A signal which can be green or red with proba | vedclass

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(C) Let $G$ be the event that the original signal is green and $R$ be the event that the original signal is red. Given $P(G) = \frac{4}{5}$ and $P(R)

Vedclass · Accepted Answer

$\frac{20}{23}$

Vedclass · Answer

(C) Let $G$ be the event that the original signal is green and $R$ be the event that the original signal is red. Given $P(G) = \frac{4}{5}$ and $P(R) = \frac{1}{5}$.Let $S_A$ and $S_B$ be the signals received by station $A$ and station $B$ respectively. The probability of correct reception is $p = \frac{3}{4}$ and incorrect is $q = \frac{1}{4}$.Let $E$ be the event that the signal received at station $B$ is green.$P(E) = P(E|G)P(G) + P(E|R)P(R)$.To receive a green signal at $B$ starting from $G$: $1$. $A$ receives $G$ correctly (prob $\frac{3}{4}$),$B$ receives $G$ correctly (prob $\frac{3}{4}$) $ightarrow \frac{3}{4} 	imes \frac{3}{4} = \frac{9}{16}$.$2$. $A$ receives $R$ incorrectly (prob $\frac{1}{4}$),$B$ receives $G$ incorrectly (prob $\frac{1}{4}$) $ightarrow \frac{1}{4} 	imes \frac{1}{4} = \frac{1}{16}$.So,$P(E|G) = \frac{9}{16} + \frac{1}{16} = \frac{10}{16} = \frac{5}{8}$.To receive a green signal at $B$ starting from $R$:$1$. $A$ receives $R$ correctly (prob $\frac{3}{4}$),$B$ receives $G$ incorrectly (prob $\frac{1}{4}$) $ightarrow \frac{3}{4} 	imes \frac{1}{4} = \frac{3}{16}$.$2$. $A$ receives $G$ incorrectly (prob $\frac{1}{4}$),$B$ receives $R$ incorrectly (prob $\frac{1}{4}$) $ightarrow \frac{1}{4} 	imes \frac{1}{4} = \frac{1}{16}$.So,$P(E|R) = \frac{3}{16} + \frac{1}{16} = \frac{4}{16} = \frac{1}{4}$.Using Bayes' Theorem,$P(G|E) = \frac{P(E|G)P(G)}{P(E|G)P(G) + P(E|R)P(R)} = \frac{\frac{5}{8} 	imes \frac{4}{5}}{\frac{5}{8} 	imes \frac{4}{5} + \frac{1}{4} 	imes \frac{1}{5}} = \frac{\frac{1}{2}}{\frac{1}{2} + \frac{1}{20}} = \frac{\frac{1}{2}}{\frac{11}{20}} = \frac{10}{11} 	imes \frac{20}{11} = \frac{20}{23}$.

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