SOLUTION: There's a faulty vending machine which, when you buy an item, has a two in three chance of giving you two of that item when you only pay for one. Annette buys a bag of chips from t
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Question 1124880: There's a faulty vending machine which, when you buy an item, has a two in three chance of giving you two of that item when you only pay for one. Annette buys a bag of chips from that machine every morning.
In the upcoming work week (Monday through Friday), what's the probability that Annette will get a 'bonus' bag of chips on Monday and Friday but not on the other days?
In the upcoming work week (Monday through Friday), what's the probability that Annette will get a 'bonus' bag of chips on any two (and only two) of the days?
Enter your answers as whole numbers or fractions in lowest terms.
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
on any given day, the probability that the vending machine will give her 2 bags when she paid for one would be 2 out of 3 = 2/3.
she buys a bag of chips every morning.l
therefore, the probability every day is 2/3.
there are 5 days in the week.
if the probability is that she will get an extra bag is 2/3, then the probability that she will not get an extra bag is 1/3.
the probability that she will get a double bag on monday and friday, but not on tuesday and wednesday and thursday, is therefore.
2/3 * 1/3 * 1/3 * 1/3 * 2/3 = .0164609053.
the probability that she will get a double bag on any two days, but not on the other three, will be:
2/3 * 2/3 * 1/3 * 1/3 * 1/3 * c(5,2) = .1646090535.
this is multiplied by c(5,2) because that's the number of ways you can get a set of 2 out of a set of 5.
c(5,2) = 5! / (2! * 3!) = (5*4*3*2*1) / (2*1*3*2*1) = (20 * 3!) / (2*1*3!) = 20 / (2*1) = 10
let 2 represent 2/3 and let 1 represent 1/3 and those 10 ways are:
2,2,1,1,1
2,1,2,1,1
2,1,1,2,1
2,1,1,1,2 ***** monday and friday only
1,2,2,1,1
1,2,1,2,1
1,2,1,1,2
1,1,2,2,1
1,1,2,1,2
1,1,1,2,2
in other words, she get 2 bags on monday and tuesday, monday and wednesday, monday and thursday, monday and friday, tuesday and wednesday, tuesday and thursday, tuesday and friday, wednesday and thursday, wednesday and friday, thursday and friday.
this is a binomial probability type of problem.
the formula for that is p(x) = p^x * q^(n-x) * c(n,x).
the total probability is shown in the following excel spreadsheet.
the sum of all probabilities is 1, as it should be.
here's what the excel spreadsheet looks like.
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