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

Step 1: Define the events and variables.
Let $B_1$ be the event of choosing the first box, and $B_2$ be the event of choosing the second box. Since a box is chosen at random: \[ P(B_1) = P(B_2) = \frac{1}{2}. \] Let $E$ be the event that a black ball is drawn.
Let $n_1$ and $n_2$ be the number of black balls in the first and second boxes, respectively. Each box contains a total of 10 balls.
The conditional probabilities are: \[ P(E|B_1) = \frac{n_1}{10}, P(E|B_2) = \frac{n_2}{10}. \]
Step 2: Apply Bayes' Theorem.
We are given: \[ P(B_2|E) = \frac{1}{5}. \] Bayes' Theorem states: \[ P(B_2|E) = \frac{P(B_2)P(E|B_2)}{P(B_1)P(E|B_1) + P(B_2)P(E|B_2)}. \]
Step 3: Substitute the known values.
\[ \frac{1}{5} = \frac{(1/2)(n_2/10)}{(1/2)(n_1/10) + (1/2)(n_2/10)}. \] Cancelling the common factors $(1/2)$ and $(1/10)$: \[ \frac{1}{5} = \frac{n_2}{n_1 + n_2}. \]
Step 4: Solve for the relationship between $n_1$ and $n_2$.
\[ n_1 + n_2 = 5 n_2 \implies n_1 = 4 n_2. \]
Step 5: Determine possible integer values.
Since $n_1, n_2$ are integers between 1 and 10:

• If $n_2 = 1$, then $n_1 = 4 \cdot 1 = 4$ (valid).
• If $n_2 = 2$, then $n_1 = 4 \cdot 2 = 8$ (valid).
• If $n_2 = 3$, then $n_1 = 4 \cdot 3 = 12$ (invalid, exceeds 10).

Hence, the possible number of black balls in the first box is: \[ \boxed{4 \text{ or } 8}. \]