It is given that the events A and B are such that P(A)=14, P(A/B)=12 and P(B/A)=23. Then, P(B) is

# It is given that the events $A$ and $B$ are such that  and $P\left(B/A\right)=\frac{2}{3}\text{.}$ Then, $P\left(B\right)$ is

1. A

$\frac{2}{3}$

2. B

$\frac{1}{2}$

3. C

$\frac{1}{6}$

4. D

$\frac{1}{3}$

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### Solution:

We have,

Now, $P\left(A/B\right)=\frac{P\left(A\cap B\right)}{P\left(B\right)}⇒\frac{1}{2}=\frac{1}{6}×\frac{1}{P\left(B\right)}⇒P\left(B\right)=\frac{1}{3}$

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