A signal which can be green or red with probability $4 / 5$ and $1 / 5$ respectively, is received by station A and then trasmitted to station B. The probability of each station receiving the signal correctly is $3 / 4$. If the signal received at station B is given, then the probability that the original signal is green, is

• A

  $\frac{3}{5}$
• B

  $\frac{6}{7}$
• Correct answer
  C

  $\frac{20}{23}$
• D

  $\frac{9}{20}$

Solution

From the tree diagram, it follows that
$\mathrm{P}\left(\mathrm{B}_{\mathrm{G}}\right)=\frac{46}{80}, \mathrm{P}(\mathrm{G})=\frac{4}{5}$ $\mathrm{P}\left(\mathrm{B}_{\mathrm{G}} / \mathrm{G}\right)=\frac{10}{16}=\frac{5}{8}$ $\therefore \mathrm{P}\left(\mathrm{B}_{\mathrm{G}} \cap \mathrm{G}\right)=\frac{5}{8} \times \frac{4}{5}=\frac{1}{2}$ $\left[\because \quad \mathrm{P}\left(\mathrm{B}_{\mathrm{G}} \cap \mathrm{G}\right)\right]$ $\left.=\mathrm{P}\left(\frac{\mathrm{B}_{\mathrm{G}}}{\mathrm{G}}\right) \times \mathrm{P}(\mathrm{G})\right]$ Now, $\mathrm{P}\left(\mathrm{G} / \mathrm{B}_{\mathrm{G}}\right)$ $=\frac{\mathrm{P}\left(\mathrm{B}_{\mathrm{G}} \cap \mathrm{G}\right)}{\mathrm{P}\left(\mathrm{B}_{\mathrm{G}}\right)}=\frac{1}{2} \times \frac{80}{40}=\frac{20}{23}$

Hence, the correct answer is option: 'C'.