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` . The probability of the drawn ball from `U_2` being white is

Options

A. `(13)/(30)`
B. `(23)/(30)`
C. `(19)/(30)`
D. `(11)/(30)`

Answer

B

Explanation

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)`