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

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., $0 \leqslant P\left( E \right) \leqslant 1$.