Question

If $A$, $B$ are two events with $P(A\cup B)=0.65$, $P(A\cap B)=0.15$, then find the value of $P\left(A^c\right)+P\left(B^c\right)$.

Answer 1

Fact: $P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$

So $P(A\cup B)+P(A\cap B)=P\left(A\right)+P\left(B\right)$

$\Rightarrow P\left(A\right)+P\left(B\right)=0.65+0.15=0.8$, substitute the given values

Therefore $P\left(A^C\right)+P\left(B^C\right)=1-P\left(A\right)+1-P\left(B\right)=2-\left[P\right(A)+P(B\left)\right]=2-0.8=1.2$