Question

A Bag I contains 5 red and 4 white balls and a Bag II contains 3 red and 3 white balls. Two balls are transferred from the Bag I to the Bag II and then one ball is drawn from the Bag II. If the ball drawn from the Bag II is red, then find the probability that one red ball and one white ball are transferred from Bag I to the Bag II.

Answer 1

Let $E_1:2$ white balls are transferred from Bag I to the Bag II, $E_2:2$ red balls are transferred from Bag I to the Bag II, $E_3:1$ red and 1 white balls are transferred from Bag I to the Bag II, E : the ball drawn from Bag II is red.
$\therefore P\left(E_1\right)=\frac{{}^4{C}_2}{{}^9{C}_2}=\frac{12}{72},P\left(E_2\right)=\frac{{}^5{C}_2}{{}^9{C}_2}=\frac{20}{72},\frac{{}^5{C}_1{\times }^4{C}_1}{{}^9{C}_2}=\frac{40}{72}\text{and},P\left(E|E_1\right)=\frac{3}{8},P\left(E|E_2\right)=\frac{5}{8},P\left(E|E_3\right)=\frac{4}{8}\text{By Baye's Theorem,}P\left(E_3|E\right)=\frac{P\left(E_3\right)P\left(E|E_3\right)}{P\left(E_1\right)P\left(E|E_1\right)+P\left(E_2\right)P\left(E|E_2\right)+P\left(E_3\right)P\left(E|E_3\right)}\Rightarrow =\frac{\frac{40}{72}\times \frac{4}{8}}{\frac{12}{72}\times \frac{3}{8}+\frac{20}{72}\times \frac{5}{8}+\frac{40}{72}\times \frac{4}{8}}=\frac{40}{9+25+40}=\frac{20}{37}$