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A signal which can be green or red with probability $4 / 5$ and $1 / 5$ respectively, is received by station A and then trasmitted 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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The correct answer is:
$\frac{20}{23}$
From the tree diagram, it follows that
$\mathrm{P}\left(\mathrm{B}_{\mathrm{G}}\right)=\frac{46}{80}, \mathrm{P}(\mathrm{G})=\frac{4}{5}$
$\mathrm{P}\left(\mathrm{B}_{\mathrm{G}} / \mathrm{G}\right)=\frac{10}{16}=\frac{5}{8}$
$\therefore \mathrm{P}\left(\mathrm{B}_{\mathrm{G}} \cap \mathrm{G}\right)=\frac{5}{8} \times \frac{4}{5}=\frac{1}{2}$
$\left[\because \quad \mathrm{P}\left(\mathrm{B}_{\mathrm{G}} \cap \mathrm{G}\right)\right]$
$\left.=\mathrm{P}\left(\frac{\mathrm{B}_{\mathrm{G}}}{\mathrm{G}}\right) \times \mathrm{P}(\mathrm{G})\right]$
Now, $\mathrm{P}\left(\mathrm{G} / \mathrm{B}_{\mathrm{G}}\right)$
$=\frac{\mathrm{P}\left(\mathrm{B}_{\mathrm{G}} \cap \mathrm{G}\right)}{\mathrm{P}\left(\mathrm{B}_{\mathrm{G}}\right)}=\frac{1}{2} \times \frac{80}{40}=\frac{20}{23}$
$\mathrm{P}\left(\mathrm{B}_{\mathrm{G}}\right)=\frac{46}{80}, \mathrm{P}(\mathrm{G})=\frac{4}{5}$
$\mathrm{P}\left(\mathrm{B}_{\mathrm{G}} / \mathrm{G}\right)=\frac{10}{16}=\frac{5}{8}$
$\therefore \mathrm{P}\left(\mathrm{B}_{\mathrm{G}} \cap \mathrm{G}\right)=\frac{5}{8} \times \frac{4}{5}=\frac{1}{2}$
$\left[\because \quad \mathrm{P}\left(\mathrm{B}_{\mathrm{G}} \cap \mathrm{G}\right)\right]$
$\left.=\mathrm{P}\left(\frac{\mathrm{B}_{\mathrm{G}}}{\mathrm{G}}\right) \times \mathrm{P}(\mathrm{G})\right]$
Now, $\mathrm{P}\left(\mathrm{G} / \mathrm{B}_{\mathrm{G}}\right)$
$=\frac{\mathrm{P}\left(\mathrm{B}_{\mathrm{G}} \cap \mathrm{G}\right)}{\mathrm{P}\left(\mathrm{B}_{\mathrm{G}}\right)}=\frac{1}{2} \times \frac{80}{40}=\frac{20}{23}$
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