Q. 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 green is
A) $\frac{3}{5}$
B) $\frac{6}{7}$
C) $\frac{20}{23}$
D) $\frac{9}{20}$
Explanation
(c)
From the three diagram it follows that
[image:16_image3.emf]
$P\left( B_{G} \right) = \frac{46}{80}$
$P(B_{G}\left| G) = \frac{10}{16} = \frac{5}{8} \right.\$
$P\left( B_{G} \cap G \right) = \frac{5}{8} \times \frac{4}{5} = \frac{1}{2}$
$P\left( G \middle| B_{G} \right) = \frac{P(B_{G} \cap G)}{P(B_{G})} = \frac{1}{2} \times \frac{80}{46} = \frac{20}{23}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.