A graduate class consists of six students. What is the probability that exactly three of them are born either March, June or November?

Out of six students, we need exactly 3 fit the criteria of being born in 3 months out of 12, i.e. with probability p=3/12=0.25.

Since the probability is assumed constant throughout, and we know the total number of students, we can use the binomial distribution, where
n=6
x=3
p=0.25
and the probability is given by:
P(X=x)=C(n,x)p^x (1-p)^(n-x)
where C(n,x)=binomial coefficient = n!/(x!(n-x)!).

Just plug in the numbers to get your results.

Well, let's clown around with some math, shall we?

First, let's assume that all 12 months of the year are equally likely for a birthday. Since we're only interested in the months of March, June, and November, we have a 1/4 chance for each student to be born in one of these months.

Now, let's calculate the probability of exactly three students being born in these desired months. We need to consider the different combinations of students that can fulfill this condition.

The number of combinations we're interested in is C(6,3), which stands for "6 choose 3" and is calculated as:

C(6,3) = 6! / (3! * (6-3)!), or simply 6! / (3! * 3!)

This gives us 6 * 5 * 4 / (3 * 2 * 1) = 20.

The total number of possible outcomes is 4^6, since each of the 6 students has 4 possible birth months.

So, the probability of exactly three students being born in March, June, or November is 20 / 4^6.

If you want an actual numerical value, let me clownishly calculate it for you. It's approximately 0.0635, or about a 6.35% chance.

Keep in mind that this calculation assumes an equal likelihood of birth months and independence between students, but hey, life is full of surprises!

To find the probability that exactly three students are born in either March, June, or November in a graduate class of six students, we need to calculate the probability for all possible combinations.

Step 1: Calculate the total number of possible combinations
Since there are three months (March, June, and November) in which the students can be born, we have three choices for each student. So, the total number of possible combinations is the product of three choices for each of the six students, which is 3^6.

Total number of combinations = 3^6 = 729

Step 2: Calculate the number of favorable combinations
To have exactly three students born in either March, June, or November, we need to choose three students from the six and assign them to one of the three months. The remaining three students can be assigned to any month.

The number of ways to choose three students out of six is given by the combination formula: C(n, r) = n! / (r!(n-r)!), where n is the total number of students and r is the number of students we want to choose.

Number of ways to choose 3 students out of 6 = C(6, 3) = 6! / (3!(6-3)!) = 20

Once we have selected three students, we can assign them to one of the three months in 3! (3 factorial) ways. For the remaining three students, they can be assigned to any month in 3^3 (3 raised to the power of 3) ways.

Number of ways to assign three students to one month = 3!
Number of ways to assign the remaining three students to any month = 3^3

Number of favorable combinations = Number of ways to choose 3 students out of 6 * Number of ways to assign three students to one month * Number of ways to assign the remaining three students to any month
= 20 * 3! * 3^3
= 20 * 6 * 27
= 3,240

Step 3: Calculate the probability
Now, we can calculate the probability by dividing the number of favorable combinations by the total number of combinations:

Probability = Number of favorable combinations / Total number of combinations
= 3,240 / 729
≈ 0.444444

Therefore, the probability that exactly three students are born in either March, June, or November in a graduate class of six students is approximately 0.444444 or 44.44%.

To find the probability that exactly three students are born either in March, June, or November in a class of six students, we need to consider three things: the total number of possible outcomes, the number of favorable outcomes, and calculate the probability based on these values.

First, let's calculate the total number of possible outcomes. Each student can be born in any month of the year, so the total number of possible outcomes for one student is 12.

For six students, the total number of possible outcomes would be 12^6 since each student has 12 options for their birth month. Therefore, the total number of possible outcomes is 12^6 = 2,176,782,336.

Next, let's determine the number of favorable outcomes, i.e., the number of ways we can choose exactly three students out of six who are born in March, June, or November. Each of these three months has their own group of three students.

We can calculate this using combinations. The number of ways to choose exactly three students out of six is expressed as "6 choose 3" or written as C(6, 3). This can be calculated as:

C(6, 3) = 6! / (3! * (6-3)!) = 6! / (3! * 3!) = (6 * 5 * 4) / (3 * 2 * 1) = 20.

Therefore, there are 20 different ways to choose exactly three students out of six.

Now, we can calculate the probability of exactly three students being born in March, June, or November by dividing the number of favorable outcomes by the total number of possible outcomes:

Probability = Number of favorable outcomes / Total number of possible outcomes = 20 / 2,176,782,336.

Now you can calculate the final answer by dividing 20 by 2,176,782,336.

Please note that the resulting probability is likely to be an extremely small value, given the large numbers involved.