A signal which can be green or red with probability $\frac{4}{5}$ and $\frac{1}{5}$ respectively, is…
- $\frac{3}{5}$
- $\frac{6}{7}$
- $\frac{20}{23}$
- $\frac{9}{20}$
Solution

$ \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} $
Asked in: JEE Advanced 2010 (Paper 2)