If $P(B)=\frac{3}{4}, P(A ∩B ∩ \overline{C})=\frac{1}{3}$ and $ P(\overline{A} ∩B ∩ \overline{C})=\frac{1}{3}$, then $P(B ∩ C)$ is
Answer & explanation
Correct answer: option 1
We have,
$ P(\overline{A} ∩B ∩ \overline{C})=\frac{1}{3}$
$⇒ P(( B ∩ \overline{C})∩ \overline{A})=\frac{1}{3}$
$⇒ P( B ∩ \overline{C})-P(( B ∩ \overline{C}) ∩ A = \frac{1}{3}$ $[∵ X ∩ \overline{Y}) = P(X) - P(X ∩ Y)]$
$⇒ P( B ∩ \overline{C}) - P(A ∩ B ∩ \overline{C})=\frac{1}{3}$
$⇒ P( B ∩ \overline{C})-\frac{1}{3}=\frac{1}{3}$
$⇒ P( B ∩ \overline{C})=\frac{2}{3}$
$⇒P(B)- P( B ∩ C)=\frac{2}{3}$
$⇒P( B ∩ C) = P(B)-\frac{2}{3}=\frac{3}{4}-\frac{2}{3}=\frac{1}{12}$