Probability Ques 114
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
(2010)
(a) $\frac{3}{5}$
(b) $\frac{6}{7}$
(c) $\frac{20}{23}$
(d) $\frac{9}{20}$
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Answer:
Correct Answer: 114.(c)
Solution:
Formula:
- From the tree diagram, it follows that
$P\left(B _G\right)=\frac{46}{80}$
$P\left(B _G \mid G\right)=\frac{10}{16}=\frac{5}{8}$
$P\left(B _G \cap G\right)=\frac{5}{8} \times \frac{4}{5}=\frac{1}{2}$
$\therefore \quad P\left(G \mid B _G\right)=\frac{\frac{1}{2}}{P\left(B _G\right)}=\frac{1}{2} \times \frac{80}{46}=\frac{20}{23}$
BETA
