Q66 / 150 · MHT CET 2009 · Mathematics

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

  • Correct answer
    A
    $0.4$
  • B
    $0.2$
  • C
    $0.6$
  • D
    $0.8$

Solution

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

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

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