Question

Bag $A$ contains $2$ white and $3$ red balls and bag $B$ contains $4$ white and $5$ red balls. One ball is drawn at random from one of the bags and is found to be red. Then the probability that it was

• A. From bag $B$ is $\frac{25}{52}$
• B. From bag $A$ is $\frac{27}{52}$
• C. From bag $B$ is $\frac{27}{52}$
• D. From bag $A$ is $\frac{25}{52}$

Answer 1

Answer: (A), (B)

From bag $A$ is $\frac{27}{52}$

Let $E_1$,$E_2$ be the events of selecting bag $A$, bag $B$ respectively.
Let '$E$' be the event of drawing red ball.
$P\left(E_1\right)=\frac{1}{2},P\left(E_2\right)=\frac{1}{2}$
$P\left(\frac{E}{E_1}\right)=\frac{3}{5}\text{and}P\left(\frac{E}{E_2}\right)=\frac{5}{9}$
$P\left(E_1/E\right)=\left[\frac{P\left(E_1\right).P\left(E/E_1\right)}{P\left(E_1\right).P\left(E/E_1\right)+P\left(E_2\right).P\left(E/E_2\right)}\right]=\frac{\frac{1}{2}\left(\frac{3}{5}\right)}{\frac{1}{2}(\frac{3}{5}+\frac{5}{9})}=\left(\frac{3\times 9}{27+25}\right)=\frac{27}{52}$
Similarly,$P\left(E_2/E\right)=\frac{\frac{1}{2}\left(\frac{5}{9}\right)}{\frac{1}{2}(\frac{3}{5}+\frac{5}{9})}=\left(\frac{5\times 5}{27+25}\right)=\frac{25}{52}$