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Math Help - Poisson as Binomial

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    Poisson as Binomial

    A textbook has n pages. The number of mistakes on a page is a poisson random variable with parameter a, and is independent of the number of mistakes on all other pages.

    Question 1: What is the expected number of pages with no mistakes?

    I get this as n*e^-a

    Question 2: When reading the book you detect each mistake with probability p, independently of other mistakes. Let M denote the number of mistakes on a particular page and let D denote the number of mistakes detected. Write down P[D = k|M = m]

    I modelled on binomial, parameters m and p

    so = m choose k * p^k(1-p)^(m-k)

    Question 3: Hence find, for each k >/= 0 P[D=k]

    I'm not sure how to start this part

    many thanks
    Last edited by mr fantastic; January 3rd 2009 at 02:09 AM. Reason: Made the questions clearer. Re-edit: There was a third question buried in there.
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    Quote Originally Posted by James0502 View Post
    A textbook has n pages. The number of mistakes on a page is a poisson random variable with parameter a, and is independent of the number of mistakes on all other pages. What is the expected number of pages with no mistakes?

    I get this as n*e^-a

    When reading the book you detect each mistake with probability p, independently of other mistakes. Let M denote the number of mistakes on a particular page and let D denote the number of mistakes detected. Write down P[D = k|M = m]

    I modelled on binomial, parameters m and p

    so = m choose k * p^k(1-p)^(m-k)

    Hence find, for each k >/= 0 P[D=k]

    I'm not sure how to start this part

    many thanks
    Calculate the probability that a single page has no mistake: p = \Pr(X = 0) = e^{-a}.

    Let Y be the random variable number of pages with no mistake.

    Y ~ Binomial \left(n, \, p = e^{-a}\right).

    E(Y) = np = n e^{-a}.
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    yeh.. I got this part.. it's the next bit that got me..

    sorry - my question was poorly written out..

    any help with question 2 would be much appreciated!
    Last edited by James0502; January 3rd 2009 at 01:40 AM.
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    Quote Originally Posted by James0502 View Post
    A textbook has n pages. The number of mistakes on a page is a poisson random variable with parameter a, and is independent of the number of mistakes on all other pages.

    Question 1: What is the expected number of pages with no mistakes?

    I get this as n*e^-a

    Question 2: When reading the book you detect each mistake with probability p, independently of other mistakes. Let M denote the number of mistakes on a particular page and let D denote the number of mistakes detected. Write down P[D = k|M = m]

    I modelled on binomial, parameters m and p

    so = m choose k * p^k(1-p)^(m-k)

    Hence find, for each k >/= 0 P[D=k]

    I'm not sure how to start this part

    many thanks
    Question 2:

    \Pr(D = k \, | \, M = m) = ^mC_k \, p^k (1 - p)^{m-k}.
    Last edited by mr fantastic; January 3rd 2009 at 02:05 AM.
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    Quote Originally Posted by James0502 View Post
    A textbook has n pages. The number of mistakes on a page is a poisson random variable with parameter a, and is independent of the number of mistakes on all other pages.

    Question 1: What is the expected number of pages with no mistakes?

    I get this as n*e^-a

    Question 2: When reading the book you detect each mistake with probability p, independently of other mistakes. Let M denote the number of mistakes on a particular page and let D denote the number of mistakes detected. Write down P[D = k|M = m]

    I modelled on binomial, parameters m and p

    so = m choose k * p^k(1-p)^(m-k)

    Question 3: Hence find, for each k >/= 0 P[D=k]

    I'm not sure how to start this part

    many thanks
    Question 3:

    If you detect k mistakes on a page then obviously m \geq k. So:

    \Pr(D = k) = \Pr(D = k \, | \, M = k) \cdot \Pr(M = k) + \Pr(D = k \, | \, M = k + 1) \cdot \Pr(M = k + 1) + \, .....


    = ^kC_k \, p^k (1 - p)^{k-k} \cdot \frac{e^{-a} a^k}{k!} + ^{k+1}C_k \, p^k (1 - p)^{k + 1 -k} \cdot \frac{e^{-a} a^{k+1}}{(k+1)!} + \, ....

    I suggest you add a couple more terms, simplify each one and take out the common factor. You're left with an infinite sum which doesn't look too bad to find I think.
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    Quote Originally Posted by mr fantastic View Post
    Question 3:

    If you detect k mistakes on a page then obviously m \geq k. So:

    \Pr(D = k) = \Pr(D = k \, | \, M = k) \cdot \Pr(M = k) + \Pr(D = k \, | \, M = k + 1) \cdot \Pr(M = k + 1) + \, .....


    = ^kC_k \, p^k (1 - p)^{k-k} \cdot \frac{e^{-a} a^k}{k!} + ^{k+1}C_k \, p^k (1 - p)^{k + 1 -k} \cdot \frac{e^{-a} a^{k+1}}{(k+1)!} + \, ....

    I suggest you add a couple more terms, simplify each one and take out the common factor. You're left with an infinite sum which doesn't look too bad to find I think.
    I will leave it to you to show that the general term is

    \Pr(D = k \, | \, M = k + n) \cdot \Pr(M = k + n) =  ^{k+n}C_k \, p^k (1 - p)^{k + n -k} \cdot \frac{e^{-a} a^{k+n}}{(k+n)!} = \frac{p^k a^k e^{-a}}{k!} \cdot \frac{[a(1 - p)]^n}{n!}

    where n = 0, 1, 2, 3, .....

    and so your sum is ~ \frac{p^k a^k e^{-a}}{k!} \sum_{n=0}^{\infty} \frac{[a(1 - p)]^n}{n!}.

    Now recall the Maclaurin Series for e^x ......
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