Q1 Problem 1
5 Points
Grading comment:
A box contains n balls, out of which exactly m are white. We select k of the balls at random, without replacement. What is the probability that i of the selected balls are white?

This is a **hypergeometric probability problem**, where we are selecting \( k \) balls from a total of \( n \) balls, of which \( m \) are white. We want to find the probability that exactly \( i \) of the selected balls are white.

### Step-by-Step Solution:

The hypergeometric probability formula gives us the probability of selecting exactly \( i \) white balls out of \( k \) total selections, without replacement, as follows:

\[
P(\text{select } i \text{ white balls}) = \frac{\binom{m}{i} \binom{n-m}{k-i}}{\binom{n}{k}}
\]

Where:

- \( \binom{m}{i} \) represents the number of ways to choose \( i \) white balls from the \( m \) white balls.
- \( \binom{n-m}{k-i} \) represents the number of ways to choose \( k-i \) non-white balls from the \( n-m \) non-white balls.
- \( \binom{n}{k} \) is the total number of ways to choose \( k \) balls from \( n \) balls.

### Formula Explanation:
- \( \binom{m}{i} \) counts how many ways we can choose \( i \) white balls from \( m \).
- \( \binom{n-m}{k-i} \) counts how many ways we can choose \( k-i \) non-white balls from the remaining \( n-m \).
- The denominator \( \binom{n}{k} \) gives the total number of ways to select \( k \) balls from \( n \), regardless of color.

### Final Probability Expression:

\[
P(i \text{ white balls}) = \frac{\binom{m}{i} \binom{n-m}{k-i}}{\binom{n}{k}}
\]

This is the required probability that exactly \( i \) of the \( k \) selected balls are white.

### Example:
If \( n = 10 \), \( m = 4 \), \( k = 5 \), and \( i = 3 \):
- \( \binom{4}{3} \) is the number of ways to choose 3 white balls from 4.
- \( \binom{6}{2} \) is the number of ways to choose 2 non-white balls from the remaining 6.
- \( \binom{10}{5} \) is the total number of ways to select 5 balls from 10.

Would you like a detailed computation for specific values?

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Here are some related questions:

1. How does the probability change if we replace the balls after selection?
2. What is the expected number of white balls if \( k \) balls are selected?
3. How would the probability change if we wanted at least \( i \) white balls?
4. How is the binomial distribution related to the hypergeometric distribution?
5. What happens to the probability if all balls are white?

**Tip:** The hypergeometric distribution applies to scenarios where sampling is without replacement, unlike the binomial distribution which assumes replacement.

A box contains n balls, out of which exactly m are white. We select k of the balls at random, without replacement. What is the probability that i of the selected balls are white?

Find the probability of selecting exactly i white balls when choosing k balls from a box containing n balls, m of which are white. Learn how to use the hypergeometric distribution formula and solve this step-by-step. Suitable for advanced high school or college-level students.

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