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 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 $\frac{20}{23}$

From the tree diagram, it follows that

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