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A bag contains $\mathbf{4}$ red and $\mathbf{4}$ black balls, another bag contains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag.
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Let $\mathrm{A}$ be the event that ball drawn is red and let $\mathrm{E}_1$ and $\mathrm{E}_2$ be the events that the ball drawn is from the first bag and second bag respectively. $P\left(E_1\right)=\frac{1}{2}, P\left(E_2\right)=\frac{1}{2}$. $\mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_1\right)=$ Probability of drawing a red ball from bag $\mathrm{I}=\frac{4}{8}=\frac{1}{2}$
$P\left(A \mid E_2\right)=$ Probability of drawing a red ball from bag II $=\frac{2}{8}=\frac{1}{4}$
Therefore by Bayes' theorem $\mathrm{P}\left(\mathrm{E}_1 \mid \mathrm{A}\right)=$ Probability that the red ball drawn is from bag $\mathrm{I}$
$$
\begin{aligned}
&=\frac{P\left(E_1\right) P\left(A \mid E_1\right)}{P\left(E_1\right) P\left(A \mid E_1\right)+P\left(E_2\right) P\left(A \mid E_2\right)} \\
&=\frac{\frac{1}{2} \times \frac{1}{2}}{\frac{1}{2} \times \frac{1}{2}+\frac{1}{2} \times \frac{1}{4}}=\frac{2}{3}
\end{aligned}
$$
$P\left(A \mid E_2\right)=$ Probability of drawing a red ball from bag II $=\frac{2}{8}=\frac{1}{4}$
Therefore by Bayes' theorem $\mathrm{P}\left(\mathrm{E}_1 \mid \mathrm{A}\right)=$ Probability that the red ball drawn is from bag $\mathrm{I}$
$$
\begin{aligned}
&=\frac{P\left(E_1\right) P\left(A \mid E_1\right)}{P\left(E_1\right) P\left(A \mid E_1\right)+P\left(E_2\right) P\left(A \mid E_2\right)} \\
&=\frac{\frac{1}{2} \times \frac{1}{2}}{\frac{1}{2} \times \frac{1}{2}+\frac{1}{2} \times \frac{1}{4}}=\frac{2}{3}
\end{aligned}
$$
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