Question

A company sells packages of cookies with the following promotion. If two cookies in a package are frostingless, you may get your money back. Suppose each cookie has a 1 100 chance of being frostingless and there are 20 cookies in a package.

(a) What is the expected number of frostingless cookies in a package? 5 packages?

(b) What is the variance in this number for one package? 5 packages?

(c) What is the probability of returning at least one package if you purchase 5?

Answer 1

Let p be the probability of a cookie being frostingless.

then, p = 1/100

a) number of trials = 20 per package

then, Using normal approximation of a binomial problem,

expected number of frostingless cookies in a package = mean of X = n*p = 20*(1/100) = 1/5 = 0.2 (answer)

and expected number of frostingless cookies in 5 package(or 100 cookies) = mean of X = n*p = 100*(1/100) = 1 (answer)

b) again, Using normal approximation of a binomial problem,

variance of number of frostingless cookie in one package = n*p*(1 - p) = 20*(1/100)*(99/100) = 0.198(answer)

and variance of number of frostingless cookie in 5 package(100cookies) = n*p*(1 - p) = 100*(1/100)*(99/100) = 0.99(answer)

c) from bnomial probability distribution using N = 20 , p = success probability = 1/100 = 0.01

probability of returning a package = probability of getting 2 or more frostingless cookies in a package(20 cookies) = P(X > 2) = 1 - P(X =0) - P(X=1) = 1 - 20C0 * (1/100)0 * (99/100)20 - 20C1 * (1/100)1 * (99/100)19 = 0.01686

Now again using binomial probability with N = 5 , p = probability of returning a package = 0.01686

P(returning at least one package if you purchase 5) = P(X > 1) = 1 - P(X =0) = 1 - 5C0 * (0.01686)0 * (1 - 0.01686)5 = 0.081505(answer)

here is a image in case you need it