A bag contains 4 what and 6 black balls. A ball is drawn at random and replaced and then again a ball is drawn. If this is replaced 4 times, what is the probability that three black and one white ball will be drawn in these 4 trials ?

Answer

To solve the problem of finding the probability of drawing three black balls and one white ball in four trials from a bag containing 4 white and 6 black balls, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Total Number of Balls**:
- The bag contains 4 white balls and 6 black balls.
- Total number of balls = 4 (white) + 6 (black) = 10 balls.

2. **Calculate the Probability of Drawing a Black Ball (P)**:
- The probability of drawing a black ball (P) is given by the formula:
\[ P = \frac{\text{Number of black balls}}{\text{Total number of balls}} = \frac{6}{10} = \frac{3}{5} \]

3. **Calculate the Probability of Drawing a White Ball (Q)**:
- The probability of drawing a white ball (Q) is given by:
\[ Q = \frac{\text{Number of white balls}}{\text{Total number of balls}} = \frac{4}{10} = \frac{2}{5} \]

4. **Determine the Required Probability**:
- We need to find the probability of drawing exactly 3 black balls and 1 white ball in 4 trials.
- This can be modeled using the binomial probability formula:
\[ P(X = k) = \binom{n}{k} p^k q^{n-k} \]
- Here, \( n = 4 \) (total trials), \( k = 3 \) (number of successes, which is drawing black balls), \( p = \frac{3}{5} \) (probability of success), and \( q = \frac{2}{5} \) (probability of failure).

5. **Calculate the Binomial Coefficient**:
- The binomial coefficient \( \binom{4}{3} \) is calculated as:
\[ \binom{4}{3} = \frac{4!}{3!(4-3)!} = \frac{4 \times 3!}{3! \times 1!} = 4 \]

6. **Substitute Values into the Formula**:
- Now substitute the values into the binomial probability formula:
\[ P(3 \text{ black}, 1 \text{ white}) = \binom{4}{3} \left(\frac{3}{5}\right)^3 \left(\frac{2}{5}\right)^1 \]
- This becomes:
\[ P(3 \text{ black}, 1 \text{ white}) = 4 \left(\frac{3}{5}\right)^3 \left(\frac{2}{5}\right) \]

7. **Calculate the Powers**:
- Calculate \( \left(\frac{3}{5}\right)^3 = \frac{27}{125} \) and \( \left(\frac{2}{5}\right) = \frac{2}{5} \).

8. **Combine the Results**:
- Now combine the results:
\[ P(3 \text{ black}, 1 \text{ white}) = 4 \times \frac{27}{125} \times \frac{2}{5} \]
- Simplifying this gives:
\[ = 4 \times \frac{54}{625} = \frac{216}{625} \]

### Final Result:
The probability of drawing three black balls and one white ball in four trials is:
\[ \frac{216}{625} \]