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    A random variable X takes integer values n=0,1,... with the following probabilites:
     P(X=n)= \frac{C}{n!2^n} , where C is a constant. Find the value of C
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    (Original post by bijesh12)
    A random variable X takes integer values n=0,1,... with the following probabilites:
     P(X=n)= \frac{C}{n!2^n} , where C is a constant. Find the value of C
    Well you know the sum of the probabilites over all the possible values X can take is "1"

    So you need to consider, 1= P(X=0) + P(X=1) + P(X=2) + ....

    If you take out the factor of C, you're left with a well known series expansion.

    If you don't recognize it, try a few basic ones you know.
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    So   C \sum_{n=0}^{\infty} \frac{1}{n!2^n} = 1

    And so :

      \sum_{n=0}^{\infty} \frac{1}{n!2^n} = 1 / C

    I'm not sure how to deal with the infinite sum.


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    Thanks I've understood how to do the summation bit. Taylor series for E^x


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    (Original post by bijesh12)
    I'm not sure how to deal with the infinite sum.
    If I rewrite the sum as \displaystyle \sum_0^\infty \dfrac{1}{n!2^n} = f(\frac{1}{2}), , where f(x) = \sum_0^\infty \dfrac{x^n}{n!}, does that help?
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    (Original post by DFranklin)
    If I rewrite the sum as \displaystyle \sum_0^\infty \dfrac{1}{n!2^n} = f(\frac{1}{2}), , where f(x) = \sum_0^\infty \dfrac{x^n}{n!}, does that help?
    Yes thanks I've figured it out to be Taylor series for e^1/2. And therefore c= 1 / sqrt (e)


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