Bag A contains 2 white and 3 red balls and bag B contains 4 white and 5 red balls. If one ball is drawn at…
Bag A contains 2 white and 3 red balls and bag B contains 4 white and 5 red balls. If one ball is drawn at random from one of the bags and is found to be red, then the probability that it was drawn from the bag B is
$\frac{23}{54}$
$\frac{25}{51}$
$\frac{25}{52}$
$\frac{27}{55}$
Solution
$E_1 \rightarrow \mathrm{~A}$ ball is drawn from bag $A$
$E_2 \rightarrow \mathrm{~A}$ ball is drawn from bag $B$
$F \rightarrow$ A ball is found to be red
$\begin{aligned}
& \therefore P\left(E_1\right)=\frac{1}{2}, P\left(E_2\right)=\frac{1}{2}, P\left(\frac{F}{E_1}\right)=\frac{3}{5}, P\left(\frac{F}{E_2}\right)=\frac{5}{9} \\
& \therefore P\left(\frac{E_2}{F}\right)=\frac{P\left(\frac{F}{E_2}\right) \cdot P\left(E_2\right)}{P\left(\frac{F}{E_1}\right) \cdot P\left(E_1\right)+P\left(\frac{F}{E_2}\right) \cdot P\left(E_2\right)} \\
& =\frac{\frac{5}{9} \times \frac{1}{2}}{\frac{3}{5} \times \frac{1}{2}+\frac{5}{9} \times \frac{1}{2}}=\frac{\frac{5}{9}}{\frac{3}{5}+\frac{5}{9}}=\frac{25}{52} .
\end{aligned}$