Question

There are two bags 1 and 2. Bag 1 contains 3 white and 4 black balls and Bag 2 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 1.

Answer 1

Hint: In this question, we need to determine the probability that a white ball has been drawn from Bag 1. For this, we will use the relation between the probabilities of the event that is dependent on the other event (Bayes Theorem).

Complete step-by-step answer:
The probability of selecting a bag with the white balls can be calculated as the ratio of the favorable condition and the total number of possibilities.
Here, the total number of bags are 2 and we need to select only 1 so, the favorable condition is 1 and the total number of possibilities are 2.
So, $ P({B_1} = {B_2}) = \dfrac{1}{2} $
Now, the favorable number of conditions for selecting white balls from Bag 1 is 3 and the total number of possibilities is 3+4=7.
So, the probability of selecting 3 white balls from the bag 1 is given as:
 $ P\left( {{W_1}} \right) = \dfrac{3}{7} $
Again, the favorable number of conditions for selecting white balls from the Bag 2 is 5 and the total number of possibilities is 5+6=11.
So, the probability of selecting 5 white balls from the bag 2 is given as:
 $ P\left( {{W_2}} \right) = \dfrac{5}{{11}} $
Now, according to Bayes Theorem, the probability of that a white ball has been drawn from Bag 1 is given as:
 $ \Rightarrow P(A/B) = \dfrac{{P(A)P(B/A)}}{{P(B)}} $ where,
 $ \Rightarrow P(A/B) $ is the probability of event A such that event B already happened.
 $ \Rightarrow P(B/A) $ is the probability of event B such that event A already happened.
Substituting the values in the above equation to determine the desired result.

\[
\Rightarrow P(E) = \dfrac{{\left( {\dfrac{1}{2} \times \dfrac{3}{7}} \right)}}{{\left( {\dfrac{1}{2} \times \dfrac{3}{7}} \right) + \left( {\dfrac{1}{2} \times \dfrac{5}{{11}}} \right)}} \\
   = \dfrac{{\left( {\dfrac{3}{{21}}} \right)}}{{\left( {\left( {\dfrac{3}{{14}}} \right) + \left( {\dfrac{5}{{22}}} \right)} \right)}} \\
   = \dfrac{3}{{21}} \times \left( {\dfrac{{14 \times 22}}{{3 \times 22 + 5 \times 14}}} \right) \\
   = \dfrac{3}{{21}} \times \dfrac{{308}}{{66 + 70}} \\
   = \dfrac{3}{{21}} \times \dfrac{{308}}{{136}} \\
   = \dfrac{{33}}{{68}} \\
 \]
Hence, the probability of drawing a white ball from the bag 1 is $ \dfrac{{33}}{{68}} $ .


Note: Bayes theorem is somehow similar to the conditional probability. Students must be careful while assigning the values to the Bayes theorem as there are many variables involved.