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 transmitted to $B$. The probability each station receives correct signal is $\frac{3}{4}$. If the signal received by $B$ is green then the probability that 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 $G=$Original signal green
$A=A$ receives correct signal
$B=B$ receives correct signal
$E$ is the event that signal received by $B$ is green.
Hence required probablity is $P\left(\frac{G}{E}\right)=\frac{P(G\cap E)}{P\left(E\right)}$
Now, $P\left(E\right)=P\left(GAB\right)+P\left(G\overline{A}\overline{B}\right)+P\left(\overline{G}A\overline{B}\right)+P\left(\overline{G}\overline{A}B\right)$
$=\frac{4}{5}.\frac{3}{4}.\frac{3}{4}+\frac{4}{5}.\frac{1}{4}.\frac{1}{4}+\frac{1}{5}.\frac{3}{4}.\frac{1}{4}+\frac{1}{5}.\frac{1}{4}.\frac{3}{4}=\frac{46}{80}$
Also, $P(G\cap E)=P\left(GAB\right)+P\left(G\overline{A}\overline{B}\right)=\frac{40}{80}$
So, we have: $P\left(\frac{G}{E}\right)=\frac{P(G\cap E)}{P\left(E\right)}=\frac{40}{46}=\frac{20}{23}$