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 April 20th, 2017, 09:59 AM #1 Newbie   Joined: Feb 2017 From: Denmark Posts: 12 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: USA Posts: 2,582 Thanks: 1427 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
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 Originally Posted by romsek 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.
Hello there, thanks for your very informative & fast reply, I just have one more question. I don't quite get how you got the "product" of the 3 fractions in the calculation above "16/24 15/23 8/22" Could you explain how you got the product and ended up with the "70/253" result, because I got a result of 40/253

Last edited by emilo0212; April 20th, 2017 at 12:38 PM. April 20th, 2017, 01:04 PM   #4
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Quote:
 Originally Posted by emilo0212 Hello there, thanks for your very informative & fast reply, I just have one more question. I don't quite get how you got the "product" of the 3 fractions in the calculation above "16/24 15/23 8/22" Could you explain how you got the product and ended up with the "70/253" result, because I got a result of 40/253
the way you get 70 is to make a typo.

The correct answer is 40/253

apologies April 20th, 2017, 01:06 PM   #5
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 Originally Posted by romsek the way you get 70 is to make a typo. The correct answer is 40/253 apologies
Noted And no problem, I do appreciate your help alot =)

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