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AP EAMCET · Maths · Probability

\(A\) 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

  1. A 0.38
  2. B 0.62
  3. C 0.72
  4. D 0.9
Verified Solution

Answer & Solution

Correct Answer

(A) 0.38

Step-by-step Solution

Detailed explanation

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