In a local cellular phone area, company A accounts for 70% of the cellular phone market, while company B accounts for the remaining 30% of the market. Of the cellular calls made with company A 2% of the calls will have some sort of interference, while 3% of the cellular calls with company B will have interference. If a cellular call is selected at random, the probability that it will NOT have interference is_________?
0.977
To find the probability that a randomly selected cellular call will not have interference, we can find the complement of the probability that it will have interference.
Let's break down the problem step by step:
1. Company A accounts for 70% of the market, so the probability of selecting a call from company A is 0.70.
2. The probability of interference in a call from company A is given as 2%, which can be written as 0.02.
3. Company B accounts for the remaining 30% of the market, so the probability of selecting a call from company B is 0.30.
4. The probability of interference in a call from company B is given as 3%, which can be written as 0.03.
Now, let's calculate the probability of interference for each company:
- For company A, the probability of interference is 0.02.
- For company B, the probability of interference is 0.03.
To find the probability that a randomly selected call will have interference, we need to consider both companies:
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)
= (0.70 * 0.02) + (0.30 * 0.03)
= 0.014 + 0.009
= 0.023
So, the probability that a randomly selected call will have interference is 0.023.
To find the probability that a randomly selected call will not have interference, we can subtract this probability from 1:
Probability of no interference = 1 - Probability of interference
= 1 - 0.023
= 0.977
Therefore, the probability that a randomly selected call will not have interference is 0.977 or 97.7%.