In a local cellular phone area, company A accounts for 60% of the cellular phone market, while company B accounts for the remaining 40% of the market. Of the cellular calls made with company A, 1% of the calls will have some sort of interference, while 2% of the cellular calls with company B will have interference. If a cellular call is selected at random, what is the probability that it will have interference?
A) 0.014
B) 0.986
C) 0.14
D) 0.028

Question

Quizplus · Accepted Answer

The probability of selecting a call from company A is 0.6, and the probability of interference for company A is 0.01. So, the probability of interference from company A is 0.6 * 0.01 = 0.006. The probability of selecting a call from company B is 0.4, and the probability of interference for company B is 0.02. So, the probability of interference from company B is 0.4 * 0.02 = 0.008. The total probability of interference is the sum of these two probabilities, which is 0.006 + 0.008 = 0.014.

In a Local Cellular Phone Area, Company a Accounts for 60