Need help with Probability Generating Functions!!

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parisnaomi
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(I’m going to post pictures of the question, mark scheme, and my working underneath)

I need help with part (ii). I have calculated the probably distribution, but am having trouble with the rest. For the amount of marks, and the wording of the question ‘stating the values of n and p’, I feel as if my method is too long-winded. I may be forgetting something, but how am I supposed to automatically recognise the distribution as terms of the expansion (2/3 + 1/3)^4 ??

I can see that n=4, since P(X=0) + ... + P(X=4) = 1, then the number of trials must be 4. I circled in orange what I think may be the way to quickly calculate p.

I’m also wondering whether the 1/81 factored out of the PGF has anything to do with it?

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Last edited by parisnaomi; 3 weeks ago
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parisnaomi
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oops the pictures are upside down haha
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DFranklin
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(Original post by parisnaomi)
(I’m going to post pictures of the question, mark scheme, and my working underneath)

I need help with part (ii). I have calculated the probably distribution, but am having trouble with the rest. For the amount of marks, and the wording of the question ‘stating the values of n and p’, I feel as if my method is too long-winded. I may be forgetting something, but how am I supposed to automatically recognise the distribution as terms of the expansion (2/3 + 1/3)^4 ??

I can see that n=4, since P(X=0) + ... + P(X=4) = 1, then the number of trials must be 4. I circled in orange what I think may be the way to quickly calculate p.

I’m also wondering whether the 1/81 factored out of the PGF has anything to do with it?

Thank you for any help Name:  9D62BD30-750E-4FCE-BB83-6D86A3228DEB.jpeg
Views: 6
Size:  33.1 KB
When you calculate the pgf you're expanding out (t^-4) (2+t^2)^4 / 81, and you should be able to see that the coefficients you get as a consequence are \left( \frac{2}{3} + \frac{1}{3}\right)^4.
If you are given that it's a binomial, then as you say, the probability P(Y = 4) must equal p^4, but you should really have been noticing the similarity of the original expansion to a binomial anyhow.
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parisnaomi
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(Original post by DFranklin)
When you calculate the pgf you're expanding out (t^-4) (2+t^2)^4 / 81, and you should be able to see that the coefficients you get as a consequence are \left( \frac{2}{3} + \frac{1}{3}\right)^4.
If you are given that it's a binomial, then as you say, the probability P(Y = 4) must equal p^4, but you should really have been noticing the similarity of the original expansion to a binomial anyhow.
So is there a problem with the way I expanded the function? I went straight into it without factoring anything out...

I also still don't see how I'm supposed to notice the similarity of the original expansion to (2/3 + 1/3)^4, I feel like there's some fundamental rule I'm forgetting :confused:.
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DFranklin
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(Original post by parisnaomi)
So is there a problem with the way I expanded the function? I went straight into it without factoring anything out...
I also still don't see how I'm supposed to notice the similarity of the original expansion to (2/3 + 1/3)^4, I feel like there's some fundamental rule I'm forgetting :confused:.
Regarding your first question: if we were talking normal probability, it would be like being told X ~ B(4, 1/3) and saying

p(X = k) = \frac {1}{81} \binom{4}{k}  2^{4-k} instead of p(X = k) = \binom{4}{k} \left(\frac{1}{3}\right)^k \left(\frac{2}{3}\right)^{4-k}.

The first expression isn't incorrect, but it's probably not the right thing to do here, particularly if you're trying to relate the expressions you get to binomial probabilities.

As for your second question, possibly knowing the pgf for a binomial would help you. Or if you think about it, the coefficient of t^k (q + pt)^n is the "probability of getting k ts" out of n trials where you have probability q of choosing the constant term in a branch and probability p of choosing the t.

At the end of the day, this is very much about being familiar with binomial expansions and pgfs; if you are, the link is obvious (in fact, you can deduce that Y is binomially distributed purely from considering pgfs if you know what you're doing).

I think at this point your best bet is to move on and try some more problems and try and improve your intuition.
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parisnaomi
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(Original post by DFranklin)
Regarding your first question: if we were talking normal probability, it would be like being told X ~ B(4, 1/3) and saying

p(X = k) = \frac {1}{81} \binom{4}{k}  2^{4-k} instead of p(X = k) = \binom{4}{k} \left(\frac{1}{3}\right)^k \left(\frac{2}{3}\right)^{4-k}.

The first expression isn't incorrect, but it's probably not the right thing to do here, particularly if you're trying to relate the expressions you get to binomial probabilities.

As for your second question, possibly knowing the pgf for a binomial would help you. Or if you think about it, the coefficient of t^k (q + pt)^n is the "probability of getting k ts" out of n trials where you have probability q of choosing the constant term in a branch and probability p of choosing the t.

At the end of the day, this is very much about being familiar with binomial expansions and pgfs; if you are, the link is obvious (in fact, you can deduce that Y is binomially distributed purely from considering pgfs if you know what you're doing).

I think at this point your best bet is to move on and try some more problems and try and improve your intuition.
Thank you, I think I understand it more now. 'Integrate' 1/81 into the expansion so it shows as a classic binomial expansion formula?

I do know the pgf of a binomial, but somehow I couldn't relate it to the original pgf. My textbook didn't have any questions like this, they were more like 'Prove the probability generating function is this', so this exam question in particular spooked me out. I'm going to try to find more questions like this!
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