Question

Let `U_1` , and `U_2`, be two urns such that `U_1`, contains `3` white and `2` red balls, and `U_2,`contains only`1` white ball. A fair coin is tossed. If head appears then `1` ball is drawn at random from `U_1`, and put into `U_2,` . However, if tail appears then `2` balls are drawn at random from `U_1,` and put into `U_2`. . Now `1` ball is drawn at random from `U_2,` .61 . The probability of the drawn ball from `U_2,` being white is A. `(13)/(30)` B. `(23)/(30)` C. `(19)/(30)` D. `(11)/(30)` Select the correct answer from above options

Answer 1

Correct Answer - B Now, probability of the drawn ball from `U_(2)` being white is `P("white"//U_(2))=P(H)*{(""^(2)C_(1))/(""^(5)C_(1))xx(""^(2)C_(1))/(""^(2)C_(1))+(""^(2)C_(1))/(""^(5)C_(1))xx(""^(1)C_(1))/(""^(2)C_(1))}` `+P(T){(""^(3)C_(2))/(""^(5)C_(2))xx(""^(3)C_(2))/(""^(3)C_(2))+(""^(2)C_(2))/(""^(5)C_(2))xx(""^(1)C_(1))/(""^(3)C_(2))+(""^(3)C_(1)*""^(2)C_(1))/(""^(5)C_(2))xx(""^(2)C_(1))/(""^(3)C_(2))}` `=(1)/(2){(3)/(5)xx1+(2)/(5)xx(1)/(2)}` `+(1)/(2){(3)/(10)xx1+(1)/(10)xx(1)/(3)+(6)/(10)xx(2)/(3)}=(23)/(30)`