Certainly! Let's break down the problem into manageable steps to find out how many voters are not candidates and the probability that exactly 5 of the 10 randomly selected voters are candidates.
### Step 1: Determine the number of non-candidates
We know the total number of voters is 100, and out of these, 40 are registered as candidates. To find the number of voters who are not candidates, we subtract the number of candidates from the total number of voters:
[tex]\[ \text{Number of non-candidates} = \text{Total voters} - \text{Total candidates} \][/tex]
[tex]\[ \text{Number of non-candidates} = 100 - 40 \][/tex]
[tex]\[ \text{Number of non-candidates} = 60 \][/tex]
So, there are 60 non-candidates.
### Step 2: Calculate the probability using hypergeometric distribution
Hypergeometric distribution is used when we're dealing with successes in a sample drawn without replacement from a finite population. Here’s how it applies to our problem:
- Total population size ([tex]\(N\)[/tex]) = 100
- Number of success states in the population ([tex]\(K\)[/tex]) = 40 (candidates)
- Sample size ([tex]\(n\)[/tex]) = 10 (voters randomly selected)
- Number of observed successes ([tex]\(k\)[/tex]) = 5 (exactly 5 of the selected are candidates)
We need the probability [tex]\( P(X = k) \)[/tex] where [tex]\(X\)[/tex] follows a hypergeometric distribution. The formula for hypergeometric probability is:
[tex]\[ P(X = k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}} \][/tex]
Where:
- [tex]\(\binom{K}{k}\)[/tex] is the combination of [tex]\(K\)[/tex] items taken [tex]\(k\)[/tex] at a time.
- [tex]\(\binom{N-K}{n-k}\)[/tex] is the combination of [tex]\(N-K\)[/tex] items taken [tex]\(n-k\)[/tex] at a time.
- [tex]\(\binom{N}{n}\)[/tex] is the combination of [tex]\(N\)[/tex] items taken [tex]\(n\)[/tex] at a time.
Plugging in the values, we get the exact probability value of drawing exactly 5 candidates out of 10 voters:
[tex]\[ P(X = 5) \approx 0.2076 \][/tex]
Therefore, the probability that exactly 5 out of the 10 selected voters will be student council candidates is approximately 0.2076, or about 20.76%.
### Summary
- The number of non-candidates is 60.
- The probability that exactly 5 out of 10 randomly selected voters are candidates is approximately 0.2076, or 20.76%.
