1. Twenty-three percent of students are from out of state. In a random sample of 6 students, what is the probability that at least 2 of them are from out of state? (round to whole number) a. 42% b. 7% c. 28% d. 64% e. none of these

2. For the problem above what is the expected value of x. a. 4.2 b. 2.7 c. 1.4 d ...5 e. none of these

Twenty-three percent of students are from out of state. In a random sample of 6 students, what is the probability that at least 2 of them are from out of state?

This is a case of binomial probability, as "success" would be "student is from out of state" and "failure" would be "student is not from out of state."

The probability, p, that a student is from out of state is 0.23. The number of experiments is n = 6.

The probability that 0 students is/are from out of state is binompdf (6, 0.23, 0), which, according to the binompdf (function of my TI-83 Plus calculator, is 0.2084. Similarly, the prob. that 1 student is from out of state is binompdf (6, 0.23, 1), or 0.3735. Adding these two probabilities together, we get 0.5819. This represents the probability that 0 or 1 student is from out of state.

To find the prob. that 2 or more are from out of state, subtract this 0.5819 from 1.0000. The correct result is 0.4181. This makes use of the "complement" property of probability. It's easier to calculate the prob. that 0 or 1 student is from out of state, as we have done here, and then subtract the result from 1, than it is to calculate the five probabilities

binompdf (6, 0.23, 2), binompdf (6, 0.23, 3), binompdf (6, 0.23, 4), binompdf (6, 0.23, 5) and binompdf (6, 0.23, 6) and add them together.