Question

In: Statistics and Probability

Health officials routinely check sanitary conditions of restaurants. Assume you visit a popular tourist spot and...

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

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?

Solutions

Expert Solution

We are given that in 3 out of every 7 restaurants checked, there were unsanitary health conditions found. This means the probability that a randomly selected restaurant is with unsanitary health conditions is 3/7.

Let X be the number of restaurants out of 10 that are with unsanitary health conditions. We can say that X has a Binomial distribution with parameters, number of trials n=10 and success probability (the probability that a restaurant is with unsanitary condition) p=3/7

The Binomial probability of X=x restaurants out of 10 are with unsanitary health conditions is given by

The probability of eating in at least one restaurant with unsanitary conditions, out of 10 times that you eat out is same as the probability of X is 1 or more.

We know that

Probability(Eat at no restaurant, which is unhealthy) is the complement of Probability(eat at 1 or more restaurant, which is unhealthy), that means

Probability(Eat at no restaurant, which is unhealthy)+Probability(eat at 1 or more restaurant, which is unhealthy)=1

or

P(X=0) +P(X>=1) =1

We can write the probability as

ans: The probability of eating in at least one restaurant with unsanitary conditions, out of 10 times that you eat out is 0.9963


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