Question

Bag-I contains 4 white and 6 black balls, Bag-II contains 4 white and 3 black balls. A ball is selected at random and it comes out to be a black ball. What is the probability that it is from Bag-I?

Hint:

For conditional probability problems like this, clearly define the events and write down all the known probabilities. Bayes' theorem provides a systematic way to find the "reverse" probability.

Answer 1

Step 1: Understanding the Question:
This is a conditional probability problem that is best solved using Bayes' theorem. We are given a final outcome (a black ball was drawn) and asked to find the probability of a specific initial condition (the ball came from Bag-I).
Step 2: Define Events and Probabilities:
Let the events be:

• \(B_1\): The ball is selected from Bag-I.
• \(B_2\): The ball is selected from Bag-II.
• \(E\): The selected ball is black.

We want to find \(P(B_1|E)\). The given probabilities are:

• Since a bag is chosen at random, \(P(B_1) = 1/2\) and \(P(B_2) = 1/2\).
• Probability of drawing a black ball from Bag-I: \(P(E|B_1) = \frac{\text{black balls in Bag-I}}{\text{total balls in Bag-I}} = \frac{6}{4+6} = \frac{6}{10} = \frac{3}{5}\).
• Probability of drawing a black ball from Bag-II: \(P(E|B_2) = \frac{\text{black balls in Bag-II}}{\text{total balls in Bag-II}} = \frac{3}{4+3} = \frac{3}{7}\).

Step 3: Apply Bayes' Theorem:
Bayes' theorem states: \[ P(B_1|E) = \frac{P(E|B_1)P(B_1)}{P(E)} \] First, we need to calculate the total probability of drawing a black ball, \(P(E)\), using the law of total probability: \[ P(E) = P(E|B_1)P(B_1) + P(E|B_2)P(B_2) \] \[ P(E) = \left(\frac{3}{5}\right)\left(\frac{1}{2}\right) + \left(\frac{3}{7}\right)\left(\frac{1}{2}\right) \] \[ P(E) = \frac{3}{10} + \frac{3}{14} = \frac{21+15}{70} = \frac{36}{70} = \frac{18}{35} \] Now we can find \(P(B_1|E)\): \[ P(B_1|E) = \frac{P(E|B_1)P(B_1)}{P(E)} = \frac{(3/5)(1/2)}{18/35} = \frac{3/10}{18/35} \] \[ P(B_1|E) = \frac{3}{10} \times \frac{35}{18} = \frac{1}{2} \times \frac{7}{6} = \frac{7}{12} \] Step 4: Final Answer:
The probability that the black ball was from Bag-I is \(7/12\).