Question

If $P(A\cup B)=0.8$ and $P(A\cap B)=0.3$, then $P\left(A^{\prime }\right)+P\left(B^{\prime }\right)$ equals

• A. $0.3$
• B. $0.5$
• C. $0.7$
• D. $0.9$

Answer 1

The correct option is D $0.9$

We know that,
$P(A^{\prime }\cap B^{\prime })=P(A\cup B{)}^{\prime }$
$=1-P(a\cup B)=1-0.8=0.2$
and $P(A^{\prime }\cup B^{\prime })=P(A\cap B{)}^{\prime }=1-P(A\cap B)=1-0.3=0.7$
But $P(A^{\prime }\cup B^{\prime })=P\left(A^{\prime }\right)+P\left(B^{\prime }\right)-P(A^{\prime }\cap B^{\prime })$
$\Rightarrow 0.7=P\left(A^{\prime }\right)+P\left(B^{\prime }\right)-0.2$
$\Rightarrow P\left(A^{\prime }\right)+P\left(B^{\prime }\right)=0.9$