The contents of three urns are 1 white, 2 red, 3 green balls; 2 white, 1 red and 1 green balls; 4 white, 5 red and 3 green balls. Two ball 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 a b , where a and b are co-prime positive integers, find (a+b) .

Step-by-step solution

The probability of drawing one white bails and one green ball from the first urn is 1 5 . The probability of drawing one white ball and one green ball from the second urn is 1 3 . The probability of drawing one white ball and one green ball from the third urn is 2 11 . Therefore, the probability that the third urn was chosen is 2 11 1 5 + 1 3 + 2 11 = 15 59 = a b Hence 15+59=74