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
- A \(0.4\)
- B \(0.2\)
- C \(0.6\)
- D \(0.8\)
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
\(P(\bar{A} \cap B)+P(A \cap \bar{B})\)
\(=P(B)-P(A \cap B)+P(A)-P(A \cap B)\)
\(=P(A \cup B)+P(A \cap B)-2 P(A \cap B)\)
\(\quad[\because P(A \cup B)=P(A)+P(B)-P(A \cap B)]\)
\(=P(A \cup B)-P(A \cap B)\)
\(=0.6-0.2\)
\(=0.4\)
Probability of exactly one of the event occurs is
\(P(\bar{A} \cap B)+P(A \cap \bar{B})\)
\(=P(B)-P(A \cap B)+P(A)-P(A \cap B)\)
\(=P(A \cup B)+P(A \cap B)-2 P(A \cap B)\)
\(\quad[\because P(A \cup B)=P(A)+P(B)-P(A \cap B)]\)
\(=P(A \cup B)-P(A \cap B)\)
\(=0.6-0.2\)
\(=0.4\)
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