Question

There are two bags , bag I and bag II . Bag I contains 4 white and 3 red balls while another bag II contains 3 white and 7 red balls . One ball is drawn at random from one of the bags and it is found to be white . Find the probability that it was drawn from Bag I .

Answer 1

Let us consider A to be the event that the ball drawn is white, ${{E}_{1}}$ to be the event that it is drawn from bag II and ${{E}_{2}}$ to be the event that it is drawn from bag II. We have to find $P\left( {{E}_{1}}/A \right)$ . We know that according to Bayes theorem, $P\left( {{E}_{i}}/A \right)=\dfrac{P\left( {{E}_{i}} \right)P\left( A/{{E}_{i}} \right)}{\sum\limits_{k=1}^{n}{P\left( {{E}_{k}} \right)P\left( A/{{E}_{k}} \right)}}$ . Thus, $P\left( {{E}_{1}}/A \right)=\dfrac{P\left( {{E}_{1}} \right)P\left( A/{{E}_{1}} \right)}{P\left( {{E}_{1}} \right)P\left( A/{{E}_{1}} \right)+P\left( {{E}_{2}} \right)P\left( A/{{E}_{2}} \right)}$ , where $P\left( {{E}_{1}} \right)$ is the probability that bag I is chosen, $P\left( {{E}_{2}} \right)$ is the probability that bag II is chosen, $P\left( A/{{E}_{1}} \right)$ is the probability that of white ball in bag I and $P\left( A/{{E}_{2}} \right)$ is the probability of white balls in bag II.

Complete step-by-step answer:
We need to find the probability that white ball is drawn from bag I out of two bags. Let us consider A to be the event that the ball drawn is white, ${{E}_{1}}$ to be the event that it is drawn from bag II and ${{E}_{2}}$ to be the event that it is drawn from bag II.
We have to find $P\left( {{E}_{1}}/A \right)$ , that is, probability that white ball is drawn from bag I out of two bags.
We know that according to Bayes theorem, $P\left( {{E}_{i}}/A \right)=\dfrac{P\left( {{E}_{i}} \right)P\left( A/{{E}_{i}} \right)}{\sum\limits_{k=1}^{n}{P\left( {{E}_{k}} \right)P\left( A/{{E}_{k}} \right)}}$ .
$\Rightarrow P\left( {{E}_{1}}/A \right)=\dfrac{P\left( {{E}_{1}} \right)P\left( A/{{E}_{1}} \right)}{\sum\limits_{k=1}^{2}{P\left( {{E}_{k}} \right)P\left( A/{{E}_{k}} \right)}}$
Let us expand the summation. We will get
$\Rightarrow P\left( {{E}_{1}}/A \right)=\dfrac{P\left( {{E}_{1}} \right)P\left( A/{{E}_{1}} \right)}{P\left( {{E}_{1}} \right)P\left( A/{{E}_{1}} \right)+P\left( {{E}_{2}} \right)P\left( A/{{E}_{2}} \right)}...\left( i \right)$
We have to first find $P\left( {{E}_{1}} \right),P\left( A/{{E}_{1}} \right),P\left( {{E}_{2}} \right)\text{ and }P\left( A/{{E}_{2}} \right)$ .
We know that $\text{Probability of an event }=\dfrac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$
We are given that there are two bags. Hence, the probability that bag I is taken is given by
$P\left( {{E}_{1}} \right)=\dfrac{1}{2}$
We will get the probability that bag II is taken in similar way.
$P\left( {{E}_{2}} \right)=\dfrac{1}{2}$
Now, we have to find $P\left( A/{{E}_{1}} \right)$ which is the probability that white balls in bag I. We are given that there are 4 white balls and 3 red balls. Hence,
Number of favourable outcomes =4
Total number of outcomes $=4+3=7$
Hence, $P\left( A/{{E}_{1}} \right)=\dfrac{4}{7}$
Now, let us find $P\left( A/{{E}_{2}} \right)$ which is the probability that white balls in bag II.
We are given that there are 3 white balls and 7 red balls. Hence,
Number of favourable outcomes =3
Total number of outcomes $=3+7=10$
Hence, $P\left( A/{{E}_{2}} \right)=\dfrac{3}{10}$
Now, let’s substitute these values in (i). We will get
$P\left( {{E}_{1}}/A \right)=\dfrac{\dfrac{1}{2}\times \dfrac{4}{7}}{\dfrac{1}{2}\times \dfrac{4}{7}+\dfrac{1}{2}\times \dfrac{3}{10}}$
Let us solve this by first cancelling the common factors. We will get
$P\left( {{E}_{1}}/A \right)=\dfrac{\dfrac{2}{7}}{\dfrac{2}{7}+\dfrac{1}{2}\times \dfrac{3}{10}}$
Let us simplify the denominator. We will get
$P\left( {{E}_{1}}/A \right)=\dfrac{\dfrac{2}{7}}{\dfrac{2}{7}+\dfrac{3}{20}}$
We will get the LCM of the 7 and 20 as 140. Hence,
$\begin{aligned} & P\left( {{E}_{1}}/A \right)=\dfrac{\dfrac{2}{7}}{\dfrac{2\times 20}{7\times 20}+\dfrac{3\times 7}{20\times 7}} \\\ & \Rightarrow P\left( {{E}_{1}}/A \right)=\dfrac{\dfrac{2}{7}}{\dfrac{40}{140}+\dfrac{21}{140}} \\\ & \Rightarrow P\left( {{E}_{1}}/A \right)=\dfrac{\dfrac{2}{7}}{\dfrac{40+21}{140}} \\\ & \Rightarrow P\left( {{E}_{1}}/A \right)=\dfrac{\dfrac{2}{7}}{\dfrac{61}{140}} \\\ \end{aligned}$
We can write this as
$P\left( {{E}_{1}}/A \right)=\dfrac{2}{7}\times \dfrac{140}{61}$
Let us cancel the common factors.
$P\left( {{E}_{1}}/A \right)=\dfrac{2\times 20}{61}=\dfrac{40}{61}$
Hence, the probability that the white balls are drawn from bag I out of two bags is $\dfrac{40}{61}$ .

Note: You may make mistake by using Bayes theorem in the form $P\left( A/B \right)=\dfrac{P\left( B/A \right)P\left( A \right)}{P\left( B \right)}$ . This cannot be used in this solution as there are multiple events. There can be a chance of error by finding $P\left( A/{{E}_{1}} \right)$ instead of $P\left( {{E}_{1}}/A \right)$ . You may make mistakes in the probability formula by writing it as $\text{Probability of an event }=\dfrac{\text{Total number of outcomes}}{\text{Number of favourable outcomes}}$ .