Question

There are two bags $I$ and $II$. Bag $I$ contains 3 white and 4 black balls and Bag $ll$ contains 5 white and 6 black balls. One ball is drawn at random from one of the bags and is found to be white. Find the probability that it was drawn from Bag $I$.

Answer 1

Let $E_1$ be the event of choosing the bag $I$ and $E_2$ be the event choosing the bag $II$ and $A$ be the event of drawing a white ball.

$=P\left(\frac{A}{E_1}\right)=\frac{3}{7}$ $P\left(E_1\right)=P\left(E_2\right)=\frac{1}{2}$

$=P\left(\frac{A}{E_2}\right)=\frac{5}{11}$

By Baye's theorem, the probability of drawing a white ball from bag $I$

$=P\left(\frac{E_1}{A}\right)=\frac{P\left(E_1\right)P\left(\frac{A}{E_1}\right)}{P\left(E_1\right)P\left(\frac{A}{E_1}\right)+P\left(E_2\right)P\left(\frac{A}{E_2}\right)}$

$=\frac{\frac{1}{2}\times \left(\frac{3}{7}\right)}{\frac{1}{2}\left(\frac{3}{7}\right)+\frac{1}{2}\left(\frac{5}{11}\right)}$

$=\frac{\frac{1}{2}\times \frac{3}{7}}{\frac{11\times 3+5\times 7}{2\times 7\times 11}}=\frac{3}{\frac{33+35}{11}}$

$=P\left(\frac{E_1}{A}\right)=\frac{33}{68}$