Question

At a high school, 30% of the students buy class rings. You
select five students at random. Find P(at least four buy
rings).
**Please include 5 decimal places in your answer**

Answer 1

Need to the find the probability of at least 4 out of 5 and 5 out of 5 since the question asks you to find at least 4 buy rings out of 5 students.
nCr(5,4)(0.30)^4(0.70)^1 + nCr(5,5)(0.30)^5(0.70)^0 = 0.03078

Answer 2

This is a binomial probability problem, where each student is a trial with a success (buying a class ring) probability of 0.3. We want to find the probability of having at least four successes in five trials, which is equal to:

P(X >= 4) = P(X = 4) + P(X = 5)

Using the binomial probability formula, we get:

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

Where n is the number of trials, k is the number of successes, and p is the probability of success.

P(X = 4) = (5 choose 4) * 0.3^4 * 0.7^1 = 0.072027
P(X = 5) = (5 choose 5) * 0.3^5 * 0.7^0 = 0.00243

Therefore,

P(X >= 4) = 0.072027 + 0.00243 = 0.074457

Rounded to 5 decimal places, the probability that at least four out of five selected students will buy class rings is 0.07446.

Answer 3

Your calculation is incorrect. The probability of at least 4 out of 5 and 5 out of 5 are not added together, but rather the probabilities are calculated separately and then added.

The probability of exactly 4 out of 5 students buying rings is:

P(X = 4) = nCr(5,4) * 0.3^4 * 0.7^1 = 0.07203

The probability of exactly 5 out of 5 students buying rings is:

P(X = 5) = nCr(5,5) * 0.3^5 * 0.7^0 = 0.00243

To find the probability of at least 4 out of 5 students buying rings, we add the probabilities of having 4 and 5 students buying rings:

P(X >= 4) = P(X = 4) + P(X = 5) = 0.07203 + 0.00243 = 0.07446

Therefore, the probability of at least 4 out of 5 students buying class rings is 0.07446. This is the final answer, rounded to 5 decimal places.