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There are 3 bags which are known to contain 2 white and 3 black balls; 4 white and 1 black balls and 3 white and 7 black balls respectively. A ball is drawn at random from one of the bags and found to be a black ball. Then the probability that it was drawn from the bag containing most black balls is
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The correct answer is:
$\frac{3}{5}$
Consider the following events :
$A \rightarrow$ Ball drawn is black; $E_1 \rightarrow$ Bag I is chosen;
$E_2 \rightarrow$ Bag II is chosen and $E_3 \rightarrow$ Bag III is chosen.
Then, $P\left(E_1\right)=\left(E_2\right)=P\left(E_3\right)=\frac{1}{3}, P\left(\frac{A}{E_1}\right)=\frac{3}{5}$.
$A \rightarrow$ Ball drawn is black; $E_1 \rightarrow$ Bag I is chosen;
$E_2 \rightarrow$ Bag II is chosen and $E_3 \rightarrow$ Bag III is chosen.
Then, $P\left(E_1\right)=\left(E_2\right)=P\left(E_3\right)=\frac{1}{3}, P\left(\frac{A}{E_1}\right)=\frac{3}{5}$.
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