A signal which can be green or red with probability $\frac{4}{5} and \frac{1}{5}$ respectively is recieved by station A and then transmitted to station B. the probability of each station receiving the signal correctly is $\frac{3}{4} .$ If the signal received at station B is green, the probability that the original signal was green is _____

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answer is 0.87.

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Detailed Solution

Original colour	Colour at A	Colour at B	Probability
G	G	G	$\frac{4}{5} \cdot \frac{3}{4} \cdot \frac{3}{4} = \frac{36}{80}$
G	R	G	$\frac{4}{5} \cdot \frac{1}{4} \cdot \frac{1}{4} = \frac{4}{80}$
R	G	G	$\frac{1}{5} \cdot \frac{1}{4} \cdot \frac{3}{4} = \frac{3}{80}$
R	R	G	$\frac{1}{5} \cdot \frac{3}{4} \cdot \frac{1}{4} = \frac{3}{80}$

The probability that signal received at B is green is
$\frac{36 + 4 + 3 + 3}{80} = \frac{23}{40}$
The probability that the case of original signal was green is
$\frac{36 + 4}{80} = \frac{1}{2}$
The required probability is $\frac{\frac{1}{2}}{\frac{23}{40}} = \frac{20}{23} = 0 .87$