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The contents of three urns are 1white, 2 red, 3 green balls; 2white, 1red and 1 green balls; 4 white, 5 red and 3 green balls. Two balls are drawn from an urn chosen at random and are found to be one white and one green. If the probability that the balls so drawn came from the third urn can be expressed as $\frac{a}{b},$ where a and b are co-prime positive integers, then the value of $(a + b) =$

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answer is 74.

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Detailed Solution

The probability of drawing one white balls and one green ball from the first urn is $\frac{1}{5}$

The probability of drawing one white ball and one green ball from the second urn is $\frac{1}{3}$

The probability of drawing one white ball and one green ball from the third urn is $\frac{2}{11},$

Therefore, the probability that the third urn was chosen is $\frac{\frac{2}{11}}{\frac{1}{5} + \frac{1}{3} + \frac{2}{11}} = \frac{15}{59} = \frac{a}{b}$

Hence $15 + 59 = 74$

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