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Question

Consider four independent trials in which an event A occurs with probability 13. The event B will occur with probability 1 if the event A occurs at least twice, it can not occur if the event A does not occur and it occurs with a probability 12 if the event A occurs once. If the probability p of the occurrence of event B can be expressed as mn, where m,nN, then the least value of m+n is

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Solution

n=4;P(A)=13;P(B)=p
Let E0= event A does not occur
E1= event A occurs once
E2= event A occurs at least twice

P(B|E0)=0
P(B|E1)=12
P(B|E2)=1
By total probability theorem,
P(B)=P(E0B)+P(E1B)+P(E2B)
=P(E0)P(B|E0)+P(E1)P(B|E1)++P(E2)P(B|E2)
=0+4C113(23)312 +(4C2(13)2(23)2+4C3(13)323+4C4(13)4)1
=16+24+8+181=4981=mn
(m+n)min=130

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