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April 20th, 2017, 09:59 AM   #1
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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.
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April 20th, 2017, 10:54 AM   #2
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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.
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April 20th, 2017, 12:14 PM   #3
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Quote:
Originally Posted by romsek View Post
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.
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April 20th, 2017, 01:04 PM   #4
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Quote:
Originally Posted by emilo0212 View Post
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
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April 20th, 2017, 01:06 PM   #5
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Originally Posted by romsek View Post
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 =)

Have a nice day/night
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