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Joe reads that 1 out of 4 eggs contains salmonella bacteria. So he never uses more than 3 eggs in cooking. If eggs do or don't contain salmonella independently of each other, the number of contaminated eggs when Joe uses 3 eggs chosen at random has the following distribution:
Joe reads that 1 out of 4 eggs contains salmonella bacteria. So he never uses more than 3 eggs in cooking. If eggs do or don't contain salmonella independently of each other, the number of contaminated eggs when Joe uses 3 eggs chosen at random has the following distribution:a. binomial; n = 4 and p = 1/4
b. binomial; n = 3 and p = 1/4
c. binomial; n = 3 and p = 1/3
d. geometric; p = 1/4
Final answer:The distribution of contaminated eggs when Joe uses 3 eggs chosen at random is binomial with n = 3 and p = 1/4, as the presence of Salmonella in each egg is independent of the others. Therefore, the correct option is B.Explanation:Understanding the Distribution of Contaminated EggsJoe's concern about the presence of Salmonella bacteria in eggs leads to a probability question. Since the probability that any egg contains Salmonella is independent of other eggs, we can use the binomial distribution to model this scenario. Joe uses 3 eggs, and if we are told 1 out of 4 eggs could be contaminated, then the probability, p, of any one egg being contaminated is 1/4.The distribution of contaminated eggs can be described as binomial; where n is the number of trials (eggs used by Joe), which is 3, and p is the probability of success (an egg being contaminated), which remains 1/4 as given in the problem. The correct distribution is therefore binomial; n = 3 and p = 1/4.It's important to observe that handwashing and cooking foods thoroughly are significant in reducing the risk of Salmonella transmission, as the bacteria cannot survive high temperatures. The use of pasteurized egg products is another safe alternative as these are treated to protect against Salmonella....