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Question

Let A,B,C be three events. If the probability of occurring exactly one event out of A and B is 1-a, out of B and C and A is 1-2a and that of occurring three events simultaneously is a2, then the probability that at least one out of A,B,C will occur is


A

12

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B

Greater than 12

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C

Less than 12

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D

Greater than 34

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Solution

The correct option is B

Greater than 12


Explanation for the correct option:

Finding the probability that at least one out of A,B,C will occur:

Given that,

Probability of getting exactly one event out of A and B =1-a

Therefore,

P[(AB)(AB)]=1-aP(AB)P(AB)=1-aP(A)+P(B)P(AB)P(AB)=1-aP(A)+P(B)2P(AB)=1-a...(i)

Similarly,

Probability of getting exactly one event out of B and C =1-2a

P(B)+P(C)2P(BC)=1-2a...(ii)

P(C)+P(A)2P(CA)=1-a..(iii)

And given that that of occurring three events simultaneously is a2,

P(ABC)=a2..(iv)

As we know,

P(ABC)=P(A)+P(B)+P(C)P(AB)P(BC)-P(CA)+P(ABC)=12[P(A)+P(B)2P(BC)+P(B)+P(C)2P(BC)+P(C)+P(A)2P(CA)]+P(ABC)=121-a+1-2a+1-a+a2

Now from (i),(ii),(iii)&(iv)

=322a+a2=12+(a-1)2>12

Hence, the correct is option(B)


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