For the three events A, B and C,P (exactly on | vedclass

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(A) We know that $P$ (exactly one of $A$ or $B$ occurs) = $P(A) + P(B) - 2P(A \cap B) = p$ ..... $(i)$Similarly,$P(B) + P(C) - 2P(B \cap C) = p$ .....

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$\frac{3p + 2p^2}{2}$

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(A) We know that $P$ (exactly one of $A$ or $B$ occurs) = $P(A) + P(B) - 2P(A \cap B) = p$ ..... $(i)$Similarly,$P(B) + P(C) - 2P(B \cap C) = p$ ..... $(ii)$and $P(C) + P(A) - 2P(C \cap A) = p$ ..... $(iii)$Adding $(i), (ii)$ and $(iii)$,we get $2[P(A) + P(B) + P(C)] - 2[P(A \cap B) + P(B \cap C) + P(C \cap A)] = 3p$Therefore,$P(A) + P(B) + P(C) - [P(A \cap B) + P(B \cap C) + P(C \cap A)] = \frac{3p}{2}$ ..... $(iv)$We are given $P(A \cap B \cap C) = p^2$ ..... $(v)$The probability of at least one of the three events $A, B$ and $C$ occurring is given by the inclusion-exclusion principle:$P(A \cup B \cup C) = P(A) + P(B) + P(C) - [P(A \cap B) + P(B \cap C) + P(C \cap A)] + P(A \cap B \cap C)$Substituting $(iv)$ and $(v)$ into the expression:$P(A \cup B \cup C) = \frac{3p}{2} + p^2 = \frac{3p + 2p^2}{2}$

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