$$y = {\left( {1 - x} \right)^2},y = 0,$$ and $$x=0$$ into two parts $${R_1}\left( {0 \le x \le b} \right)$$ and
$${R_2}\left( {b \le x \le 1} \right)$$ such that $${R_1} - {R_2} = {1 \over 4}.$$ Then $$b$$ equals
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 $${U_2}$$ being white is
The probability of the drawn ball from $${U_2}$$ being white is
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 $${U_2}$$ being white is
Given that the drawn ball from $${U_2}$$ is white, the probability that head appeared on the coin is