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(C) Given that $A, B$ and $C$ are independent events with $P(A)=P(B)=P(C)=P$. The probability that at least two of $A, B$ and $C$ occur is the sum of

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$3 P^{2}-2 P^{3}$

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(C) Given that $A, B$ and $C$ are independent events with $P(A)=P(B)=P(C)=P$. The probability that at least two of $A, B$ and $C$ occur is the sum of the probabilities of the following mutually exclusive cases: $1$. Exactly two events occur: $(A \cap B \cap C') \cup (A \cap B' \cap C) \cup (A' \cap B \cap C)$ $2$. All three events occur: $(A \cap B \cap C)$ Since the events are independent, $P(A \cap B \cap C') = P(A)P(B)P(C') = P 	imes P 	imes (1-P) = P^{2}(1-P)$. Similarly, $P(A \cap B' \cap C) = P^{2}(1-P)$ and $P(A' \cap B \cap C) = P^{2}(1-P)$. Also, $P(A \cap B \cap C) = P 	imes P 	imes P = P^{3}$. Thus, $P(	ext{at least two occur}) = 3 	imes P^{2}(1-P) + P^{3}$. $= 3P^{2} - 3P^{3} + P^{3} = 3P^{2} - 2P^{3}$.

If $A, B$ and $C$ are three independent events such that $P(A)=P(B)=P(C)=P$, then $P$ (at least two of $A, B$ and $C$ occur) is equal to

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