A) = \frac{2}{3}$ and $P(B) = \frac{3}{4}$.
Assuming the events are independent,the probability of passing both is $P(A \cap B) = P(A) \times P(B) = \frac{2}{3} \times \frac{3}{4} = \frac{1}{2}$.
The probability of passing at least one is $P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{2}{3} + \frac{3}{4} - \frac{1}{2} = \frac{8+9-6}{12} = \frac{11}{12}$.
We need to find the conditional probability $P(A \cap B | A \cup B) = \frac{P((A \cap B) \cap (A \cup B))}{P(A \cup B)} = \frac{P(A \cap B)}{P(A \cup B)}$.
Substituting the values,we get $\frac{1/2}{11/12} = \frac{1}{2} \times \frac{12}{11} = \frac{6}{11}$.

(A) Given: $P(\overline{A}) = 0.3 \implies P(A) = 1 - 0.3 = 0.7$.
$P(B) = 0.4 \implies P(\overline{B}) = 1 - 0.4 = 0.6$.
$P(A \cap \overline{B}) = 0.5$.
Since $P(A) = P(A \cap B) + P(A \cap \overline{B})$,we have $0.7 = P(A \cap B) + 0.5$,so $P(A \cap B) = 0.2$.
We need to find $P(B | (A \cup \overline{B})) = \frac{P(B \cap (A \cup \overline{B}))}{P(A \cup \overline{B})}$.
Numerator: $P(B \cap (A \cup \overline{B})) = P((B \cap A) \cup (B \cap \overline{B})) = P((B \cap A) \cup \emptyset) = P(A \cap B) = 0.2$.
Denominator: $P(A \cup \overline{B}) = P(A) + P(\overline{B}) - P(A \cap \overline{B}) = 0.7 + 0.6 - 0.5 = 0.8$.
Therefore,$P(B | (A \cup \overline{B})) = \frac{0.2}{0.8} = \frac{1}{4} = 0.25$.

(D) Using the definition of conditional probability,$P(E/F) = \frac{P(E \cap F)}{P(F)}$.
Given expression: $P(A / A \cap B) + P(B / A \cap B)$
$= \frac{P(A \cap (A \cap B))}{P(A \cap B)} + \frac{P(B \cap (A \cap B))}{P(A \cap B)}$
Since $A \cap (A \cap B) = A \cap B$ and $B \cap (A \cap B) = A \cap B$,we have:
$= \frac{P(A \cap B)}{P(A \cap B)} + \frac{P(A \cap B)}{P(A \cap B)}$
$= 1 + 1 = 2$

(B) The sample space $S$ for rolling two dice has $6 \times 6 = 36$ outcomes.
Event $A$ is the event that the same number shows on each die: $A = \{(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)\}$. Thus,$n(A) = 6$ and $P(A) = \frac{6}{36} = \frac{1}{6}$.
Event $B$ is the event that the sum of the numbers is greater than $7$: $B = \{(2,6), (3,5), (3,6), (4,4), (4,5), (4,6), (5,3), (5,4), (5,5), (5,6), (6,2), (6,3), (6,4), (6,5), (6,6)\}$. Thus,$n(B) = 15$ and $P(B) = \frac{15}{36} = \frac{5}{12}$.
The intersection $A \cap B$ is the set of outcomes where the numbers are the same and the sum is greater than $7$: $A \cap B = \{(4,4), (5,5), (6,6)\}$. Thus,$n(A \cap B) = 3$ and $P(A \cap B) = \frac{3}{36} = \frac{1}{12}$.
Now,calculate the conditional probabilities:
$P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{1/12}{5/12} = \frac{1}{5}$.
$P(B \mid A) = \frac{P(A \cap B)}{P(A)} = \frac{1/12}{1/6} = \frac{1}{2}$.

(C) Let $E_1$ be the event that either of the dice shows $2$.
Let $E_2$ be the event that the sum of the numbers on the dice is $6$.
The sample space for the sum being $6$ is $E_2 = \{(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)\}$.
Thus,the number of outcomes in $E_2$ is $n(E_2) = 5$.
The outcomes where either of the dice shows $2$ given the sum is $6$ are $E_1 \cap E_2 = \{(2, 4), (4, 2)\}$.
Thus,the number of outcomes in $E_1 \cap E_2$ is $n(E_1 \cap E_2) = 2$.
The conditional probability is given by $P(E_1 | E_2) = \frac{n(E_1 \cap E_2)}{n(E_2)} = \frac{2}{5}$.

(A) Since $A$ and $B$ are mutually exclusive events,we have $P(A \cap B) = 0$.
By the definition of conditional probability,$P(A \mid B^c) = \frac{P(A \cap B^c)}{P(B^c)}$.
We know that $P(A \cap B^c) = P(A) - P(A \cap B)$.
Since $P(A \cap B) = 0$,it follows that $P(A \cap B^c) = P(A)$.
Also,$P(B^c) = 1 - P(B)$.
Therefore,$P(A \mid B^c) = \frac{P(A)}{1 - P(B)}$.

(B) By the definition of conditional probability,we have:
$P(A | B^c) = \frac{P(A \cap B^c)}{P(B^c)}$
Since $B^c$ is the complement of $B$,$P(B^c) = 1 - P(B)$.
Also,$A \cap B^c$ represents the event where $A$ occurs but $B$ does not. This can be written as $A \setminus (A \cap B)$.
Therefore,$P(A \cap B^c) = P(A) - P(A \cap B)$.
Substituting these into the formula:
$P(A | B^c) = \frac{P(A) - P(A \cap B)}{1 - P(B)}$

(A) We know that by the definition of conditional probability:
$P(\frac{B}{A}) = \frac{P(A \cap B)}{P(A)}$
$\Rightarrow P(A) P(\frac{B}{A}) = P(A \cap B)$
Now,using De Morgan's Law,we have:
$P(\bar{A} \cup \bar{B}) = P(\overline{A \cap B})$
Using the property of complement events,$P(\bar{E}) = 1 - P(E)$:
$P(\overline{A \cap B}) = 1 - P(A \cap B)$
Substituting the value of $P(A \cap B)$ from the first step:
$P(\bar{A} \cup \bar{B}) = 1 - P(A) P(\frac{B}{A})$
Thus,option $A$ is the correct statement.

(A) Given: $P(P) = 1/4$,$P(P|Q) = 1/2$,and $P(Q|P) = 1/3$.
We need to find $P(\bar{P}|\bar{Q})$.
First,we find $P(P \cap Q)$ using the definition of conditional probability:
$P(Q|P) = \frac{P(P \cap Q)}{P(P)} \implies P(P \cap Q) = P(Q|P) \times P(P) = \frac{1}{3} \times \frac{1}{4} = \frac{1}{12}$.
Next,we find $P(Q)$:
$P(P|Q) = \frac{P(P \cap Q)}{P(Q)} \implies P(Q) = \frac{P(P \cap Q)}{P(P|Q)} = \frac{1/12}{1/2} = \frac{1}{6}$.
Now,calculate the probabilities of the complements:
$P(\bar{P}) = 1 - P(P) = 1 - 1/4 = 3/4$.
$P(\bar{Q}) = 1 - P(Q) = 1 - 1/6 = 5/6$.
Using De Morgan's Law,$P(\bar{P} \cup \bar{Q}) = 1 - P(P \cap Q) = 1 - 1/12 = 11/12$.
Then,$P(\bar{P} \cap \bar{Q}) = P(\bar{P}) + P(\bar{Q}) - P(\bar{P} \cup \bar{Q}) = \frac{3}{4} + \frac{5}{6} - \frac{11}{12} = \frac{9 + 10 - 11}{12} = \frac{8}{12} = \frac{2}{3}$.
Finally,the conditional probability is:
$P(\bar{P}|\bar{Q}) = \frac{P(\bar{P} \cap \bar{Q})}{P(\bar{Q})} = \frac{2/3}{5/6} = \frac{2}{3} \times \frac{6}{5} = \frac{4}{5}$.

(A) Given that $P(\overline{F}) = 0.7$ and $P(E \cap F) = 0.2$.
First,we find $P(F)$ using the complement rule: $P(F) = 1 - P(\overline{F}) = 1 - 0.7 = 0.3$.
The conditional probability formula is $P(E \mid F) = \frac{P(E \cap F)}{P(F)}$.
Substituting the known values: $P(E \mid F) = \frac{0.2}{0.3} = \frac{2}{3}$.

(C) We are given $P(A) = \frac{1}{7}$,$P(A|B) = \frac{2}{3}$,and $P(B) = \frac{2}{7}$.
Using the definition of conditional probability,$P(A|B) = \frac{P(A \cap B)}{P(B)}$.
Therefore,$P(A \cap B) = P(A|B) \times P(B) = \frac{2}{3} \times \frac{2}{7} = \frac{4}{21}$.
Now,we need to find $P(B|A)$,which is defined as $P(B|A) = \frac{P(A \cap B)}{P(A)}$.
Substituting the values,we get $P(B|A) = \frac{4/21}{1/7} = \frac{4}{21} \times \frac{7}{1} = \frac{4}{3}$.
Wait,let us re-evaluate the calculation. Given $P(A|B) = \frac{2}{3}$,$P(B) = \frac{2}{7}$,$P(A) = \frac{1}{7}$.
$P(A \cap B) = \frac{2}{3} \times \frac{2}{7} = \frac{4}{21}$.
$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{4/21}{1/7} = \frac{4}{21} \times 7 = \frac{4}{3}$.
Since $\frac{4}{3} > 1$,this is impossible for a probability. Let us re-check the input values. If $P(A|B) = \frac{2}{3}$ was intended to be $\frac{2}{5}$,then $P(A \cap B) = \frac{2}{5} \times \frac{2}{7} = \frac{4}{35}$.
Then $P(B|A) = \frac{4/35}{1/7} = \frac{4}{35} \times 7 = \frac{4}{5}$.
Thus,the correct option is $C$.

(C) Given $P(A)=0.5, P(B)=0.4$,and $P(A \cap B)=0.3$.
We need to find $P(A^{\prime} / B^{\prime})$.
By the definition of conditional probability,$P(A^{\prime} / B^{\prime}) = \frac{P(A^{\prime} \cap B^{\prime})}{P(B^{\prime})}$.
Using De Morgan's Law,$A^{\prime} \cap B^{\prime} = (A \cup B)^{\prime}$,so $P(A^{\prime} \cap B^{\prime}) = P((A \cup B)^{\prime}) = 1 - P(A \cup B)$.
First,calculate $P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.5 + 0.4 - 0.3 = 0.6$.
Then,$P(A^{\prime} \cap B^{\prime}) = 1 - 0.6 = 0.4$.
Next,calculate $P(B^{\prime}) = 1 - P(B) = 1 - 0.4 = 0.6$.
Finally,$P(A^{\prime} / B^{\prime}) = \frac{0.4}{0.6} = \frac{2}{3}$.

(B) When two dice are thrown,the total number of outcomes is $n(S) = 6 \times 6 = 36$.
Event $A$ is the event of getting a sum $\ge 8$. The outcomes are:
$A = \{(2,6), (3,5), (3,6), (4,4), (4,5), (4,6), (5,3), (5,4), (5,5), (5,6), (6,2), (6,3), (6,4), (6,5), (6,6)\}$.
Thus,$n(A) = 15$.
Event $B$ is the event of getting a number $\le 3$ on at least one die.
Event $A \cap B$ is the event of getting a sum $\ge 8$ $AND$ at least one die showing a number $\le 3$.
Looking at the elements of $A$,the outcomes where at least one die is $\le 3$ are:
$A \cap B = \{(2,6), (3,5), (3,6), (5,3), (6,2), (6,3)\}$.
Thus,$n(A \cap B) = 6$.
The conditional probability is given by $P(B / A) = \frac{n(A \cap B)}{n(A)} = \frac{6}{15} = \frac{2}{5}$.

(C) Given,$P(E_1) = \frac{1}{4}$,$P(E_2 / E_1) = \frac{1}{2}$,and $P(E_1 / E_2) = \frac{1}{4}$.
$(A)$ Since $P(E_2 / E_1) = \frac{P(E_1 \cap E_2)}{P(E_1)}$,we have $\frac{1}{2} = \frac{P(E_1 \cap E_2)}{1/4}$,so $P(E_1 \cap E_2) = \frac{1}{8}$.
Also,$P(E_1 / E_2) = \frac{P(E_1 \cap E_2)}{P(E_2)} = \frac{1}{4}$.
Substituting $P(E_1 \cap E_2) = \frac{1}{8}$,we get $\frac{1/8}{P(E_2)} = \frac{1}{4}$,which implies $P(E_2) = \frac{1}{2}$. Thus,$(A)$ matches with $(iv)$.
$(B)$ $P(E_1 \cup E_2) = P(E_1) + P(E_2) - P(E_1 \cap E_2) = \frac{1}{4} + \frac{1}{2} - \frac{1}{8} = \frac{2+4-1}{8} = \frac{5}{8}$. Thus,$(B)$ matches with $(ii)$.
$(C)$ $P(\bar{E}_1 / \bar{E}_2) = \frac{P(\bar{E}_1 \cap \bar{E}_2)}{P(\bar{E}_2)} = \frac{1 - P(E_1 \cup E_2)}{1 - P(E_2)} = \frac{1 - 5/8}{1 - 1/2} = \frac{3/8}{1/2} = \frac{3}{4}$. Thus,$(C)$ matches with $(vi)$.
$(D)$ $P(E_1 / \bar{E}_2) = \frac{P(E_1 \cap \bar{E}_2)}{P(\bar{E}_2)} = \frac{P(E_1) - P(E_1 \cap E_2)}{1 - P(E_2)} = \frac{1/4 - 1/8}{1 - 1/2} = \frac{1/8}{1/2} = \frac{1}{4}$. Thus,$(D)$ matches with $(i)$.

(B) Given: $P(E_1) = \frac{1}{2}$ and $P(E_1 \cup E_2) = \frac{2}{3}$.
Since $E_1$ and $E_2$ are independent events,$P(E_1 \cap E_2) = P(E_1)P(E_2) = \frac{1}{2}P(E_2)$.
Using the formula $P(E_1 \cup E_2) = P(E_1) + P(E_2) - P(E_1 \cap E_2)$:
$\frac{2}{3} = \frac{1}{2} + P(E_2) - \frac{1}{2}P(E_2)$
$\frac{2}{3} - \frac{1}{2} = \frac{1}{2}P(E_2)$
$\frac{1}{6} = \frac{1}{2}P(E_2) \implies P(E_2) = \frac{1}{3}$. (Matches $A-iii$)
Now,$P(E_1 \cap E_2) = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6}$.
$P(\frac{E_1}{E_2}) = \frac{P(E_1 \cap E_2)}{P(E_2)} = \frac{1/6}{1/3} = \frac{1}{2}$. (Matches $B-i$)
$P(\frac{\bar{E}_2}{E_1}) = \frac{P(\bar{E}_2 \cap E_1)}{P(E_1)} = \frac{P(E_1) - P(E_1 \cap E_2)}{P(E_1)} = \frac{1/2 - 1/6}{1/2} = \frac{1/3}{1/2} = \frac{2}{3}$. (Matches $C-v$)
$P(\bar{E}_1 \cup \bar{E}_2) = P(\overline{E_1 \cap E_2}) = 1 - P(E_1 \cap E_2) = 1 - \frac{1}{6} = \frac{5}{6}$. (Matches $D-ii$)
Thus,the correct match is $A-iii, B-i, C-v, D-ii$.

(D) Given that $A$ and $B$ are independent events.
$\therefore P(A \cap B) = P(A) \cdot P(B)$.
By definition of conditional probability,$P(A \mid B) = \frac{P(A \cap B)}{P(B)}$.
Substituting the independence condition: $P(A \mid B) = \frac{P(A) \cdot P(B)}{P(B)} = P(A)$ ... $(i)$.
Now,consider $P(A \mid B^C) = \frac{P(A \cap B^C)}{P(B^C)}$.
Since $A$ and $B$ are independent,$A$ and $B^C$ are also independent.
Thus,$P(A \cap B^C) = P(A) \cdot P(B^C)$.
Therefore,$P(A \mid B^C) = \frac{P(A) \cdot P(B^C)}{P(B^C)} = P(A)$ ... $(ii)$.
From equations $(i)$ and $(ii)$,we get $P(A \mid B) = P(A \mid B^C)$.

(A) Given that $A$ and $B$ are independent events,we have $P(A) = \frac{1}{3}$ and $P(B) = \frac{2}{7}$.
We need to find $P\left(\frac{A}{B^C}\right)$,where $B^C$ is the complement of $B$.
By the definition of conditional probability,$P\left(\frac{A}{B^C}\right) = \frac{P(A \cap B^C)}{P(B^C)}$.
Since $A$ and $B$ are independent,$A$ and $B^C$ are also independent.
Therefore,$P(A \cap B^C) = P(A) \times P(B^C)$.
Substituting this into the formula,we get $P\left(\frac{A}{B^C}\right) = \frac{P(A) \times P(B^C)}{P(B^C)} = P(A)$.
Given $P(A) = \frac{1}{3}$,the value is $\frac{1}{3}$.

(B) Given: $P(A)=0.6$,$P(B)=0.3$,and $P(A \mid B)=0.5$.
First,find $P(A \cap B)$ using the formula $P(A \cap B) = P(A \mid B) \times P(B)$.
$P(A \cap B) = 0.5 \times 0.3 = 0.15$.
We need to find $P(\bar{B} \mid \bar{A}) = \frac{P(\bar{B} \cap \bar{A})}{P(\bar{A})}$.
Using De Morgan's Law,$P(\bar{B} \cap \bar{A}) = P(\overline{A \cup B}) = 1 - P(A \cup B)$.
$P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.6 + 0.3 - 0.15 = 0.75$.
So,$P(\bar{B} \cap \bar{A}) = 1 - 0.75 = 0.25$.
Also,$P(\bar{A}) = 1 - P(A) = 1 - 0.6 = 0.4$.
Therefore,$P(\bar{B} \mid \bar{A}) = \frac{0.25}{0.4} = \frac{25}{40} = 0.625$.

(D) Given that $P(E_1) = \frac{1}{4}$ and $P(E_2 \mid E_1) = \frac{1}{2}$.
Using the definition of conditional probability,$P(E_1 \cap E_2) = P(E_1) \times P(E_2 \mid E_1) = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8}$.
We are also given $P(E_1 \mid E_2) = \frac{1}{4}$.
By definition,$P(E_1 \mid E_2) = \frac{P(E_1 \cap E_2)}{P(E_2)}$,so $\frac{1}{4} = \frac{1/8}{P(E_2)}$.
This implies $P(E_2) = \frac{1/8}{1/4} = \frac{1}{2}$.
We need to find $P(\bar{E}_1 \mid E_2)$.
Using the property of conditional probability,$P(\bar{E}_1 \mid E_2) = 1 - P(E_1 \mid E_2)$.
Substituting the given value,$P(\bar{E}_1 \mid E_2) = 1 - \frac{1}{4} = \frac{3}{4}$.

(A) Total number of girls $= 30$.
Given that $40 \%$ of the girls are good at Mathematics.
Number of girls good at Mathematics $= \frac{40}{100} \times 30 = 12$.
Number of girls not good at Mathematics $= 30 - 12 = 18$.
Since the student selected is already known to be a girl,we only consider the sample space of girls.
Required probability $= \frac{\text{Number of girls not good at Mathematics}}{\text{Total number of girls}} = \frac{18}{30} = \frac{3}{5}$.

(B) Let $B$ be the event that Bill hits the target and $G$ be the event that George hits the target.
Given $P(B) = 0.7$ and $P(G) = 0.4$.
The probability that Bill misses is $P(B') = 1 - 0.7 = 0.3$.
The probability that George misses is $P(G') = 1 - 0.4 = 0.6$.
Since the events are independent,the probability that exactly one shot hits the target is the sum of the probabilities of two mutually exclusive cases: (Bill hits and George misses) or (Bill misses and George hits).
$P(\text{Exactly one hit}) = P(B \cap G') + P(B' \cap G) = P(B)P(G') + P(B')P(G)$
$= (0.7 \times 0.6) + (0.3 \times 0.4) = 0.42 + 0.12 = 0.54$.
We want to find the conditional probability that it was George's shot,given that exactly one hit occurred.
$P(G \text{ hit} | \text{Exactly one hit}) = \frac{P(B' \cap G)}{P(\text{Exactly one hit})} = \frac{0.12}{0.54} = \frac{12}{54} = \frac{2}{9}$.
Thus,the correct option is $B$.