Two events A and B are such that P(A)=frac{1} | vedclass

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(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

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only $(I)$ is correct

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(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 $P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{1}{2} \implies P(A \cap B) = \frac{1}{2}P(A) = \frac{1}{2} 	imes \frac{1}{4} = \frac{1}{8}$.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 $P(A \cap B) = \frac{1}{8} 
eq 0$, $A$ and $B$ are not mutually exclusive. Statement $(II)$ is incorrect.$(III)$ $P(A|B)+P(A|\bar{B}) = \frac{P(A \cap B)}{P(B)} + \frac{P(A \cap \bar{B})}{P(\bar{B})} = \frac{1/8}{1/2} + \frac{P(A)-P(A \cap B)}{1-P(B)} = \frac{1}{4} + \frac{1/4-1/8}{1/2} = \frac{1}{4} + \frac{1/8}{1/2} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} 
eq 1$. Statement $(III)$ is incorrect.Thus, only $(I)$ is correct.

Two events $A$ and $B$ are such that $P(A)=\frac{1}{4}$, $P(A|B)=\frac{1}{4}$ and $P(B|A)=\frac{1}{2}$. Consider the following statements:
$(I) P(\bar{A}|\bar{B})=\frac{3}{4}$
$(II) A$ and $B$ are mutually exclusive
$(III) P(A|B)+P(A|\bar{B})=1$
Then,

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Two events $A$ and $B$ are such that $P(A)=\frac{1}{4}$, $P(A|B)=\frac{1}{4}$ and $P(B|A)=\frac{1}{2}$. Consider the following statements:$(I) P(\bar{A}|\bar{B})=\frac{3}{4}$$(II) A$ and $B$ are mutually exclusive$(III) P(A|B)+P(A|\bar{B})=1$Then,