Binomial Probability (1 Viewer)

[ND]

Originally posted by Affinity
ND.. why didn't you take it to a 3rd year MATHS student with straight HDs?

yeah I am at UNSW.. by the way, you wouldn't happen to be a very good chess player would you?

I don't know any 3rd year maths students, i think the maths dudes are all over at NSW.

Good chess player? i'm not bad but i never really play. I'm not the person you're thinking of (if that's why you're asking).

Not to be arrogant but that question wasn't hard.. took me about 5 mins..
the Actuarial science students at MQ probably just couldn't be bothered to think about the question.

I've either gotten worse at maths, or OLDMAN's q's are getting harder (though after seeing the solution it doesn't appear too hard), cos last year i could actually do OLDMAN's questions. (or maybe it's just the fact that i really suck at probability. kinda ironic that i'm doin actuarial eh?)

Hi ND. Its the old class back. Lets see, we've got the actuarials, the mathematicians, the Year 12 wannabes . ,but we don't have the medicos yet.

Heheh, give it some time, i'm sure the medics will come.

 

[Grey Council]

oo, fair enough Affinity.

And what I meant by typing it up was typing it up in plain sight. ^_^

what i'm surprised at is Xayma. He quoted archman, but didn't see what archman had written. hmph

 

[Xayma]

Originally posted by Affinity
Archman typed his answer up if you haven't noticed.. it's hidden in his other post

Yeah It was just sifting through all that working I couldnt figure out which was the actual answer.

Originally posted by Grey Council
what i'm surprised at is Xayma. He quoted archman, but didn't see what archman had written. hmph

I saw what he had written, going through it I couldnt figure out which one was the answer.

 

[OLDMAN]

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Throw a set of N biased coins, with the probability(head) for nth coin being 1/(2n+1) (for n=1,2,...N). What is the probability of an odd number of heads. Note: no longer a binomial distribution.
___________________________________________________

The negative terms in the expansion of
(2/3-1/3)(4/5-1/5)(6/7-1/7)(8/9-1/9)...(2N/(2N+1)-1/(2N+1)) correspond to an odd number of heads.
The expansion "telescopes" to 1/(2n+1).
Since P(even)+P(odd)=1 and P(even)-P(odd)=1/(2n+1)
then P(odd)=n/(2n+1)

btw the first easier problem could also be done this way, if the Yr 12s care to practise.

 

[Xayma]

Ahh ok is that the answer, I forgot that it was 1/(2n+1) not n/(2n+1)

 

[Affinity]

OLDerMAN always have some odd tricks up his sleeves.. I feel so inferior

 

[OLDMAN]

Affinity : I am not saying this as balm, because I was about to say it before I posted this sol. I have noticed your superior (to me, that is ) adeptness in handling difficult series in previous threads together with this one.

 

[turtle_2468]

Affinity is a good chess player.
And, archman, I will see you soon

 

[Archman]

Originally posted by turtle_2468
And, archman, I will see you soon

Indeed. See you very soon

To oldman: very nice solution.

 

[flyin']

Originally posted by ND
I don't know any 3rd year maths students, i think the maths dudes are all over at NSW.

I'll be studying third year Mathematics in two years time. In the meantime, I think the 3rd year acst people have other things to worry about.

If I weren't studying 3rd year acst next year, I'd be doing 3rd year maths which doesn't involve probability. That's stat, which I'm (spose to be) doing now.



turtle is a medic.