Bob sells home insurance policies by visiting potential clie

Bob sells home insurance policies by visiting potential clients door to door. At each home, the probability of making a sale (given his track record) is 20%.

a)If Bob visits 5 homes, what is the probability that he will sell exactly 3 policies?

b)If Bob visits 5 homes, what is the probability that he will sell 3 policies or more? (5 points)

c)Bob has a quota of 10 policies for a month. How many homes he needs to visit to have a 90% chance of making his quota

Solution

a)If Bob visits 5 homes, what is the probability that he will sell exactly 3 policies?

Given X follows Binomial distribution with n=5 and p=0.2

P(X=x)=5Cx*(0.2^x)*(0.8^(5-x)) for x=0,1,2,3,4,5

So P(X=3) =5C3*(0.2^3)*(0.8^(5-3)) =0.0512

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b)If Bob visits 5 homes, what is the probability that he will sell 3 policies or more? (5 points)

P(X>=3) = P(X=3)+P(X=4)+P(X=5)

=5C3*(0.2^3)*(0.8^(5-3))+...+5C5*(0.2^5)*(0.8^(5-5))

=0.05792

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c)Bob has a quota of 10 policies for a month. How many homes he needs to visit to have a 90% chance of making his quota

P(X<=2) =0.94208

So n=2