Question

If $A$ and $B$ are two independent events such that $P(A^{\prime }\cap B)=2/15$ and $P(A\cap B^{\prime })=1/6$, the $P\left(B\right)$ is

• A. $1/5$
• B. $1/6$
• C. $4/5$
• D. $5/6$

Answer 1

Answer: (B), (C)

$1/6$
$4/5$

Since $A$ and $B$ are independent
$2/15=P(A^{\prime }\cap B)=P\left(A^{\prime }\right)P\left(B\right)=[1-P(A\left)\right]P\left(B\right)$ (1)
and $1/6=P(A\cap B^{\prime })=P\left(A\right)P\left(B^{\prime }\right)=P\left(A\right)[1-P(B\left)\right]$ (2)
Subtracting (2) from (1) we get
$P\left(A\right)-P\left(B\right)=1/30$ or $P\left(A\right)=P\left(B\right)+1/30$
Putting this value in (2), we get
$\left[P\right(B)+1/30][1-P(B\left)\right]=1/6$
$\Rightarrow 30P(B{)}^2-29P(B)+4=0$
$\Rightarrow P\left(B\right)=1/6,4/5$.