Question

lt is given that the events $A$ and $B$ are such that $P\left(A\right)=\frac{1}{4},P(\frac{A}{B})=\frac{1}{2}$ and $P(\frac{B}{A})=\frac{2}{3}$ then $P\left(B\right)=$

• A. $\frac{1}{6}$
• B. $\frac{1}{3}$
• C. $\frac{2}{3}$
• D. $\frac{1}{2}$

Answer 1

The correct option is B $\frac{1}{3}$

$P\left(\frac{B}{A}\right)=\frac{P(A\cap B)}{P\left(A\right)}$
Using this formula, we get $P(A\cap B)=\frac{1}{6}$
Now $P\left(\frac{A}{B}\right)=\frac{P(A\cap B)}{P\left(B\right)}$
then, $P\left(B\right)=\frac{1}{3}$