Question

If 5% of the cars coming off the assembly line have some defect in them, what is the probability that out of three cars chosen at random, exactly one car will be defective? Assume that the number of defective cars has a Poisson distribution.

A. 0.129
B. 0.135
C. 0.151
D. 0.174

Answer 1

The probability that exactly one out of three cars chosen at random will be defective, given a 5% defect rate and assuming a Poisson distribution, is 0.135.

Step-by-step explanation:

The question asks for the probability that exactly one car will be defective if 5% of the cars are defective and we are choosing three cars. Assuming a Poisson distribution, we can calculate this using the Poisson probability formula P (X = x) = (e-μμx / x!), where μ is the average number of successes (defective cars) and x is the exact number, we are looking for (one defective car).

Since 5% of the cars are defective, the average number μ for three cars is 3 * 5% = 0.15. Plugging into the Poisson formula, we get P (X = 1) = e-0.15 * 0.151 / 1! = 0.135 (Option B).