The J.O.Supplies Company buys calculators from a Korean supplier.The probability of a defective calculator is 10 percent.If 3 calculators are selected at random,what is the probability that one of the calculators will be defective?
A).0702
B).0010
C).2430
D).7290

Question

Quizplus · Accepted Answer

To solve the problem, we can use the binomial distribution formula:

P(X = k) = nCk * p^k * (1-p)^(n-k)

where:
- P(X = k) is the probability of getting exactly k defective calculators
- nCk is the number of ways to choose k items out of n
- p is the probability of getting a defective calculator
- (1-p) is the probability of getting a non-defective calculator
- n is the total number of calculators selected (in this case, n=3)

To find the probability of getting one defective calculator, we need to calculate the probability of getting one defective calculator and two non-defective calculators, and then multiply by the number of ways we can choose which calculator is defective.

P(one defective calculator) = 3C1 * 0.1^1 * 0.9^2 = 0.243

Therefore, the answer is C) 0.2430.

The JOSupplies Company Buys Calculators from a Korean Supplier