439 views
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) = $p^2$, where $0 < p < \frac{1}{2}$. Then the probability of at least one of the three events A, B and C occuring is

Let $P(A^o)$ denote probability of only $A$ and similarly for $B,C$.

$P(A^o) + P(B^o) = p = P(A) - P(A \cap B) + P(B) - P(A \cap B) = P(A) + P(B) - 2P(A\cap B)$

Similarly,

$P(B^o) + P(C^o) =P(B) + P(C) - 2P(B\cap C) = p$

$P(C^o) + P(A^o) =P(C) + P(A) - 2P(C\cap A) = p$

Also, $P(A \cap B \cap C) = p^2$.

We want $P\overline{(A \cup B \cup C)}$

$P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(C \cap A) + P(A \cap B \cap C) \\= 3p/2 + p^2$

So, required probability $= 1 - \frac{3p}{2} - p^2$
by

1 vote