Question

A signal which can be green or red with probability \[\dfrac{4}{5}\] and \[\dfrac{1}{5}\] respectively, is received by station A and then transmitted to station B. The probability of each station receiving the signal correctly is \[\dfrac{3}{4}\]. If the signal received at station B is green, then the probability that the original signal was green is
A. \[\dfrac{3}{5}\]
B. \[\dfrac{6}{7}\]
C. \[\dfrac{{20}}{{23}}\]
D. \[\dfrac{9}{{20}}\]

Answer 1

Hint: In this problem we will proceed by naming the given events and finding their probabilities. Then find the probability that the signal received by B is green and then use conditional probability to get the final answer.

Complete step-by-step answer:
Let us consider the events as
\[G\]: original signal is green
\[{E_1}\]: A receives the signal correctly
\[{E_2}\]: B receives the signal correctly
\[E\]: signal received by B is green
Given that \[P\left( G \right) = \dfrac{4}{5},P\left( {{E_1}} \right) = \dfrac{3}{4},P\left( {{E_2}} \right) = \dfrac{3}{4}\]
Now, the probability of the event that does not receives the original signal as green is given by
\[P\left( {\bar G} \right) = P\left( {1 - G} \right) = 1 - \dfrac{4}{5} = \dfrac{1}{5}\]
The probability of the event that does not receives the signal correctly is given by
\[P\left( {{{\bar E}_1}} \right) = P\left( {1 - {E_1}} \right) = 1 - \dfrac{3}{4} = \dfrac{1}{4}\]
And the probability of the event that does not receives the signal correctly is given by
\[P\left( {{{\bar E}_2}} \right) = P\left( {1 - {E_2}} \right) = 1 - \dfrac{3}{4} = \dfrac{1}{4}\]
So, the probability that the signal was received B is green is given by
\[
   \Rightarrow P\left( E \right) = P\left( {G{E_1}{E_2}} \right) + P\left( {G{{\bar E}_1}{{\bar E}_2}} \right) + P\left( {\bar G{E_1}{{\bar E}_2}} \right) + P\left( {\bar G{{\bar E}_1}{E_2}} \right) \\
   \Rightarrow P\left( E \right) = \left( {\dfrac{4}{5} \times \dfrac{3}{4} \times \dfrac{3}{4}} \right) + \left( {\dfrac{4}{5} \times \dfrac{1}{4} \times \dfrac{1}{4}} \right) + \left( {\dfrac{1}{5} \times \dfrac{3}{4} \times \dfrac{1}{4}} \right) + \left( {\dfrac{1}{5} \times \dfrac{1}{4} \times \dfrac{3}{4}} \right) \\
   \Rightarrow P\left( E \right) = \left( {\dfrac{{36}}{{80}}} \right) + \left( {\dfrac{4}{{80}}} \right) + \left( {\dfrac{3}{{80}}} \right) + \left( {\dfrac{3}{{80}}} \right) \\
   \Rightarrow P\left( E \right) = \dfrac{{36 + 4 + 3 + 3}}{{80}} \\
  \therefore P\left( E \right) = \dfrac{{46}}{{80}} \\
\]
And the
Now, the probability that the original signal was green when the signal received at station B is green is given by
\[
   \Rightarrow P\left( {G\left| E \right.} \right) = \dfrac{{P\left( {G \cap E} \right)}}{{P\left( E \right)}} = \dfrac{{P\left( {G{E_1}{E_2}} \right) + P\left( {G{{\bar E}_1}{{\bar E}_2}} \right)}}{{P\left( E \right)}} \\
   \Rightarrow P\left( {G\left| E \right.} \right) = \dfrac{{\left( {\dfrac{4}{5} \times \dfrac{3}{4} \times \dfrac{3}{4}} \right) + \left( {\dfrac{4}{5} \times \dfrac{1}{4} \times \dfrac{1}{4}} \right)}}{{\dfrac{{46}}{{80}}}} \\
   \Rightarrow P\left( {G\left| E \right.} \right) = \dfrac{{\left( {\dfrac{{36}}{{80}}} \right) + \left( {\dfrac{4}{{80}}} \right)}}{{\dfrac{{46}}{{80}}}} \\
   \Rightarrow P\left( {G\left| E \right.} \right) = \dfrac{{\dfrac{{36 + 4}}{{80}}}}{{\dfrac{{46}}{{80}}}} = \dfrac{{40}}{{46}} = \dfrac{{20}}{{23}} \\
  \therefore P\left( {G\left| E \right.} \right) = \dfrac{{20}}{{23}} \\
\]
Thus, the correct option is C. \[\dfrac{{20}}{{23}}\]

Note: The intersection of two sets \[A\] and \[B\], denoted by \[A \cap B\], is the set containing all elements of \[A\] that also belong to \[B\] (or equivalently, all elements of \[B\] that also belong to \[A\]). The probability of an event is always lying between 0 and 1 i.e., \[0 \leqslant P\left( E \right) \leqslant 1\].