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MHT CET · Maths · Probability

Given \(P(A \cup B)=0.6, P(A \cap B)=0.2\), the probability of exactly one of the event occurs is

  1. A \(0.4\)
  2. B \(0.2\)
  3. C \(0.6\)
  4. D \(0.8\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(0.4\)

Step-by-step Solution

Detailed explanation

Given, \(P(A \cup B)=0.6, P(A \cap B)=0.2\)
Probability of exactly one of the event occurs is
\(P(\bar{A} \cap B)+P(A \cap \bar{B})\)
\(=P(B)-P(A \cap B)+P(A)-P(A \cap B)\)
\(=P(A \cup B)+P(A \cap B)-2 P(A \cap B)\)
\(\quad[\because P(A \cup B)=P(A)+P(B)-P(A \cap B)]\)
\(=P(A \cup B)-P(A \cap B)\)
\(=0.6-0.2\)
\(=0.4\)