Question

There are two bags 1 and 2, containing 3 red and 4 white balls and 2 red and 3 white balls respectively. A bag is selected at random and a ball is drawn from it. If it is found to be a red ball, find the probability that it is drawn from the first bag.

Answer 1

Hint: We start solving the problem by finding the total no. of balls in the bag. We then find the probability to draw a red ball from each bag. We then find the probability of choosing a bag from the two given bags. We then recall the definition of Bayes theorem in which we use all these values and make necessary calculations to get the required probability.

Complete step by step answer:
We have been told that there are 2 bags which are marked as 1 and 2.
Bag 1 contains 3 red and 4 white balls.
\[\therefore \] Total number of balls in bag 1 = 3 + 4 = 7 balls.
Bag 2 contains 2 red and 3 white balls.
\[\therefore \] Total number of balls in bag 2 = 2 + 3 = 5 balls.
It is said that a ball is selected at random and a red ball is chosen.
Let A be the event of selecting Bag 1. Let B be the event of picking a red ball.
From Bayes theorem, P (A / B), the probability that event A happened, given that test B was positive.
\[P\left( A/B \right)=\dfrac{P\left( A \right)P\left( B/A \right)}{P\left( B \right)}\].
P (A / B) = How often A happens given that B happens.
P (B / A) = How often B happens given that A happens.
Here, P (A) = number of favorable outcomes / Total number of possible outcomes.
P (A) = probability of choosing from Bag 1 / total of 2 Bags = \[\dfrac{1}{2}\]
P (A) = \[\dfrac{1}{2}\].
P (B) = Probability of picking up a red ball.
This can be from either bag 1 or bag 2.
P (B) = probability of choosing from bag 1 + probability of choosing from bag 2.
P (B) = \[\dfrac{1}{2}\times \] (number of red balls in bag 1/ total number of balls in bag 1) + \[\dfrac{1}{2}\times \] (number of red balls in bag 2/ total number of balls in bag 2).
P (B) = \[\dfrac{1}{2}\times \dfrac{3}{7}+\dfrac{1}{2}\times \dfrac{2}{5}\].
Now, P (B / A) = probability of choosing a red ball from bag 1.
\[P\left( B/A \right)=\dfrac{3}{7}\].
Thus, by Bayes theorem,
\[P\left( A/B \right)=\dfrac{P\left( A \right)P\left( B/A \right)}{P\left( B \right)}\].
Now put the values we got in the above expression and simplify it.
\[\Rightarrow P\left( A/B \right)=\dfrac{\dfrac{1}{2}\times \dfrac{3}{7}}{\dfrac{1}{2}\times \dfrac{3}{7}+\dfrac{1}{2}\times \dfrac{2}{5}}\].
\[\Rightarrow P\left( A/B \right)=\dfrac{\dfrac{1}{2}\times \dfrac{3}{7}}{\dfrac{1}{2}\left( \dfrac{3}{7}+\dfrac{2}{5} \right)}\].
\[\Rightarrow P\left( A/B \right)=\dfrac{\dfrac{3}{7}}{\left( \dfrac{3}{7}+\dfrac{2}{5} \right)}\].
\[\Rightarrow P\left( A/B \right)=\dfrac{\dfrac{3}{7}}{\dfrac{15+14}{35}}\].
\[\Rightarrow P\left( A/B \right)=\dfrac{\dfrac{3}{7}}{\dfrac{29}{35}}\].
\[\Rightarrow P\left( A/B \right)=\dfrac{3}{7}\times \dfrac{35}{29}\].
\[\Rightarrow P\left( A/B \right)=\dfrac{15}{29}\].
We have found the probability that the red ball is drawn from bag 1 as \[\dfrac{15}{29}\].
∴ The probability that the red ball is drawn from bag 1 is \[\dfrac{15}{29}\].

Note:
We should know that Bayes’s theorem is similar to the conditional probability. We can make mistakes while calculating the probability of choosing a red ball from each bag as $\dfrac{3}{4}$ or $\dfrac{2}{3}$. Whenever we get this type of problem, we should start by assigning variables to the events to get a better view of calculations. We should not make calculation mistakes while solving this problem. We can also find the probability that the black ball is drawn from one of the given bags.