A box contains 24 identical balls of which 12 are white and 12 are black. The balls are drawn at random from the box one at a time with replacement. The probability that a white ball is drawn for the 4th time on the 7th draw is `5//64` b. `27//32` c. `5//32` d. `1//2`

Options

A. `5//64`
B. `27//32`
C. `5//32`
D. `1//2`

Explanation

To solve the problem, we need to find the probability that a white ball is drawn for the 4th time on the 7th draw, given that there are 12 white balls and 12 black balls in the box. The draws are made with replacement.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We need to find the probability that the 4th white ball is drawn on the 7th draw.
- This means that in the first 6 draws, we must have exactly 3 white balls drawn.

2. **Defining the Variables**:
- Let \( n = 7 \) (total draws),
- Let \( r = 4 \) (the 4th success),
- Let \( k = 3 \) (the number of successes in the first 6 draws).

3. **Calculating the Probability of Success**:
- The probability of drawing a white ball (success) is:
\[ P(\text{white}) = \frac{12}{24} = \frac{1}{2} \]
- The probability of drawing a black ball (failure) is:
\[ P(\text{black}) = 1 - P(\text{white}) = \frac{1}{2} \]

4. **Using the Binomial Distribution**:
- We can use the binomial probability formula to calculate the probability of getting exactly 3 white balls in the first 6 draws:
\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]
- Here, \( n = 6 \), \( k = 3 \), and \( p = \frac{1}{2} \):
\[ P(X = 3) = \binom{6}{3} \left(\frac{1}{2}\right)^3 \left(\frac{1}{2}\right)^{6-3} \]
\[ = \binom{6}{3} \left(\frac{1}{2}\right)^6 \]

5. **Calculating \( \binom{6}{3} \)**:
- The binomial coefficient \( \binom{6}{3} \) is calculated as:
\[ \binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \]

6. **Calculating the Probability**:
- Now substituting back:
\[ P(X = 3) = 20 \cdot \left(\frac{1}{2}\right)^6 = 20 \cdot \frac{1}{64} = \frac{20}{64} = \frac{5}{16} \]

7. **Finding the Final Probability**:
- The probability that the 4th white ball is drawn on the 7th draw is simply the probability of drawing a white ball on the 7th draw, which is \( \frac{1}{2} \):
\[ P(\text{4th white on 7th}) = P(X = 3) \cdot P(\text{white on 7th}) = \frac{5}{16} \cdot \frac{1}{2} = \frac{5}{32} \]

### Final Answer:
The probability that a white ball is drawn for the 4th time on the 7th draw is:
\[ \frac{5}{32} \]