Question

A bag X contain 4 white balls and 2 black balls, while another bag Y contain 3 white balls and 3 black balls. Two balls are drawn (without replacement) at random from one of the bags and were found to be one white and one black. Find the probability that the balls were drawn from bag Y.

Answer 1

Hint: In this question we have given two bags from which balls are drawn. So first find the probability that balls are drawn from any bag. since choosing of bags is equiprobable so $p({{E}_{1}})=\dfrac{1}{2}$,$p({{E}_{2}})=\dfrac{1}{2}$.
After that find the conditional probability that balls are drawn from bags X and Y that is $p\left( {E}/{{{E}_{1}}}\; \right)$ and $p\left( {E}/{{{E}_{2}}}\; \right)$. After that apply Bayes’ theorem in order to find the required probability.

Complete step by step answer:
Let ${{E}_{1}}$= Event of choosing bag X
       ${{E}_{2}}$= Event of choosing bag Y
        $E$= Event of drawing of two balls (without replacement) one white and one black
Then $p({{E}_{1}})=\dfrac{1}{2}$ and $p({{E}_{2}})=\dfrac{1}{2}$
Now $p\left( {E}/{{{E}_{1}}}\; \right)$= Event of drawing one white and one black from bag X
As drawing of second ball is dependent on the drawing of the first ball, hence the events of drawing balls are dependent.
Here it should be clear that sequence of drawing ball is either white then black or first black then white. As ball is drawn without replacement so sample space is changed for second drawing so, we can write
\[\begin{align}
  & \left( {E}/{{{E}_{1}}}\; \right)=wb\text{ or bw} \\
 & p\left( {E}/{{{E}_{1}}}\; \right)=\left( \dfrac{4}{6} \right)\left( \dfrac{2}{5} \right)+\left( \dfrac{2}{6} \right)\left( \dfrac{4}{5} \right) \\
 & p\left( {E}/{{{E}_{1}}}\; \right)=\dfrac{16}{30} \\
\end{align}\]
Similarly, we can write
$\begin{align}
  & \left( {E}/{{{E}_{2}}}\; \right)=wb\text{ or }bw \\
 & p\left( {E}/{{{E}_{2}}}\; \right)=\left( \dfrac{3}{6} \right)\left( \dfrac{3}{5} \right)+\left( \dfrac{3}{6} \right)\left( \dfrac{3}{5} \right) \\
 & p\left( {E}/{{{E}_{2}}}\; \right)=\dfrac{18}{30} \\
\end{align}$
Now we have to find the probability that balls are drawn from bag Y that is $P\left( \dfrac{{{E}_{2}}}{E} \right)$. Now from Bayes’ theorem we have
\[\begin{align}
  & P\left( \dfrac{{{E}_{2}}}{E} \right)=\dfrac{p({{E}_{2}})P({E}/{{{E}_{2}})}\;}{P({{E}_{1}})P({E}/{{{E}_{1}})+P({{E}_{2}})P({E}/{{{E}_{2}})}\;}\;} \\
 & P\left( \dfrac{{{E}_{2}}}{E} \right)=\dfrac{\left( \dfrac{1}{2} \right)\left( \dfrac{18}{30} \right)}{\left( \dfrac{1}{2} \right)\left( \dfrac{16}{30} \right)+\left( \dfrac{1}{2} \right)\left( \dfrac{18}{30} \right)} \\
\end{align}\]
Upon simplification we can write
\[\begin{align}
  & P\left( \dfrac{{{E}_{2}}}{E} \right)=\dfrac{18}{16+18} \\
 & P\left( \dfrac{{{E}_{2}}}{E} \right)=\dfrac{18}{34} \\
 & P\left( \dfrac{{{E}_{2}}}{E} \right)=\dfrac{9}{17} \\
\end{align}\]
Hence probability that balls are drawn from bag Y is $\dfrac{9}{16}$.

Note:
It is important to note that if ${{E}_{1}},{{E}_{2}},{{E}_{3}}....{{E}_{n}}$be mutually exclusive and exhaustive event associated with a random experiment , and let $E$be any event that occurs with some ${{E}_{i}}$Then
\[P\left( \dfrac{{{E}_{i}}}{E} \right)=\dfrac{p({{E}_{i}})P({E}/{{{E}_{i}})}\;}{\sum{p({E}/{{{E}_{i}})p({{E}_{i}})}\;}}\]
And $p(E)=\sum\limits_{i=1}^{n}{p\left( {E}/{{{E}_{i}}}\; \right)}p({{E}_{i}})$
The first formula is for bayes’ theorem and the second one is for total probability.