Question

If two events A and B such that $P\left(A^{\prime }\right)=0.3,P\left(B\right)=0.5$ and $P(A\cap B)=0.3$, then $P(B|A\cup B^{\prime })$ is

• A. $3/8$
• B. $2/3$
• C. $5/6$
• D. $1/4$

Answer 1

Answer: (A)

$3/8$

We have $P(A\cup B^{\prime })$
$=P\left(A\right)+P\left(B^{\prime }\right)-P(A\cap B^{\prime })$
$=[1-P(A^{\prime }\left)\right]+[1-P(B\left)\right]-\left[P\right(A)-P(A\cap B\left)\right]$
$=(1-0.3)+(1-0.5)-(0.7-0.3)=0.8$.

Now, $P(B|A\cup B^{\prime })=\frac{P[B\cap A(\cup B^{\prime }\left)\right]}{P(A\cup B^{\prime })}$

$=\frac{P\left[\right(B\cap A)\cup (B\cap B^{\prime }\left)\right]}{P(A\cup B^{\prime })}$

$=\frac{P(A\cap B)}{P(A\cup B^{\prime })}=\frac{0.3}{0.8}=\frac{3}{8}$