Question

There are 3 boxes each containing 3 red and 5 green balls. Also, there are 2 boxes each containing 4 red and 2 green balls. A green ball is selected at random. Find the probability that this green ball is from a box of the first group.

• A. $\frac{54}{61}$
• B. $\frac{45}{61}$
• C. $\frac{8}{31}$
• D. None of these

Answer 1

The correct option is B $\frac{45}{61}$

Let $E_1,E_2$ and A be the events defined as follows:
$E_1$=selecting a box from the first group
$E_2$=selecting a box from the second group and
A=ball drawn is green
Since there are 5 boxes out of which 3 boxes belong to the first group and 2 boxes belong to the second group. Therefore,
$P\left(E_1\right)=\frac{3}{5},P\left(E_2\right)=\frac{2}{5}$
If $E_1$ has already occurred, then a box from the first group is chosen. The box chosen contains 5 green balls and 3 red balls.
Therefore the probability of drawing a green ball from it is $\frac{5}{8}$.
So, $P\left(\frac{A}{E_1}\right)=\frac{5}{8}$
Similarly, $P\left(\frac{A}{E_2}\right)=\frac{2}{6}=\frac{1}{3}$
Now, we have to find $P\left(\frac{E_1}{A}\right)$
By Baye's rule, we have
$P\left(\frac{E_1}{A}\right)=\frac{P\left(E_1\right)P\left(\frac{A}{E_1}\right)}{P\left(E_1\right)P\left(\frac{A}{E_1}\right)+P\left(E_2\right)P\left(\frac{A}{E_2}\right)}$
$=\frac{\frac{3}{5}\times \frac{5}{8}}{\frac{3}{5}\times \frac{5}{8}+\frac{2}{5}\times \frac{1}{3}}=\frac{45}{61}$