In a local mobile phone area,company A accounts for 60% of the mobile phone market,while company B accounts for the remaining 40% of the market.Of the mobile calls made with company A,1% of the calls will have some sort of interference,while 2% of the mobile calls with company B will have interference.If a mobile call is selected at random,the probability that it will not have interference is
A)0.14.
B)0.014.
C)0.028.
D)0.986.

Question

Quizplus · Accepted Answer

We need to find the probability that a call selected at random will not have interference. Let's use the law of total probability to find this probability:

P(interference) = P(A) * P(interference|A) + P(B) * P(interference|B)
P(interference) = 0.6 * 0.01 + 0.4 * 0.02
P(interference) = 0.014

The probability of interference is 0.014. Therefore, the probability of no interference is:

P(no interference) = 1 - P(interference)
P(no interference) = 1 - 0.014
P(no interference) = 0.986

Thus, the best choice is (D) 0.986.

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