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

Question

Quizplus · Accepted Answer

We can use Bayes' theorem to solve this problem. Let A represent the event that the call is with company A, and B represent the event that the call has interference. We want to find P(A|B), the probability that the call was with company A given that it has interference.

Using Bayes' theorem:

P(A|B) = P(B|A) * P(A) / P(B)

P(A) = 0.6 (given) 
P(B|A) = 0.01 (given)
P(B|B) = 0.02 (given)

To find P(B), we can use the law of total probability:

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

Now we can substitute all of these values into Bayes' theorem:

P(A|B) = 0.01 * 0.6 / 0.014 
P(A|B) = 0.429

Therefore, the probability that the call was with company A given that it has interference is 0.429, or approximately 42.9%.

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