Question

If probability that exactly one of events A B .C occurs is 0.6 and probability that none of A B C occur is 0.2 then probability that ateast two of A B C occur is (A) 0.6 (B) 0.4 (C) 0.8 (D) 0.2​

Answer 1

p (AUB)=0.6,P (A intersection B)=0.2

P (AUB)=P (A)+P (B)-P (A intersection B)

0.6=P (A)+P (B)-0.2

0.6+0.2=P (A)+P (B)

P (A)+P (B)=0.8

P (A bar )+P (Bbar)=(1-p (A))+(1-p (B))

=2-(P(A)+P (B))

=2-(0.8)

=1.2e

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