Q77 / 160 · AP EAMCET 2018 (22 Apr Shift 2) · Mathematics

$A$ and $B$ are two events such that $P(A)=0.58$, $P(B)=0.32$ and $P(A \cap B)=0.28$. Then the probability that neither $A$ nor $B$ occurs is

  • Correct answer
    A
    0.38
  • B
    0.62
  • C
    0.72
  • D
    0.9

Solution

$\begin{aligned} & P(\bar{A} \cap \bar{B})=P(A \cup B)=1-P(A \cup B) \\ & =1-\{P(A)+P(B)-P(A \cap B)\} \\ & =1-\{0.58+0.32-0.28\}=1-0.62=0.38\end{aligned}$

Hence, the correct answer is option: 'A'.

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