If A, B, and C can hit a target 4 out of 5 ti | vedclass

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Question

(A) Let $P(A) = 4/5$,$P(B) = 3/4$,and $P(C) = 2/3$. The probabilities of missing the target are $P(\overline{A}) = 1/5$,$P(\overline{B}) = 1/4$,and $P

Vedclass · Accepted Answer

$13/30$

Vedclass · Answer

(A) Let $P(A) = 4/5$,$P(B) = 3/4$,and $P(C) = 2/3$. The probabilities of missing the target are $P(\overline{A}) = 1/5$,$P(\overline{B}) = 1/4$,and $P(\overline{C}) = 1/3$.The event that exactly two hit the target occurs in three mutually exclusive cases: $(A \cap B \cap \overline{C})$,$(A \cap \overline{B} \cap C)$,and $(\overline{A} \cap B \cap C)$.Since $A, B,$ and $C$ are independent events:$P(A \cap B \cap \overline{C}) = P(A) 	imes P(B) 	imes P(\overline{C}) = (4/5) 	imes (3/4) 	imes (1/3) = 12/60 = 1/5$.$P(A \cap \overline{B} \cap C) = P(A) 	imes P(\overline{B}) 	imes P(C) = (4/5) 	imes (1/4) 	imes (2/3) = 8/60 = 2/15$.$P(\overline{A} \cap B \cap C) = P(\overline{A}) 	imes P(B) 	imes P(C) = (1/5) 	imes (3/4) 	imes (2/3) = 6/60 = 1/10$.The total probability is the sum of these probabilities:$P = 12/60 + 8/60 + 6/60 = 26/60 = 13/30$.

If $A, B,$ and $C$ can hit a target $4$ out of $5$ times,$3$ out of $4$ times,and $2$ out of $3$ times respectively,what is the probability that exactly two of them hit the target?

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