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

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$p^2(2-p)$

Vedclass · Answer

(A) Let $S$ denote success (passing) and $F$ denote failure. 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 $(S, S, S), (S, S, F), (S, F, S), (F, S, S)$.$P(S, S, S) = P(S_1) 	imes P(S_2|S_1) 	imes P(S_3|S_2) = p 	imes p 	imes p = p^3$.$P(S, S, F) = P(S_1) 	imes P(S_2|S_1) 	imes P(F_3|S_2) = p 	imes p 	imes (1-p) = p^2(1-p)$.$P(S, F, S) = P(S_1) 	imes P(F_2|S_1) 	imes P(S_3|F_2) = p 	imes (1-p) 	imes \frac{p}{2} = \frac{p^2(1-p)}{2}$.$P(F, S, S) = P(F_1) 	imes P(S_2|F_1) 	imes P(S_3|S_2) = (1-p) 	imes \frac{p}{2} 	imes p = \frac{p^2(1-p)}{2}$.Summing these probabilities: $p^3 + p^2(1-p) + \frac{p^2(1-p)}{2} + \frac{p^2(1-p)}{2} = p^3 + p^2(1-p) + p^2(1-p) = p^3 + 2p^2(1-p) = p^3 + 2p^2 - 2p^3 = 2p^2 - p^3 = p^2(2-p)$.

$X$	$0$	$1$	$2$	$3$	$4$
$P(X)$	$k$	$2k$	$4k$	$6k$	$8k$

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