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 and has interference,what is the probability that it was with company A?
A)0.071
B)0.429
C)0.571
D)It cannot be determined.

Question

Quizplus · Accepted Answer

To find the probability that a call with interference was made with company A, we use Bayes' theorem. The probability of interference from company A is (0.60 * 0.01) and from company B is (0.40 * 0.02). The total probability of interference is the sum of these, which is 0.006 + 0.008 = 0.014. The probability that an interfered call is from company A is (0.006 / 0.014) = 0.429.

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