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.
Answered in a previous post.
To find the probability that at least 3 out of 5 randomly chosen schools will be closed, we can use the concept of combinations and probabilities.
Step 1: Calculate the total number of ways to choose 5 schools out of 680.
We can use the formula for combinations, denoted as "nCr" or "C(n, r)":
nCr = n! / (r! * (n - r)!)
In this case, we have n = 680 (total number of schools) and r = 5 (number of schools chosen).
Using the formula, we can calculate C(680, 5):
C(680, 5) = 680! / (5! * (680 - 5)!)
Step 2: Calculate the number of ways to choose at least 3 closed schools out of 130.
To calculate this, we need to consider two cases: choosing exactly 3 closed schools and choosing all 5 closed schools.
Case 1: Exactly 3 closed schools
The number of ways to choose 3 closed schools out of 130 is given by C(130, 3).
Case 2: All 5 closed schools
The number of ways to choose all 5 closed schools out of 130 is given by C(130, 5).
Step 3: Calculate the probability
To calculate the probability, we divide the number of favorable outcomes (choosing at least 3 closed schools) by the total number of possible outcomes (choosing any 5 schools).
The probability of choosing at least 3 closed schools out of 5 can be calculated as:
P(at least 3 closed schools) = (C(130, 3) + C(130, 5)) / C(680, 5)
Now, let's calculate the probability:
C(680, 5) = 680! / (5! * (680 - 5)!) = 680! / (5! * 675!) = 680 * 679 * 678 * 677 * 676 / (5 * 4 * 3 * 2 * 1)
C(130, 3) = 130! / (3! * (130 - 3)!) = 130! / (3! * 127!) = 130 * 129 * 128 / (3 * 2 * 1)
C(130, 5) = 130! / (5! * (130 - 5)!) = 130! / (5! * 125!) = 130 * 129 * 128 * 127 * 126 / (5 * 4 * 3 * 2 * 1)
P(at least 3 closed schools) = (C(130, 3) + C(130, 5)) / C(680, 5)
Now, calculate the values and perform the division to find the final probability.