Suppose that 20% of the students in a college are smokers. A sample of 10 students is randomly selected from this college. What is the probability that at least 3 students in this sample are smokers?
s = smoker ... p(s) = .20
n = non-smoker ... p(n) = .80
this is a binomial probability ... p(s) + p(n) = 1
(n + s)^10 = n^10 + 10 n^9 s + 45 n^8 s^2 + 120 n^7 s^3 + ... + s^10
the probability is one, minus the sum of the 1st four terms
... 1 - [.8^10 + (10 * .8^9 * .2) + (45 * .8^8 * .2^2) + (120 * .8^7 * .2^3)]
To find the probability that at least 3 students in the sample are smokers, we need to calculate the probability of 3, 4, 5, 6, 7, 8, 9, or 10 students being smokers.
First, we need to calculate the probability of exactly "k" students being smokers in the sample. The probability of "k" successes in a binomial distribution can be calculated using the binomial probability formula:
P(k) = (nCk) * p^k * (1-p)^(n-k)
Where:
- n is the sample size (10 in this case)
- k is the number of successes (3, 4, 5, 6, 7, 8, 9, or 10 students being smokers)
- p is the probability of success (e.g., the probability of a student being a smoker, which is 0.2 in this case)
- nCk is a combination formula representing the number of ways to choose "k" successes from "n" trials, calculated as n! / (k!(n-k)!)
To find the probability of at least 3 students being smokers, we need to sum up the individual probabilities for 3, 4, 5, 6, 7, 8, 9, or 10 students.
P(at least 3) = P(3) + P(4) + P(5) + P(6) + P(7) + P(8) + P(9) + P(10)
You can use a calculator or statistical software to perform these calculations easily.