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HSC 2015 MX2 Permutations & Combinations Marathon (archive) (4 Viewers)

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braintic

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Re: 2015 permutation X2 marathon

Wot... so he's wrong?
No, I'm saying that if the colour of the first ball had been noted, then the answer would be different.
It wasn't noted, so he is correct.
 

Drsoccerball

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Re: 2015 permutation X2 marathon

No, I'm saying that if the colour of the first ball had been noted, then the answer would be different.
It wasn't noted, so he is correct.
What would the answer be if it was noted?
 

seanieg89

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Re: 2015 permutation X2 marathon

A woman has two children, one of whom is a girl born on a Friday.

What is the probability that her other child is also female?

Bonus: Consider this same problem with "born on a Friday" replaced by another random condition that holds with probability q. Explain the results of how the original question depends on the parameter q.
 

Drsoccerball

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Re: 2015 permutation X2 marathon

A woman has two children, one of whom is a girl born on a Friday.

What is the probability that her other child is also female?

Bonus: Consider this same problem with "born on a Friday" replaced by another random condition that holds with probability q. Explain the results of how the original question depends on the parameter q.
its 1/2 ? assuming the first female is locked in
 

InteGrand

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Re: 2015 permutation X2 marathon

its 1/2 ? assuming the first female is locked in
Nope, there's lots of famous probability Q's like these (famous because the answers are counterintuitive for most people).

(E.g. Search up the Monty Hall Problem.)
 

kawaiipotato

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Re: 2015 permutation X2 marathon

A woman has two children, one of whom is a girl born on a Friday.

What is the probability that her other child is also female?

Bonus: Consider this same problem with "born on a Friday" replaced by another random condition that holds with probability q. Explain the results of how the original question depends on the parameter q.
I don't think the condition matters does it?
Since theres four possibilities (boy girl) (girl boy) (boy boy) (girl girl)
Then its 1/4
But id still think it is 1/2....
 
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braintic

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Re: 2015 permutation X2 marathon

I don't think the condition matters does it?
Since theres four possibilities (boy girl) (girl boy) (boy boy) (girl girl)
Then its 1/4
But id still think it is 1/2....
It would be 1/2 if you were told that a particular child (say the first-born) was a girl born on a Friday.
It is not 1/2 because you are not told which child - only that one of them satisfies this condition.

Not sure we needed the Friday part though - it would still have been a valid and equally unintuitive question without that.
 
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Drsoccerball

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Re: 2015 permutation X2 marathon

It would be 1/2 if you were told that a particular child (say the first-born) was a girl born on a Friday.
It is not 1/2 because you are not told which child - only that one of them satisfies this condition.

Not sure we needed the Friday part though - it would still have been a valid and equally unintuitive question without that.
But we are told that she was born on friday so the first part would be 1/2 right?
 

seanieg89

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Re: 2015 permutation X2 marathon

The first part is not 1/2, although as braintic said, it would be if the statement was something like "my eldest child is a girl born on a Friday".

Part of the difficulty in this classic question is posed by potentially ambiguous wording.

Consider the alternate wording which may be a little less tricky to interpret:
"I have two children, at least one of whom is a girl that is born on a Friday. What is the probability that my other child is also a girl?"
 

braintic

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Re: 2015 permutation X2 marathon

But we are told that she was born on friday so the first part would be 1/2 right?
No it's not. We are not told that "she" is born on a Friday, we are told that one of them is born on a Friday.
Being told that one is a "girl Friday" is not enough to distinguish between the children, because the other could be also be a girl born on a Friday.
Whereas there is only one first-born.

I believe the answer is 1/3 if we are not given the Friday info, and 13/27 if we are.
 

seanieg89

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Re: 2015 permutation X2 marathon

Yep: 1/3 and 13/27 are the correct numbers. The role of the friday information I find the most interesting, as if you replace it by a random property with probability q, then the answer is near 1/3 for large q and near 1/2 for small q (like the q=1/7 in this question). It is fun to think about why this should be the case.

(I recommend students actually calculate this still as an exercise.)
 

braintic

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Re: 2015 permutation X2 marathon

If anyone is having difficulty, you would probably have a better chance of seeing the concept if the question was rephrased in terms of counting.

Focusing on the case without the Friday info:

"Of all the two child families with at least one girl, what fraction actually have two girls?"
 

Drsoccerball

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Re: 2015 permutation X2 marathon

If anyone is having difficulty, you would probably have a better chance of seeing the concept if the question was rephrased in terms of counting.

Focusing on the case without the Friday info:

"Of all the two child families with at least one girl, what fraction actually have two girls?"
 

braintic

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Re: 2015 permutation X2 marathon

Well - yes - and that probably helps with the latter part of the question.
But here its just a matter of listing out the cases: GG, GB, BG, BB
and then eliminating the BB case.
 

Drsoccerball

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Re: 2015 permutation X2 marathon

Well - yes - and that probably helps with the latter part of the question.
But here its just a matter of listing out the cases: GG, GB, BG, BB
and then eliminating the BB case.
So whats the second part of the question asking us to find :p...
 

braintic

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Re: 2015 permutation X2 marathon

So whats the second part of the question asking us to find :p...
Well for starters, that wasn't the actual question. Can you answer the question with the Friday info included?
 

Kaido

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Re: 2015 permutation X2 marathon

Dont know if this question is hard enough to be X2, but anyways...

A classic b-day problem:
30 people in a room, what is the probability that at least 2 people have the same b-day

Answer: 71%
 
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