Question

A signal which can be green or red with probability $\frac{4}{5}$ and $\frac{1}{5}$ respectively, is received 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, then the probability that the original signal was green is

• A. $\frac{3}{5}$
• B. $\frac{6}{7}$
• C. $\frac{20}{23}$
• D. $\frac{9}{20}$

Answer 1

The correct option is C $\frac{20}{23}$

Let Event $G=$ original signal is green
$E_1=A$ receives the signal correct
$E_2=B$ receives the signal correct
$E=$ signal received by $B$ is green
$P$(signal received by $B$ is green)
$=P\left(G{E}_1{E}_2\right)+P\left(G{\overline{E}}_1{\overline{E}}_2\right)+P\left(\overline{G}{E}_1{\overline{E}}_2\right)+P\left(\overline{G}{\overline{E}}_1{E}_2\right)$

$=(\frac{4}{5}\times \frac{3}{4}\times \frac{3}{4})+(\frac{4}{5}\times \frac{1}{4}\times \frac{1}{4})+(\frac{1}{5}\times \frac{3}{4}\times \frac{1}{4})+(\frac{1}{5}\times \frac{1}{4}\times \frac{3}{4})$

$P\left(E\right)=\frac{46}{80}$
$P\left(G/E\right)=\frac{(\frac{4}{5}\times \frac{3}{4}\times \frac{3}{4})+(\frac{4}{5}\times \frac{1}{4}\times \frac{1}{4})}{\frac{46}{80}}=\frac{20}{23}$