There are 3 bags, each containing 5 white and 3 black balls. Also, there are 2 bags, each containing 2 white and 4 black balls. A white ball is drawn at random. Find the probability that this ball is from a bag of the first group.

Home » There are 3 bags, each containing 5 white and 3 black balls. Also, there are 2 bags, each containing 2 white and 4 black balls. A white ball is drawn at random. Find the probability that this ball is from a bag of the first group.

There are 3 bags, each containing 5 white and 3 black balls. Also, there are 2 bags, each containing 2 white and 4 black balls. A white ball is drawn at random. Find the probability that this ball is from a bag of the first group.

Let $A$ : the set of first 3 bags
$B$ : a set of next 2 bags
WB : White ball
BB : Black ball
Now we can change the problem to two bags, i.e. bag A containing 15 white and 9 black balls( 5 white and 3 black in each bag) and bag $B$ containing 4 white and 8 black balls( 2 white and 4 black balls in each bag)
Probability of selecting bag $A$ is $\frac{3}{5}$ ( 3 bags are in $A$ ) and selecting $B$ is $\frac{2}{5}(2$ bags are in $B)$
We want to find the probability of selected white ball is from bag A
$\begin{array}{l} \mathrm{P}(\mathrm{A} \mid \mathrm{WB})=\frac{\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{WB} \mid \mathrm{A})}{\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{WB} \mid \mathrm{A})+\mathrm{P}(\mathrm{B}) \cdot \mathrm{P}(\mathrm{WB} \mid \mathrm{B})} \\ =\frac{\left(\frac{3}{5}\right)\left(\frac{15}{24}\right)}{\left(\frac{3}{5}\right)\left(\frac{15}{24}\right)+\left(\frac{2}{5}\right)\left(\frac{4}{12}\right)} \\ =\frac{45}{61} \end{array}$
Therefore, the probability of selected white ball is from the first group is $\frac{45}{61}$