A manufacturer makes mobile phone. The manufacturer employs are inspector to check the quality of its product. The inspector tested a random sample of the phone being defective as 0.049. Ashmita buys three of the phones made by the manufacturer. Find the probability that.
1) At least one of the mobile phone is defective.
2) Exactly one of the mobile phone is defective.
Thank you.
Prob(nogood) = .049
prob(good) = 1-.049 = .951
1) at least one defective--> 1 - all three are good
= 1 - .951^3 = .1399
2) exactly one nogood = C(3,1) (.049)(.951^2) = .1329
To solve these probability problems, we can use the concept of the binomial distribution. In this case, we have a sample of three phones, and we want to find the probability of certain outcomes regarding the number of defective phones.
First, let's calculate the probability of a single phone being defective. We are given that the inspector tested a random sample of the phones, and the probability of a phone being defective is 0.049. Therefore, the probability of a phone being non-defective is 1 - 0.049 = 0.951.
1) Probability that at least one phone is defective:
To find the probability of at least one phone being defective, we need to calculate the complement probability of none of the phones being defective and subtract it from 1.
P(at least one defective) = 1 - P(none defective)
The probability that none of the phones are defective can be calculated by multiplying the probability of each phone being non-defective together since the smartphones are assumed to be independent.
P(none defective) = (0.951) * (0.951) * (0.951)
P(none defective) = 0.951^3
Therefore,
P(at least one defective) = 1 - P(none defective)
P(at least one defective) = 1 - 0.951^3
2) Probability that exactly one phone is defective:
To find the probability of exactly one phone being defective, we need to consider the different ways in which this can occur. It can either be the first phone, the second phone, or the third phone that is defective.
We use the binomial distribution formula to calculate the probability of exactly one success (defective) in three trials (phones).
P(exactly one defective) = (3 choose 1) * (probability of defective) * (probability of non-defective)^(number of non-defective)
The number of ways to choose one defective out of three is given by (3 choose 1) = 3.
Therefore,
P(exactly one defective) = (3 choose 1) * (0.049) * (0.951)^(2)
P(exactly one defective) = 3 * 0.049 * (0.951)^(2)
To calculate these probabilities, you can substitute the values into the formulas and evaluate the expressions using a calculator.