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$A signal which can be green or red with probability (\frac{4}{5}) and (\frac{1}{5}) respectively is received$ $by the station A and Transmitted to B . The probability of each station receiving signal$ $correctly is \frac{3}{4} . If signal in received in B is green. The probability original signal was$ green is

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$\frac{3}{5}$

$\frac{6}{7}$

$\frac{20}{23}$

$\frac{9}{20}$

answer is C.

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Detailed Solution

$G = Original signal green. A = A receive correct signal, B = B receive signal correct. E$ $is signal received by B is green.$

$P (\frac{G}{E}) = \frac{P (G \cap E)}{P (E)}$

$P (E) = P (G A B) + P (G \bar{A} \bar{B}) + P (\bar{G} A \bar{B}) + P (\bar{G} \bar{A} B)$

$= \frac{4}{5} . \frac{3}{4} . \frac{3}{4} + \frac{4}{5} . \frac{1}{4} . \frac{1}{4} + \frac{1}{5} . \frac{3}{4} . \frac{1}{4} + \frac{1}{5} . \frac{1}{4} . \frac{3}{4} = \frac{46}{80}$

$P (G \cap E) = P (G A B) + P (G \bar{A} \bar{B}) = \frac{40}{80}$

$∴ P (\frac{G}{E}) = \frac{P (G \cap E)}{P (E)} = \frac{20}{23}$

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