Probablility questions

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.)

A. Consider randomly selecting 2 devices from the supply. In how many ways can any two devices be selected? (You might imagine a family with 2 students are being given devices to work at home. Assume that order does not matter.)

B. In how many ways can you select two Chromebooks? (Again, order does not matter)

C. What is the probability of selecting two devices and both are Chromebooks? Round the answer to 4 digits after the decimal point.

D. 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.

Answer 1

NOTE: I'm not sure which format you have for choosing, but due to formatting constraints, I will use _nC_r format in this solution.

A) First, list the possible outcomes: 2 of the same type (3 ways that can happen) or 2 different types (3 ways that can occur. Then calculate the number of possible outcomes in each case and add these together:

N = ₁₀C₂ + ₂₀C₂ + ₅C₂ + ₁₀C₁•₂₀C₁ + ₁₀C₁•₅C₁ + ₂₀C₁•₅C₁

= 45 + 190 + 10 + 200 + 50 + 100

= 595

B) This is presented above as ₂₀C₂ = 190.

C) Divide the possible options desired (B) by the total options available (A):

P = 190/595 ≈ 0.3193 or 31.9328%

D) Subtract the probability of getting only Chromebooks (C) from 100%

P = 1 - 0.3193 = 0.6807 or 68.0672%