Question

The probability that aleast one of $A$ and $B$ occurs is $0.6$. If $A$ and $B$ occur simultaneously with probability $0.3$, then $P\left(A^{\prime }\right)+P\left(B^{\prime }\right)$ is equal to

• A. $0.9$
• B. $0.5$
• C. $1.1$
• D. $1.2$

Answer 1

Answer: (C)

$1.1$

Given, $P(A\cup B)=0.6$ and $P(A\cap B)=0.3$

Therefore, $P\left(A^{\prime }\right)+P\left(B^{\prime }\right)=1-P\left(A\right)+1-P\left(B\right)$

$=2-\left[P\right(A)+P(B\left)\right]$
$=2-\left[P\right(A\cup B)+P(A\cap B)$

$=2-(0.6+0.3)$

$=2-0.9=1.1$