unknow: A group of 8 friends (5 girls and 3 boys) plans to watch a movie, but they have only 5 tickets. If they randomly decide who will watch the movie, what is the probability that there are at least 3 girls in the group that watch the movie?
First, the number of ways you can choose 5 people from 8 is given by 8!/(5!*3!) which is
8!/(5!*3!)
There will be at least three girls except in the situation when all three boys are chosen. The number of ways can you choose 3 boys from 5 people is 5!/(3!*2!) which is
5!/(3!*2!)
So the probability that there are at least 3 girls is (56 - 10)/56 or 46/56 = 0.821
I've assumed you know that the exclamation mark stands for "factorial". That is n! = n*(n-1)*(n-2)*....*2*1
Just in case you are uncertain about the above formulae, I'll illustrate all the possible ways of choosing 3 boys and 2 girls below.
Each column below represents the tickets in the order in which they are drawn.
b b b g g ... 1
b b g b g ... 2
b b g g b ... 3
b g b b g ... 4
b g b g b ... 5
b g g b b ... 6
g b b b g ... 7
g b b g b ... 8
g b g b b ... 9
g g b b b ... 10
+33616 
+118613 
