Question

Alan reads that one out of four eggs contains salmonella bacteria. Thus, he never uses more

than three eggs in cooking. If eggs do or don't contain salmonella independent of each
other, the number of contaminated eggs when Alan uses three chosen at random has the
distribution.

A) binomial with N = 4 and P = 1/3
B) binomial with N = 3 and P = 1/4
C) normal with μ = 3/4 and σ = √(3)(1/4)(1 - 1/4)
D) normal with μ = 4/3 and σ = √(4)(1/3)(1 - 1/3)

Answer 1

Alan's scenario fits a binomial distribution with three trials and a 25% success rate. Thus, the number of contaminated eggs when Alan uses three has the distribution B) binomial with N = 3 and P = 1/4.

Step-by-step explanation:

The question revolves around the concept of probability distributions in mathematics, specifically the binomial distribution. Alan reads that there is a chance that one out of four eggs can contain salmonella bacteria, and so he chooses to only use three eggs in cooking to minimize his risk of contamination. Since the presence of salmonella in each egg is independent of the others, the number of contaminated eggs follows a binomial distribution. With three eggs chosen at random and one out of four likely to contain salmonella, his scenario matches a binomial distribution with N = 3, representing the number of trials, and P = 1/4, representing the probability of success (finding a contaminated egg).The correct answer to the distribution in question is B) binomial with N = 3 and P = 1/4.This question is a clear application of binomial probability, which uses specific criteria: There should be a fixed number of trials (N), each trial should have two possible outcomes (success or failure), and the probability of success (P) should be the same for each trial. The trials also need to be independent of one another.