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Suppose that on urn $B_{1}$ contains 2 white and 3 black balls and another urn $B_{2}$ contains 3 white and 4 black balls one urn is selected at random and a ball is selected at random and a ball is drawn from it. If a ball is drawn is found black. Find the probability that the urn choosen was $B_{1}$ .

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Detailed Solution

Given

Let $B_{1}$ & $B_{2}$ are the probability of selecting
Box $- I & I I$

$P (B_{1}) = 1 / 2; P (B_{2}) = 1 / 2$

E be the event of selecting balck ball

$P (E / B_{1}) = \frac{3}{5}$ $P (E / B_{2}) = \frac{4}{7}$

$P (B_{1} / E) = \frac{P (E / B_{1}) P (B_{1})}{P (E / B_{1}) P (B_{1}) + P (E / B_{2}) P (B_{2})}$

$= \frac{(\frac{3}{5}) (\frac{1}{2})}{(\frac{3}{5}) (\frac{1}{2}) + (\frac{4}{7}) (\frac{1}{2})} = \frac{\frac{3}{5} \frac{1}{2}}{(\frac{3}{5} + \frac{4}{7}) \frac{1}{2}}$

$= \frac{\frac{3}{5}}{\frac{21 + 20}{35}} = \frac{3}{5} \times \frac{\overset{7}{35}}{41} = \frac{21}{41}$

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