Question

Bag $I$ contains $2$ blacks and $8$ red balls, bag $II$contains $7$ black and $3$ red balls and bag $III$ contains $5$ black and $5v$ red balls. One bage is chosen at random and a ball is drawn from it which is found to bered. Find the probability that the ball is drawn from bag $II$

Answer 1

Let $R$ be the event of drawing a red ball.

Let $E_1,E_2$ and $E_3$ be the event that bag $I,II$ or $III$ is chosen.

Therefore,

$P\left(E_1\right)=\frac{1}{3},P\left(E_2\right)=\frac{1}{3},P\left(E_3\right)=\frac{1}{3}$

$P\left(R|E_1\right)=\frac{8}{10}$

$P\left(R|E_2\right)=\frac{3}{10}$

$P\left(R|E_3\right)=\frac{5}{10}$

Now,

Probability that the red ball drawn is from bag $II$-

$P\left(E_2|R\right)=\frac{P\left(E_2\right)\cdot P\left(R|E_2\right)}{P\left(E_1\right)\cdot P\left(R|E_1\right)+P\left(E_2\right)\cdot P\left(R|E_2\right)+P\left(E_3\right)\cdot P\left(R|E_3\right)}$

$\Rightarrow P\left(E_2|R\right)=\frac{\frac{1}{3}\times \frac{3}{10}}{\frac{1}{3}\times \frac{8}{10}+\frac{1}{3}\times \frac{3}{10}+\frac{1}{3}\times \frac{5}{10}}=\frac{3}{8+3+5}=\frac{3}{16}$

Hence probability that the ball is drawn from bag $II$ is $\frac{3}{16}$.