JEE Advanced · Mathematics · 32. Probability
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 & Solution
Correct Answer
(C)
\(\frac{20}{23}\)
Step-by-step Solution
Detailed explanation
From the tree-diagram it follows that

\[
\begin{gathered}
P\left(B_G \mid G\right)=\frac{10}{16}=\frac{5}{8} \\
\therefore \quad P\left(B_G \cap G\right)=\frac{5}{8} \times \frac{4}{5}=\frac{1}{2} \\
P\left(G \mid B_G\right)=\frac{\frac{1}{2}}{P\left(B_G\right)}=\frac{1}{2} \times \frac{80}{46}=\frac{20}{23}
\end{gathered}
\]

\[
\begin{gathered}
P\left(B_G \mid G\right)=\frac{10}{16}=\frac{5}{8} \\
\therefore \quad P\left(B_G \cap G\right)=\frac{5}{8} \times \frac{4}{5}=\frac{1}{2} \\
P\left(G \mid B_G\right)=\frac{\frac{1}{2}}{P\left(B_G\right)}=\frac{1}{2} \times \frac{80}{46}=\frac{20}{23}
\end{gathered}
\]
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