There are 3 bags A, B and C. Bag A contains 2 white and 3 black balls, bag B contains 4 white and 2 black balls and Bag C contains 3 white and 2 black balls. If a ball is drawn at random from a randomly chosen bag, then the probability that the ball drawn is black is

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$\frac{2}{3}$

$\frac{1}{9}$

$\frac{4}{9}$

$\frac{5}{9}$

answer is B.

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Detailed Solution

	A	B	C
White	2	4	3
Black	3	2	2

$E_{1} =$ Selected Bag A $P (E_{1}) = \frac{1}{3}$

$E_{2} =$ Selected Bag B $P (E_{2}) = \frac{1}{3}$

$E_{3} =$ Selected Bag C $P (E_{3}) = \frac{1}{3}$

A=a ball in selected from selected it is black

$\begin{array}{rcl} P (\frac{A}{E_{1}}) & = & \frac{3}{5} P (\frac{A}{E_{2}}) = \frac{2}{6} P (\frac{A}{E_{3}}) = \frac{2}{5} \\ P (A) & = & \frac{1}{3} \times \frac{3}{5} + \frac{1}{3} \times \frac{2}{6} + \frac{1}{3} \times \frac{2}{5} \\ = & \frac{1}{5} + \frac{1}{9} + \frac{2}{15} = \frac{5}{15} + \frac{1}{9} = \frac{3 + 1}{9} \\ = & \frac{4}{9} \end{array}$