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April 20th, 2017, 09:59 AM  #1 
Newbie Joined: Feb 2017 From: Denmark Posts: 11 Thanks: 1  Math question I can't figure out
So my teacher asked the class to figure out this question: A classroom with 24 students, containing: 8 girls and 16 boys, are going to a party together, in this party there's gonna be a princess and two princes selected at the end of the night by lottery. What would the chance be for 1 girl & 2 boys being selected? This might be simple for most of you, but please don't criticize me, I just struggle when it comes to probability, chances etc. I'd love an explanation & answer if possible. 
April 20th, 2017, 10:54 AM  #2 
Senior Member Joined: Sep 2015 From: CA Posts: 1,105 Thanks: 576 
we have to assume the selections are all independent of one another. Assume we chose 2 boys and then a girl Choosing the first boy has probability $\dfrac{16}{24}$ as there are 16 boys to choose from 24 students. Similarly the probability of choosing the second boy is $\dfrac{15}{23}$ as there are now 15 boys and 23 students to choose from. Finally the probability of choosing the girl is $\dfrac{8}{22}$ The total probability is the product of these since the selections are independent, so $p = \dfrac{16}{24}\dfrac{15}{23}\dfrac{8}{22} = \dfrac {70}{253}$ one can observe that this number remains unchanged if we change the order that the students were selected. For example if we chose the girl first we'd have $p = \dfrac{8}{24} \dfrac{16}{23}\dfrac{15}{22}$ and this is the same value as that calculated by assuming the selection was bbg. 
April 20th, 2017, 12:14 PM  #3  
Newbie Joined: Feb 2017 From: Denmark Posts: 11 Thanks: 1  Quote:
Last edited by emilo0212; April 20th, 2017 at 12:38 PM.  
April 20th, 2017, 01:04 PM  #4  
Senior Member Joined: Sep 2015 From: CA Posts: 1,105 Thanks: 576  Quote:
The correct answer is 40/253 apologies  
April 20th, 2017, 01:06 PM  #5 
Newbie Joined: Feb 2017 From: Denmark Posts: 11 Thanks: 1  

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