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: Prior Probabilities.
$P(B_1) = 1/2$, $P(B_2) = 1/2$. Likelihood of Black given Bag-I: $P(E|B_1) = 6/10 = 0.6$. Likelihood of Black given Bag-II: $P(E|B_2) = 3/7 \approx 0.428$.
Step 2: Total Probability of Black.
$P(E) = P(E|B_1)P(B_1) + P(E|B_2)P(B_2) = (0.6 \times 0.5) + (3/7 \times 0.5)$ $P(E) = 0.3 + 3/14 = 21/70 + 15/70 = 36/70 = 18/35$.
Step 3: Bayes' Theorem.
$P(B_1|E) = \frac{P(E|B_1)P(B_1)}{P(E)} = \frac{0.3}{18/35} = \frac{3/10}{18/35} = \frac{3 \times 35}{10 \times 18} = \frac{105}{180} = 7/12$.