A graduate class consist of six students. what is the probability that exactly three of then are born in either in april or in october?

the size of the class does not matter. Assuming a random population,

P(April)=P(Oct) = 1/12

P of 3 students all in same month is thus (1/12)^3

Given two months, p is twice that.

so, the probability here is 2*(1/12)^3

To find the probability that exactly three students in the graduate class are born in either April or October, we need to determine the total number of possible outcomes and the number of favorable outcomes.

First, let's calculate the total number of possible outcomes. Since there are six students in the class, the total number of ways to arrange their birth months is 12P6 (permutations of 12 items taken 6 at a time). We can use the formula for permutations:

nPr = n! / (n - r)!

Therefore, the total number of possible outcomes is:
12P6 = 12! / (12 - 6)!
= 12! / 6!
= (12 * 11 * 10 * 9 * 8 * 7) / (6 * 5 * 4 * 3 * 2 * 1)
= 924

Next, let's calculate the number of favorable outcomes, which are the combinations of three students being born in either April or October.

For favorable outcomes, we have two cases:
Case 1: Three students born in April and no students born in October.
- Number of ways to choose three students from six: 6C3
- Number of ways to arrange three students among three places: 3P3 (permutations of 3 items taken 3 at a time, which is equal to 3!)
- Total number of favorable outcomes for Case 1: 6C3 * 3P3 = (6! / (3! * (6 - 3)!) * 3P3 = (6 * 5 * 4 * 3 * 2 * 1) / (3 * 2 * 1) * (3 * 2 * 1) = 20 * 6 = 120

Case 2: Three students born in October and no students born in April.
- Number of ways to choose three students from six: 6C3
- Number of ways to arrange three students among three places: 3P3 (permutations of 3 items taken 3 at a time, which is equal to 3!)
- Total number of favorable outcomes for Case 2: 6C3 * 3P3 = (6! / (3! * (6 - 3)!) * 3P3 = (6 * 5 * 4 * 3 * 2 * 1) / (3 * 2 * 1) * (3 * 2 * 1) = 20 * 6 = 120

Finally, we add the favorable outcomes from both cases:
Total number of favorable outcomes = Favorable outcomes from Case 1 + Favorable outcomes from Case 2
= 120 + 120
= 240

Therefore, the probability that exactly three students are born in either April or October is:
Probability = Favorable outcomes / Total outcomes
= 240 / 924
≈ 0.2594 (rounded to four decimal places) or 25.94%