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 For the three events A,B and C,P( exactly one of the  events A or B occurs )=P( exactly one of the two events 

B or C occurs )=P( exactly one of the events C or A occurs )=p and P( all the three events occur simultaneously )=p2

 where 0<p<1/2. Then the probability of  at least one of the three events A,B and C occurring is 

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a
3p+2p22
b
p+3p24
c
p+3p22
d
3p+2p24

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detailed solution

Correct option is A

We know thatP(exactly one of A or B occurs) =P(A)+P(B)−2P(A∩B)Therefore,P(A)+P(B)−2P(A∩B)=p------1Similarly,P(B)+P(C)−2P(B∩A)=p----2andP(C)+P(A)−2P(C∩A)=p-----3Adding Eqs. (1), (2) and (3) we get2[P(A)+P(B)+P(C)−P(A∩B)−P(B∩C)−P(C∩A)]=3p⇒ P(A)+P(B)+P(C)−P(A∩B)−P(B∩C)−P(C∩A)=3p/2---4It is also given that P(A∩B∩C)=p2----5Now,P(at least one of A, B and Q)=P(A)+P(B)+P(C)−P(A∩B)−P(B∩C)−P(C∩A)+P(A∩B∩C)=3p2+p2=3p+2p22          [Using Eqs. (4) and (5)]


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The odd against an event are 5 to 2 and the odds in favour of another disjoint event are 3 to 5. Then the probability that atleast one of the events will happen is


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