Binomial Probability (2 Viewers)

[OLDMAN]

An interesting binomial probability question with a twist.

A die is thrown n times. What is the probability of getting an odd number of sixes.

Here's the twist : use (p-q)^n rather than the usual binomial distribution of (p+q)^n.

 

[Archman]

is it:
(1-(2/3)^n)/2 ?

(highlight to see clearly)

edit: damn the colour didnt come out too well.

 

[McLake]

Originally posted by Archman
(highlight to see clearly)

My highlight colour is gray, it dosn't really work ...

TIP: To hide answers put them in fake html tags (to see what I mean "quote" my post)

<A fake Tag can older answers like this>
<Answer = 42>

 

[Grey Council]

lol, nice mclake. ^__^
and is Archman right?

 

[Affinity]

I think ..<I agree with archman> <-- fake html tag</I>

anyway, extension question
find the probability such that the number of sixes rowed is a multiple of 6

 

[McLake]

Originally posted by Affinity
I think ..<I> <-- fake html tag</I>

Your "fake" tag has turned into an italics tag. Oh well ...

 

[Xayma]

Originally posted by McLake
Your "fake" tag has turned into an italics tag. Oh well ...

Wasn't he indicating that there is a fake html tag after I think...

 

[OLDMAN]

Too easy for archman. Raises the bar a touch.

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.

 

[ND]

Originally posted by OLDMAN
Too easy for archman. Raises the bar a touch.

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.

I couldn't think of a decent way of doing it, so i took it to a 3rd year actuarial student for ideas - he couldn't find a decent way of doing it, so i took it to a 3rd year actuarial with straight HD's - he couldn't find a decent way of doing it. Let's see if archman can do it.

 

[Affinity]

ND.. missed you, where are you now?

 

[ND]

At macq doing actuarial. I haven't been posting much, but occasionally come here and look for probability questions (cos the ones in the stat textbook are all trivial). You doing actuarial too? (i'm asking cos of your sig)

 

[Archman]

Originally posted by McLake
My highlight colour is gray, it dosn't really work ...

TIP: To hide answers put them in fake html tags (to see what I mean "quote" my post)

woah, nice!.. <It is 42 indeed>

Originally posted by Affinity
anyway, extension question
find the probability such that the number of sixes rowed is a multiple of 6

Heres what i got, havent been able to simplify it yet. not sure if its right tho..
let w be the 6th root of unity, probability = SUM m from 0 to 5 (5/6+1/6w^m)^n /6

Originally posted by OLDMAN
Too easy for archman. Raises the bar a touch.

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.

Here we go:
n/(2n+1). Proof by induction: obviously true for n=1, now assume true for n=k-1, hence the probability for k-1 coins is (k-1)/(2k-1). now for n=k, if the last coin is tails, then there is an odd number of heads in the first k-1 coins, hence the probability for this is (k-1)/(2k-1) * 2k/(2k+1). if the last coin is heads, then there must be an even number of heads in the first k-1 coins, the probablity for that is (1 - (k-1)/(2k-1)) * 1/(2k+1) = k/(2k-1) * 1/(2k+1)
Adding the two:
(k-1)/(2k-1) * 2k/(2k+1) + k/(2k-1) * 1/(2k+1)
= ((2k-2)k + k)/(2k-1)(2k+1)
= (2k-1)k/(2k-1)(2k+1)
= k/(2k+1)
yay!

 

[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?

 

[Xayma]

Originally posted by Archman
In case you havent noticed yet, i've hidden quite a few stuff here... jeez this business can get addictive

Hmm so what exactly is the answer I cant seem to find it hidden there?

 

[Archman]

there, all fixed

 

[Grey Council]

jeeezaa! And i have to compete against archman at school. And he tells me that i MIGHT come first in 4u maths.
*mutter mutter*
fat chance.

Is Archman right? And Archman, you might as well just type the answer up, i mean, if 3rd year actuarial can't do it, i REALLY doubt any of us are gonna even try. :-\

 

[Giant Lobster]

is archman the king of 4u in ur school?
if so, whoa.. ive heard about this guy

 

[Affinity]

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.

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

 

[Affinity]

here's one without induction, bit sketchy though:

[(x+2)(x+4)...(x+2n) + (-1^n)(x-2)(x-4)...(x-2n)]/2
=sum of the terms involving odd powers of x in the expansion of
[(x+2)(x+4)...(x+2n)]

this can be proved by expansion (remember the relation between sum of zeroes and coefficients)

if x = 1 in the above equality, LHS = 3*5*...*(2n-1) * [n]

RHS = sum of coefficients of the odd powers

to determine the probability we need to find the sum of the
coefficients of odd power of x in the expansion of
(x/3 + 2/3) * (x/5+4/5) * ... *(x/(2n+1) + 2n/(1+2n))

= [(x+2)(x+4)...(x+2n)]/(3*5*...*(2n+1)]

and using the result shown earlier

= n/(2n+1)

 

[OLDMAN]

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.

Archman and Affinity have the right answer. But it should only be a three line affair -though admittedly, I can't count up to 3 .