Question 480841: A restaurant has 10 choices for appetizers, 20 choices for the main course, and 12 choices for the dessert. Of the 12 desserts half of them are sugar-free.
What is the probability that a person chooses 3 appetizers, 2 main course items, and 2 desserts in such a way that exactly both are sugar-free?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! without worrying about sugar-free desserts, the number of possible combinations are:
10C3 * 20C2 * 12C2
with worrying about sugar-free desserts, the number of possible combinations are:
10C3 * 20C2 * 6C2
the probability will be:
(10C3 * 20C2 * 6C2) / (10C3 * 20C2 * 12C2)
the 10C3 and the 20C2 in the numerator and the denominator cancel out and you are left with:
6C2 / 12C2
this becomes 15/66 which is equal to .227272727
that's the probability that you will get 2 desserts that are sugar-free with your meal.
this assumes that the order in which you receive the appetizers and the main courses and the desserts is not important.
if order is important, then you would use the permutations formula to get:
6P2 / 12P2 = .227272727
the probability is the same.
6C2 = 15
12C2 = 66
6P2 = 30
12P2 = 132
15/66 is the same ratio as 30/132
the combination formula is:
nCx = n! / ((n-x)! * x!)
the permutation formula is:
nPx = n! / (n-x)!
|
|
|
