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  4. A signal which can be green or red with probability 4 / 5 and 1 / 5 respectively, is received by station A and then transmitted 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

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Q. A signal which can be green or red with probability $4 / 5$ and $1 / 5$ respectively, is received by station $A$ and then transmitted 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

Probability - Part 2

Solution:

From the tree diagram, it follows that
image
$\therefore P\left(B_G \cap G\right)=\frac{5}{8} \times \frac{4}{5}=\frac{1}{2}\left[\because P\left(B_G \cap G\right) =P\left(\frac{B_G}{G}\right) \times P(G)\right]$
$ \text { Now, } P\left(G / B_G\right)=\frac{P\left(B_G \cap G\right)}{P\left(B_G\right)}=\frac{1}{2} \times \frac{80}{46}=\frac{20}{23}$