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VCE Stuff => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematics => Topic started by: /0 on September 02, 2009, 11:13:24 pm

Title: Probability of a coin
Post by: /0 on September 02, 2009, 11:13:24 pm: A coin is tossed $n$ times. Let X denote the number of heads in the first $p$ tosses and Y denote the number of heads in the last $q$ tosses.

How is the probability distribution for (X,Y) generated?

(any ideas, even speculations would be great)

I've tried using PIE but it doesn't seem to apply too well...
Title: Re: Probability of a coin
Post by: Mao on September 02, 2009, 11:47:25 pm: $Pr[(X,Y) = (x,y)] = ^pC_x \cdot ^{q}C_y \cdot 2^{-n}$ (the product of the two respective binomial dist, as they are independent events)
Title: Re: Probability of a coin
Post by: /0 on September 03, 2009, 12:00:41 am: thanks, mao but unfortunately I don't think they are independent events... if 'p' and 'q' overlap (p+q > n), then the value X takes will affect the probability of what value Y takes.

Sorry if it's an impractical question... I got it from the special case n = 4, p = 3, q = 3 (in an actuarial text where they provided the P.D. without proof), and I figured that if there was a systematic way of generating a P.D. then it could be easily generalised with variables.
Title: Re: Probability of a coin
Post by: Mao on September 03, 2009, 12:27:18 am: Quote from: /0 on September 03, 2009, 12:00:41 am
thanks, mao but unfortunately I don't think they are independent events... if 'p' and 'q' overlap (p+q > n), then the value X takes will affect the probability of what value Y takes.

Sorry if it's an impractical question... I got it from the special case n = 4, p = 3, q = 3 (in an actuarial text where they provided the P.D. without proof), and I figured that if there was a systematic way of generating a P.D. then it could be easily generalised with variables.
ooh THAT's what you mean...

Break it down into three independent events?

suppose there is overlap, i.e. overlap = p + q - n > 0, hence the three independant events occurs over n-q, p + q - n, and n-p (if there were no overlap, they can be treated as two individual events)
now suppose N heads were achieved overall, then N-Y heads were achieved in the first, X + Y - N in the middle, N - X in the last.
The restrictions are hard to deal with. To start, the minimum N occurs when as much of X and Y fit in the first and last section. the maximum N occurs when as much of X and Y fit in the middle section.

Then the probability becomes $Pr[(X,Y) = (x,y)] = \sum_{N} \binom{n-q}{N-y}\binom{p+q-n}{x + y - N}\binom{n-p}{N-x}2^{-n}$
And by brute force [considering all cases of how things can distribute across the three sets], a minimum and maximum of N can be determined.
This is not elegant, but it is one way to work it out, and especially easy with some logical statements and a processor.
Title: Re: Probability of a coin
Post by: /0 on September 03, 2009, 12:51:24 am: Wow Mao, I didn't quite expect the answer to be so complicated, thanks though
Title: Re: Probability of a coin
Post by: Mao on September 03, 2009, 08:36:04 am: I think there should be a simpler and more elegant way. I just brute-forced it. (and the restrictions on N are horrible, and the expression changes like a billion times depending on what X, Y, p, q and n are...)
Title: Re: Probability of a coin
Post by: TrueTears on September 03, 2009, 09:35:59 pm: Quote from: /0 on September 02, 2009, 11:13:24 pm
A coin is tossed $n$ times. Let X denote the number of heads in the first $p$ tosses and Y denote the number of heads in the last $q$ tosses.

How is the probability distribution for (X,Y) generated?

(any ideas, even speculations would be great)

I've tried using PIE but it doesn't seem to apply too well...
Oh, this question in the actuarial book.