Question

If $A$ and $B$ are two events such that $P(A\cup B)=0.65$ and $P(A\cap B)=0.15$, then $P\left(\overline{A}\right)+P\left(\overline{B}\right)=$

• A. $0.6$
• B. $0.8$
• C. $1.2$
• D. $1.4$

Answer 1

Answer: (C)

$1.2$

$P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$
$P(A\cup B)=1-P\left(\overline{A}\right)-1-P\left(\overline{B}\right)-P(A\cap B)$
$0.65=2-P\left(\overline{A}\right)-P\left(\overline{B}\right)-0.15$ Given: $[P(A\cup B)=0.65;P(A\cap B)=0.15]$
$P\left(\overline{A}\right)+P\left(\overline{B}\right)=2-0.65-0.15=1.2$