Question

The probability that at least one of $A$ and $B$ occurs is $0.6$ and probability that they occur simultaneously is $0.3$, then $P\left(A^{\prime }\right)+P\left(B^{\prime }\right)$ is:

• A. $0.9$
• B. $1.15$
• C. $1.1$
• D. $1.2$

Answer 1

Answer: (C)

$1.1$

We have have $P(A\cup B)=0.6$ and $P(A\cap B)=0.3$.
We know that
$P\left(A\right)+P\left(B\right)=P(A\cup B)+(A\cap B)=0.6+0.3=0.9$
$\therefore P\left(A^{\prime }\right)+P\left(B^{\prime }\right)=1-P\left(B\right)+1-P\left(A\right)=2-0.9=1.1$