Let $P(E)$ denote probability of an event $E$. Given $P(A) = 1$, $P(B) = \dfrac{1}{2}$, find the values of $P(A|B)$ and $P(B|A)$.

Let $P(E)$ denote probability of an event $E$. Given $P(A) = 1$, $P(B) = \dfrac{1}{2}$. The values of $P(A|B)$ and $P(B|A)$ respectively are 

(A) $\dfrac{1}{4}$, $\dfrac{1}{2}$ 

(B) $\dfrac{1}{2}$, $\dfrac{1}{4}$ 

(C) $\dfrac{1}{2}$, 1 

(D) 1, $\dfrac{1}{2}$ 

(GATE 2003) 

Answer: (D) 1, $\dfrac{1}{2}$ 

Explanation: 

Since, $P(A) = 1$ and $P(B) = \dfrac{1}{2}$, hence, $P(A \cap B) = P(B) = \dfrac{1}{2}$. 

Now, $P(A|B) = \dfrac{P(A \cap B)}{P(B)} = \dfrac{\dfrac{1}{2}}{\dfrac{1}{2}} = 1$, 

and $P(B|A) = \dfrac{P(A \cap B)}{P(A)} = \dfrac{\dfrac{1}{2}}{1} = \dfrac{1}{2}$.