Question

If for two events $A$ and $B$,$P(A\cup B)=\frac{1}{2},P(A\cap B)=\frac{2}{5}$and then $P\left(A^c\right)+P\left(B^c\right)$equals

• A. $\frac{4}{5}$
• B. $\frac{5}{4}$
• C. $\frac{11}{10}$
• D. None of these

Answer 1

The correct option is C $\frac{11}{10}$

Given $P(A\cup B)=\frac{1}{2}$ and $P(A\cap B)=\frac{2}{5}$
Fact $P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$
$\Rightarrow P\left(A\right)+P\left(B\right)=P(A\cup B)+P(A\cap B)$ $=\frac{1}{2}+\frac{2}{5}=\frac{9}{10}$
$\Rightarrow 1-P\left(A^c\right)+1-P\left(B^c\right)=\frac{9}{10}$
$\Rightarrow P\left(A^c\right)+P\left(B^c\right)=2-\frac{9}{10}=\frac{11}{10}$