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Q.
It is given that the events $A$ and $B$ are such that $P(A) = \frac{1}{4}, P(A|B) = \frac{1}{2}$ and $P(B|A) = \frac{2}{3}$. Then $P(B)$ is
Probability - Part 2
Solution:
From the definition of independence of events
$P(A|B) = \frac{P(A \cap B)}{P(B)}$
Then $P(B) \cdot P(A|B) = P(A \cap B ) \,...(1)$
Interchanging the role of $A$ and $B$ in (1)
$P (A )P (B|A ) = P (B \cap A)\, ...(2)$
As $A \cap B = B \cap A$, we have from $(1)$ and $(2)$
$P (A )P (B |A ) = P (B )P (A |B)$
$\Rightarrow \frac{1}{4} \cdot \frac{2}{3} = P(B) \cdot \frac{1}{2}$
$\Rightarrow P(B) = \frac{1}{4} \cdot \frac{2}{3} \cdot 2 = \frac{1}{3}$