Certainly! Let's solve this problem using the principles of hypergeometric distribution step by step.
Given:
- Total number of voters = 100
- Number of student council candidates = 40
- Number of non-candidates = 100 - 40 = 60
- Number of voters to be selected = 10
- Number of student council candidates in the selected group = 5
We need to find the probability that in a randomly selected group of 10 voters, exactly 5 are student council candidates.
Hypergeometric distribution is perfect for this scenario, as it deals with successes and failures in draws without replacement. The formula for the hypergeometric probability [tex]\(P(X = k)\)[/tex] is:
[tex]\[ P(X = k) = \frac{\binom{C}{k} \binom{N - C}{n - k}}{\binom{N}{n}} \][/tex]
where:
- [tex]\( N \)[/tex] is the total number of voters (100)
- [tex]\( C \)[/tex] is the number of student council candidates (40)
- [tex]\( n \)[/tex] is the number of voters selected (10)
- [tex]\( k \)[/tex] is the number of student council candidates in the selected group (5)
First, let's find these binomial coefficient values:
1. Binomial coefficient for selecting 5 student council candidates from 40 candidates:
[tex]\[ \binom{40}{5} = \frac{40!}{5! \cdot (40 - 5)!} \][/tex]
2. Binomial coefficient for selecting the remaining 5 non-candidates from 60 voters:
[tex]\[ \binom{60}{5} = \frac{60!}{5! \cdot (60 - 5)!} \][/tex]
3. Binomial coefficient for selecting 10 voters from the total 100 voters:
[tex]\[ \binom{100}{10} = \frac{100!}{10! \cdot (100 - 10)!} \][/tex]
Now we can place these into the formula:
[tex]\[ P(X = 5) = \frac{\binom{40}{5} \cdot \binom{60}{5}}{\binom{100}{10}} \][/tex]
By calculating the values (not shown for brevity and complexity), it turns out:
[tex]\[ P(X = 5) \approx 0.20760568129294876 \][/tex]
So the final probability that exactly 5 voters out of the 10 selected are student council candidates is approximately 0.2076, or in percentage terms, about 20.76%.
