Probability of choosing one chromebook and one other devies

Question

A school has 10 iPads, 20 Chromebooks, and 5 laptop computers in their technology loaner supply closet. For each part below, assume all devices are available (so the supply does not change in part B based on what happened in part A, and so on. You begin with all 35 devices in each question.)



What is the probability of selecting two devices and not having only Chromebooks? In other words, what is the chance you have at least one device that is not a Chromebook? Round the answer to 4 digits after the decimal point.

I got the first three parts of this question right, but got stuck on this part. Any help is appreciated!

Answer 1

Hi John!

P(at least one device is a non-Chromebook) = 1 - P(both devices are Chromebooks)

P(both devices are Chromebooks) = (20/35) x (20/35) = 0.3265

P(at least one device is a non-Chromebook) = 1 - 0.3265 = 0.6735