Question

A signal swhich can be green or red with probability $\frac{4}{5}$ and $\frac{1}{5}$ rexpectively, is received by station $A$ and then transmitted to station $B$. The probability of each staation receiving the signal correctly is $\frac{3}{4}$. If the signal received at station $B$ is green, then the probability that the origonal signal was green is

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

Answer 1

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


$P\left(B_G\right)=\frac{46}{80}$
$P\left(B_G|G\right)=\frac{10}{16}=\frac{5}{8}$
$\therefore P(B_G\cap G)=\frac{5}{8}\times \frac{4}{5}=\frac{1}{2}$
$P\left(G|B_G\right)=\frac{\frac{1}{2}}{P\left(B_G\right)}=\frac{1}{2}\times \frac{80}{46}=\frac{20}{23}$