In a high school, 35% of the 3500 students participate in sports. The school board is proposing a participation fee to help with costs. To get student opinion, the school board randomly selects a committee of 6 students to join a discussion of the proposal. That is the probability that exactly 4 of the students participate in sports?

To find the probability that exactly 4 students out of the 6 randomly selected participate in sports, we can use the binomial probability formula.

The probability of success (a student participating in sports) is given as 35% or 0.35, while the probability of failure (a student not participating in sports) is 1 - 0.35 = 0.65.

The formula to calculate the probability of getting exactly k successes in n trials is:

P(X = k) = (n C k) * p^k * (1 - p)^(n - k)

where (n C k) is the combination formula, also known as "n choose k".

Using this formula, we can substitute the given values into the equation:

P(X = 4) = (6 C 4) * 0.35^4 * (1 - 0.35)^(6 - 4)

Since (n C k) = n! / (k! * (n - k)!), we can calculate it as follows:

(6 C 4) = 6! / (4! * (6 - 4)!) = 6! / (4! * 2!)

Calculating this:

(6 C 4) = (6 * 5 * 4!) / (4! * 2) = (6 * 5) / 2 = 15

Substituting all the values into the formula:

P(X = 4) = 15 * 0.35^4 * (1 - 0.35)^(6 - 4)

Calculating this:

P(X=4) = 15 * 0.35^4 * 0.65^2

P(X=4) = 15 * 0.01540375 * 0.4225

P(X=4) ≈ 0.0924

Therefore, the probability that exactly 4 of the 6 students participate in sports is approximately 0.0924 or 9.24%.

To find the probability that exactly 4 out of the 6 randomly selected students participate in sports, we can use the binomial probability formula:

P(X = k) = (n C k) * p^k * (1 - p)^(n - k)

Where:
- P(X = k) is the probability of exactly k successes,
- (n C k) is the number of ways to choose k items from a set of n,
- p is the probability of success (in this case, the probability that a randomly selected student participates in sports),
- k is the number of successes, and
- n is the total number of trials or selections.

In this case:
- n = 6 (since we are selecting 6 students),
- p = 0.35 (since 35% of the students participate in sports), and
- k = 4 (since we are interested in exactly 4 students who participate in sports).

Now, let's substitute these values into the formula:

P(X = 4) = (6 C 4) * 0.35^4 * (1 - 0.35)^(6 - 4)

Calculating the values inside the formula:

(6 C 4) = 6! / (4!(6 - 4)!) = 6! / (4! * 2!)
= (6 * 5 * 4 * 3 * 2 * 1) / (4 * 3 * 2 * 1 * 2 * 1)
= 15

0.35^4 = 0.00543

(1 - 0.35)^(6 - 4) = 0.4225

Now, substituting these values back into the formula:

P(X = 4) = 15 * 0.00543 * 0.4225
≈ 0.0368

Therefore, the probability that exactly 4 out of the 6 randomly selected students participate in sports is approximately 0.0368, or about 3.68%.