Aspire Faculty ID #3652 · Topic: NIMCET 2012 · Just now
NIMCET 2012

Let $P(E)$ denote the probability of event $E$. Given $P(A) = 1$, $P(B) = \frac{1}{2}$, the values of $P(A \mid B)$ and $P(B \mid A)$ respectively are

Solution

Since $P(A)=1$, event $A$ is a sure event. 
Therefore for any event $B$, 
 $P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{P(B)}{P(B)}=1$ 
 Also, $P(B \mid A)=\frac{P(A \cap B)}{P(A)}=\frac{P(B)}{1}=\frac{1}{2}$ 
 Thus the correct pair is: 
 $P(A \mid B)=1$ and $P(B \mid A)=\frac{1}{2}$

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