For three events A, B and C, P (exactly one of the events A or B occurs) = P( exactly one of the events B or C occurs) = P (exactly one of the events C or A occurs) = p and P (all the three events occur simultaneously) = $p^2 $, where $ 0 < p < \frac{1}{2}$. Then the probability of atleast one if the three events A, B and C occuring is
Answer & explanation
Correct answer: option 1
We have,
$P(A) +P(B) -2P( A ∩ B) = p $ ...........(i)
$P(B) +P(C) -2P( B ∩ C) = p $ ...........(ii)
$P(C) +P(A) -2P( C ∩ A) = p $ ...........(iii)
and, $ P(A ∩ B ∩ C) = p^2 $ ..............(iv)
Adding (i), (ii) and (iii), we get
$2[P(A) +P(B) +P(C)+P(A ∩ B) - P (B ∩ C) - P( A ∩ C) ]= 3p $
$⇒ P(A) +P(B) +P(C) - P(A ∩ B) - P(B ∩ C) $
$P( A ∩ C)=\frac{3p}{2}$
∴ Required probability
$ = P( A ∪ B ∪ C)$
$ = P(A) + P(B) + P(C) - P(A ∩ B) - P( B ∩C) - P( A ∩ C) + P( A ∩ B ∩C) $
$=\frac{3p}{2} + p^2 = \frac{3p+2p^2}{2}$