8. The Chicago Public School system has about 680 schools. About 130 of the schools will be closed. If five schools are chosen at random find the probability that at least 3 of them will be closed.

0.051367

To find the probability that at least 3 of the 5 schools chosen at random will be closed out of a total of 680 schools with 130 closed, we can use the concept of combinations and the binomial probability formula.

Step 1: Find the probability of selecting 3 closed schools and 2 open schools.
To calculate this probability, we need to use the following formula for a binomial distribution:
P(X = k) = C(n, k) * p^k * (1 - p)^(n - k)
where P(X = k) is the probability of getting exactly k successes, C(n, k) is the number of combinations, p is the probability of success, and n is the number of trials.

In this case, n (number of trials) is 5 (as we are selecting 5 schools), k (number of closed schools) is 3, p (probability of selecting a closed school) is 130/680 (as there are 130 closed schools out of a total of 680 schools), and the probability of selecting an open school is 1 - p.

So, P(X = 3) = C(5, 3) * (130/680)^3 * (1 - 130/680)^2

Step 2: Find the probability of selecting 4 closed schools and 1 open school.
Using the same formula with k = 4, we can calculate P(X = 4).

P(X = 4) = C(5, 4) * (130/680)^4 * (1 - 130/680)^1

Step 3: Find the probability of selecting 5 closed schools.
Similarly, P(X = 5) = C(5, 5) * (130/680)^5 * (1 - 130/680)^0

Step 4: Calculate the probability of at least 3 closed schools.
To calculate the probability of at least 3 closed schools, we need to find the sum of P(X = 3), P(X = 4), and P(X = 5).

P(at least 3 closed schools) = P(X = 3) + P(X = 4) + P(X = 5)

Finally, substitute the values into the equations above and calculate the probability.