Question

There are two boxes each containing 10 balls. In each box, few of them are black balls and rest are white. A ball is drawn at random from one of the boxes and found that it is black. If the probability that the black ball drawn is from the second box is $\frac{1}{5}$, then number of black balls in the first box is

• A. 5 or 10
• B. 2 or 7
• C. 4 or 8
• D. 3 or 6 or 9

Hint:

This is a classic application of Bayes' Theorem. When you are given a conditional probability like $P(A|B)$ and asked to find the reverse conditional probability $P(B|A)$, Bayes' Theorem is the tool to use.

Answer 1

The Correct Option is C

We need to find the number of black balls in the first box. Let's denote the number of black balls in the first box as \(x\) and the number of black balls in the second box as \(y\). Here is how we can solve this problem:

1. Each box contains a total of 10 balls.
2. The problem states that the probability of drawing a black ball from the second box is \(\frac{1}{5}\).
3. The total probability of drawing a black ball from one of the boxes can be given by considering both boxes:
4. The probability that a black ball is drawn from the second box is given by:
5. According to the given condition:
6. Simplifying this gives:
7. Cross-multiplying results in:
8. Rearranging gives:
9. We know both boxes can contain a total of 10 balls, hence:
10. Solving \(4y = x\) under integer constraints, possible values for \(x\) (number of black balls in the first box) can be:

The correct answer is \(4\) or \(8\) based on calculations.