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(A) Let $S$ denote success (pass) and $F$ denote failure (fail). The probability of passing the first test is $P(S_1) = p$,so $P(F_1) = 1-p$.For subse

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$p^2(2-p)$

Vedclass · Answer

(A) Let $S$ denote success (pass) and $F$ denote failure (fail). The probability of passing the first test is $P(S_1) = p$,so $P(F_1) = 1-p$.For subsequent tests,$P(S_{n+1} | S_n) = p$ and $P(S_{n+1} | F_n) = \frac{p}{2}$.The candidate is selected if they pass at least two tests. The possible favorable outcomes are:$1$. Pass $I$,Pass $II$,Fail $III$: $p 	imes p 	imes (1-p) = p^2(1-p)$$2$. Pass $I$,Fail $II$,Pass $III$: $p 	imes (1-p) 	imes \frac{p}{2} = \frac{p^2(1-p)}{2}$$3$. Fail $I$,Pass $II$,Pass $III$: $(1-p) 	imes \frac{p}{2} 	imes p = \frac{p^2(1-p)}{2}$$4$. Pass $I$,Pass $II$,Pass $III$: $p 	imes p 	imes p = p^3$Summing these probabilities:$P = p^2(1-p) + \frac{p^2(1-p)}{2} + \frac{p^2(1-p)}{2} + p^3$$P = p^2(1-p) + p^2(1-p) + p^3$$P = 2p^2 - 2p^3 + p^3 = 2p^2 - p^3 = p^2(2-p)$.

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