On the freshman-sophomore board, there are 10 freshman members and 13 sophomore board members. If a committee of 12 members is randomly chosen, what is the probability that the committee has exactly 6 freshman?

To solve this probability problem, we need to determine the probability of selecting exactly 6 freshmen from a group of 10 freshmen and 13 sophomores for a committee of 12 members.

### Step-by-Step Approach:

1. **Total number of ways to select 12 members from 23 members**:
   The total number of ways to choose 12 members from 23 is given by the combination formula:
   \[
   \text{Total ways} = \binom{23}{12}
   \]
   where \(\binom{n}{r}\) is the number of combinations of \(n\) items taken \(r\) at a time and is calculated as:
   \[
   \binom{n}{r} = \frac{n!}{r!(n-r)!}
   \]

2. **Number of ways to select 6 freshmen from 10 freshmen**:
   The number of ways to select exactly 6 freshmen from 10 is:
   \[
   \text{Ways to choose 6 freshmen} = \binom{10}{6}
   \]

3. **Number of ways to select 6 sophomores from 13 sophomores**:
   After selecting 6 freshmen, the remaining 6 members of the committee must be sophomores. The number of ways to choose 6 sophomores from 13 is:
   \[
   \text{Ways to choose 6 sophomores} = \binom{13}{6}
   \]

4. **Probability calculation**:
   The probability is the ratio of the number of favorable outcomes to the total number of outcomes, so:
   \[
   \text{Probability} = \frac{\binom{10}{6} \times \binom{13}{6}}{\binom{23}{12}}
   \]

Now, let's calculate each combination.

\[
\binom{10}{6} = \frac{10!}{6!(10-6)!} = \frac{10!}{6!4!} = 210
\]
\[
\binom{13}{6} = \frac{13!}{6!(13-6)!} = \frac{13!}{6!7!} = 1716
\]
\[
\binom{23}{12} = \frac{23!}{12!(23-12)!} = \frac{23!}{12!11!} = 1352078
\]

Finally, the probability is:
\[
\text{Probability} = \frac{210 \times 1716}{1352078} = \frac{360360}{1352078} \approx 0.2665
\]

### Conclusion:
The probability of selecting exactly 6 freshmen from the group of 12 members is approximately \(0.2665\), or about **26.65%**.

---

Would you like further details or clarification on any step? Here are some related questions:

1. What is the general formula for combinations, and how is it derived?
2. How would the problem change if we wanted at least 6 freshmen?
3. What if the committee had fewer members, say 10 instead of 12?
4. How can the result be interpreted in terms of odds rather than probability?
5. What if the problem was to find the probability of choosing exactly 5 sophomores?

**Tip:** Combination problems often involve factorials, so reviewing factorial rules and shortcuts can help speed up these calculations.

On the freshman-sophomore board, there are 10 freshman members and 13 sophomore board members. If a committee of 12 members is randomly chosen, what is the probability that the committee has exactly 6 freshmen?

Learn how to calculate the probability of selecting exactly 6 freshmen from a committee of 12 members, chosen from 10 freshmen and 13 sophomores. This step-by-step solution involves key combinatorial concepts, including the combination formula and probability calculation. Ideal for students in Grades 10-12.

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