Probability of trains A, B and C arriving  on times is 1/3, 2/5 and 2/3 respectively. What is the probability that almost one of the trains arrive on time?

a) 7/15        b) 22/45       c) 23/45       d) None of these 

Dear Student,Probability of train A arriving on time=P(A)=13Thus, Probability of train A not arriving on time=P(Ac)=1-13=23Similarly,P(B)=25 and P(Bc)=35P(C)=23 and P(Cc)=13Probability that ATMOST one of the trains arrive on time=Probability that none of them arrive on time+Probability that only train A arrive on time+Probability that only train B arrive on time+Probability that only train C arrive on time=P(Ac)×P(Bc)×P(Cc)+P(A)×P(Bc)×P(Cc)+P(Ac)×P(B)×P(Cc)+P(Ac)×P(Bc)×P(C)=23×35×13+13×35×13+23×25×13+23×35×23=645+345+445+1245=2545=59Regards

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