Question

A signal which can be green or red with probability $\frac{4}{5}$ and $\frac{1}{5}$, respectively, is received at 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 $\frac{20}{23}$

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

$P\left(\text{signal received by B is green}\right)=P\left(G{E}_1{E}_2\right)+P(G\cap E_1\cap E_2)+P(\cap G{E}_1\cap E_2)+P(\cap G\cap E_1{E}_2)$

$P\left(E\right)=\frac{46}{5\times 16}$

$P\left(G/E\right)=\frac{\frac{40}{5}\times 16}{\frac{46}{5}\times 16}=\frac{20}{23}.$