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

Answer: (C)

$\frac{ 20}{23}$

Answer 2

From the tree diagram, it follows that $P(B_G) = \frac{46}{80}$ $P(B_G /G) =\frac{10}{16} = \frac{5}{8}$ $P(B_G \cap G) =\frac{5}{8} \times \frac{4}{5} = \frac{1}{2}$ $P(G / B_G) = \frac{2}{P(B_G)} = \frac{1}{2} \times \frac{ 80}{46} = \frac{20}{23}$