Ask your own question, for FREE!
Mathematics 20 Online
OpenStudy (anonymous):

FAN+MEDAL PLEASE HELP! 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?

OpenStudy (anonymous):

@Hero

OpenStudy (anonymous):

@ganeshie8

OpenStudy (anonymous):

@goformit100

OpenStudy (anonymous):

@freckles

OpenStudy (anonymous):

@Robert136

OpenStudy (anonymous):

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.

OpenStudy (anonymous):

there u go @alyssagoodm

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!