If 3 balls are ”randomly drawn” from a bowl containing 6 white and 5 black balls. What is the probability that one of the balls is white and the other tow black?

To solve this problem, we will use the **hypergeometric distribution** again since we are dealing with a scenario of drawing items (balls) without replacement from a finite population.

### Step-by-Step Solution:

1. **Understand the Parameters:**
   - Total number of white balls: \( W = 6 \)
   - Total number of black balls: \( B = 5 \)
   - Total number of balls: \( N = W + B = 6 + 5 = 11 \)
   - Number of balls drawn: \( n = 3 \)
   - We want to find the probability of drawing exactly one white ball and two black balls.

2. **Apply the Hypergeometric Distribution Formula:**
   The hypergeometric probability formula is:

   \[
   P(X = k) = \frac{\binom{W}{k} \cdot \binom{B}{n - k}}{\binom{N}{n}}
   \]

   Here:
   - \( \binom{W}{k} \) is the number of ways to choose \( k = 1 \) white ball from \( W = 6 \) white balls.
   - \( \binom{B}{n - k} \) is the number of ways to choose \( n - k = 3 - 1 = 2 \) black balls from \( B = 5 \) black balls.
   - \( \binom{N}{n} \) is the total number of ways to choose \( n = 3 \) balls from \( N = 11 \) balls.

3. **Calculate the Individual Components:**

   - \( \binom{6}{1} = \frac{6!}{1!(6-1)!} = 6 \)
   - \( \binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10 \)
   - \( \binom{11}{3} = \frac{11!}{3!(11-3)!} = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165 \)

4. **Compute the Probability:**

   \[
   P(\text{1 white and 2 black}) = \frac{\binom{6}{1} \cdot \binom{5}{2}}{\binom{11}{3}} = \frac{6 \times 10}{165}
   \]

   \[
   P(\text{1 white and 2 black}) = \frac{60}{165} = \frac{12}{33} = \frac{4}{11} \approx 0.3636
   \]

### Conclusion:

The probability that exactly one of the balls is white and the other two are black is **approximately 0.3636**, or **36.36%**.

Would you like more details on this calculation, or do you have any questions?

Here are 5 related questions that may interest you:
1. What is the probability of drawing exactly two white balls and one black ball?
2. How would the probability change if we increased the number of black balls in the bowl?
3. What is the expected number of white balls when drawing 3 balls from the bowl?
4. How does the probability change if we draw 4 balls instead of 3?
5. What is the probability of drawing no white balls at all?

**Tip:** When calculating probabilities involving draws without replacement, always consider the hypergeometric distribution to determine the exact outcomes.

Learn how to calculate the probability of drawing exactly one white ball and two black balls from a bowl containing 6 white and 5 black balls. Detailed step-by-step solution using the hypergeometric distribution formula. Suitable for advanced high school students.

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