Question

Bag $I$ contains $3$ white and $3$ red balls and bag $II$ contains $5$ white and $4$ red balls. $A$ bag is taken at random and a ball is drawn from it. If the ball drawn is white, then find the probability that it was drawn from bag $II$.

• A. $\frac{11}{19}$
• B. $\frac{14}{19}$
• C. $\frac{12}{19}$
• D. $\frac{10}{19}$

Answer 1

Answer: (D)

$\frac{10}{19}$

Answer 2

Let $E_1$, $E_2$ and $A$ be the events defined as follows : $E_1 =$ bag $I$ is taken $E_2 =$ bag $II$ is taken $A =$ ball drawn is white Then, $P(E_1) = \frac{1}{2} = P(E_2)$ $P(A|E_1) = P$(drawing a white ball from bag $I$) $= \frac{3}{6} = \frac{1}{2}$ $P(A|E_2) = P$(drawing a white ball from bag $II$) $= \frac{5}{9}$ We want to find $P(E_2|A)$ By using Bayes' theorem, we have $P\left(E_{2}|A\right) = \frac{P\left(E_{2}\right)P\left(A |E_{2}\right)}{P \left(E_{1}\right)P\left(A|E_{1}\right) + P\left(E_{2}\right)P\left(A|E_{2}\right)}$ $= \frac{\frac{1}{2}\cdot\frac{5}{9}}{\frac{1}{2}\cdot\frac{5}{9}+\frac{1}{2}\cdot\frac{1}{2}} = \frac{\frac{5}{18}}{\frac{5}{18}+\frac{1}{4}}$ $= \frac{5}{18}\times\frac{36}{19} = \frac{10}{19}$