Question

1) Health officials routinely check sanitary conditions of restaurants. Assume you visit a popular tourist spot and read in the newspaper that in 3 of every 7 restaurants checked, there were unsanitary health conditions found. Assuming you are planning to eat out 10

times while you are there on vacation, answer the following questions:
a) How likely is it that you will eat at three restaurants with unsanitary conditions?
b) How likely is it that you will eat at 4 or 5 restaurants with unsanitary conditions?
c) Explain how you would compute the probability of eating in at least one restaurant with unsanitary conditions. Could you use the complement to solve this problem?
d) What is the most likely number to occur in this experiment?
e) How variable will the data be around the most likely number?
f) Is this is a binomial distribution?
g) If it is a binomial distribution, does that mean that the likelihood of success is always
50% since there are only two possible outcomes?

Answer 1

a) The probability of eating at a restaurant with unsanitary conditions is 3/7. If you plan to eat at 10 different restaurants, the probability of eating at exactly 3 restaurants with unsanitary conditions is given by the binomial probability formula:

P(X = 3) = (10 choose 3) * (3/7)^3 * (4/7)^7 = 120 * (0.42857)^3 * (0.57143)^7 = 0.1437

So, the likelihood of eating at three restaurants with unsanitary conditions is 14.37%.

b) The probability of eating at 4 or 5 restaurants with unsanitary conditions can be found by summing the binomial probabilities for each case:

P(X = 4) = (10 choose 4) * (3/7)^4 * (4/7)^6 = 210 * (0.42857)^4 * (0.57143)^6 = 0.1735

P(X = 5) = (10 choose 5) * (3/7)^5 * (4/7)^5 = 252 * (0.42857)^5 * (0.57143)^5 = 0.1428

P(X = 4 or X = 5) = P(X = 4) + P(X = 5) = 0.1735 + 0.1428 = 0.3163

So, the likelihood of eating at 4 or 5 restaurants with unsanitary conditions is 31.63%.

c) The probability of eating in at least one restaurant with unsanitary conditions can be found by subtracting the probability of eating in no restaurants with unsanitary conditions from 1. The probability of eating in no restaurants with unsanitary conditions is given by the binomial probability formula:

P(X = 0) = (10 choose 0) * (3/7)^0 * (4/7)^10 = 1 * (1) * (0.57143)^10 = 0.0139

P(X >= 1) = 1 - P(X = 0) = 1 - 0.0139 = 0.9861

So, the probability of eating in at least one restaurant with unsanitary conditions is 98.61%.

d) The most likely number is 3, as it has the highest probability of occuring.

e) The data is not very variable around the most likely number, as the probability of the other numbers is relatively low.

f) Yes, this is a binomial distribution as the experiment has only two possible outcomes, "success" (eating at a restaurant with unsanitary conditions) or "failure" (eating at a restaurant with sanitary conditions) and the number of trials is fixed (10)

g) No, the likelihood of success is not always 50% in a binomial distribution. The probability of success is given by the proportion of successful trials in the population, which in this case is 3/7.