Question
Download Solution PDFIn a local cellular phone area, companies A and B account for 60% and 40% of the cellular phone market. 1% of the calls made with company A and 2% of the cellular calls with company B will have interference. If a cellular call is selected at random, the probability that it will have interference is:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFThe correct answer is 0.014.
Key Points
To calculate the probability that a randomly selected cellular call will have interference, the probabilities of selecting a call from each company and the probability of interference for each company are to be considered.
Given that Company A accounts for 60% (0.60) of the cellular phone market and 1% (0.01) of their calls have interference, and Company B accounts for 40% (0.40) of the market and 2% (0.02) of their calls have interference, we can calculate the overall probability of interference as follows:
Probability of interference
= (Probability of selecting Company A) × (Probability of interference in Company A) + (Probability of selecting Company B) × (Probability of interference in Company B)
=
= 0.006 + 0.008
= 0.014
Hence, the probability that a randomly selected cellular call will have interference is 0.014
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