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

The correct option is C

$\frac{20}{23}$

Explanation for the correct option:

Step 1. Let G is the event that original signal is green

E1 be the event that A receives correct signal

E2 be the event that B receives correct signal

Step 2. Let E be the event in which signal received by B is green

Therefore, Probability of the received signal by B is green

$=P\left(G{E}_1{E}_2\right)+P\left(G{\overline{E}}_1{\overline{E}}_2\right)+P\left(\overline{G}{E}_1{\overline{E}}_2\right)+P\left(\overline{G}{\overline{E}}_1{E}_2\right)$

$=\left(\frac{4}{5}\times \frac{3}{4}\times \frac{3}{4}\right)+\left(\frac{4}{5}\times \frac{1}{4}\times \frac{1}{4}\right)+\left(\frac{1}{5}\times \frac{3}{4}\times \frac{1}{4}\right)+\left(\frac{1}{5}\times \frac{1}{4}\times \frac{3}{4}\right)$

$\Rightarrow$ $P\left(E\right)=\frac{46}{80}$

Step 3. the required probability:

$\Rightarrow P\left(\frac{G}{E}\right)=\frac{\left({\displaystyle \frac{4}{5}}\times {\displaystyle \frac{3}{4}}\times {\displaystyle \frac{3}{4}}\right)+\left({\displaystyle \frac{4}{5}}\times {\displaystyle \frac{1}{4}}\times {\displaystyle \frac{1}{4}}\right)}{{\displaystyle \frac{46}{80}}}$

$\Rightarrow P\left(\frac{G}{E}\right)=\frac{\mathbf{20}}{\mathbf{23}}$

Hence, Option ‘C’ is Correct.