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Math Help - Probability distribution question

  1. #1
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    Probability distribution question

    Note: Sorry if this is considered a uni question, I'm not sure, so you'll have to tell me.

    The question
    A busy switchboard receives 150 calls an hour on average. Assume that the probability, p_k, of getting k calls in a given minute is:

    p_n = e^{-\lambda}\frac{\lambda^{k}}{k!}

    where \lambda = s the average number of calls per minute.

    a) Find the probability of getting exactly 3 calls in a given minute.
    b) Find th probability of getting at least 2 calls in a given minute.


    I'm not sure how to attempt this. I'm sure I'm missing something obvious, but there's no examples in this text to guide me. Any assistance would be great.
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  2. #2
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    Quote Originally Posted by Glitch View Post
    a) Find the probability of getting exactly 3 calls in a given minute.

    \displaystyle P(k=3) = e^{-150}\frac{150^{3}}{3!}

    Quote Originally Posted by Glitch View Post
    b) Find th probability of getting at least 2 calls in a given minute.
    \displaystyle P(k\geq 2) =  1-P(k=1)-P(k=0) = 1-e^{-150}\frac{150^{1}}{1!}-e^{-150}\frac{150^{0}}{0!}

    Recall 0!=1
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  3. #3
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    Your answer to a) doesn't match the textbook solution of 0.214

    Even by intuition (as bad as it tends to be for probability) it looks much too small.

    I'm yet to find a solution myself. :/
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  4. #4
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    Ok, reading it a second time, you 150 calls per hour, which is the same as 2.5 per minute....

    So

    \displaystyle P(k=3) = e^{-2.5}\frac{2.5^{3}}{3!} = 0.214


    Sorry for the confusion! Do you follow?

    Same applies to part b)
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  5. #5
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    Ahh, that's better.

    With part b, I got this far:

    e^{-2.5}\sum\limits_{k = 2}^{\infty} \frac{(2.5)^k}{k!}

    I know this converges by the ratio test, but I can't remember how to find the value it converges to. :/
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  6. #6
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    Why are you wanting to know what this converges to?

    Just find your calculator!

    \displaystyle P(k\geq K) = 1-P(k<K)

    so

    \displaystyle P(k\geq 2) = 1-P(k=1)-P(k=0) = 1-e^{-2.5}\frac{2.5^{1}}{1!}-e^{-2.5}\frac{2.5^{0}}{0!}
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  7. #7
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    It's confirmed, my brain is officially fried in this heat.

    Thanks. :P
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  8. #8
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    Quote Originally Posted by Glitch View Post
    my brain is officially fried in this heat.
    You in Melbourne as well?
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  9. #9
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    Quote Originally Posted by pickslides View Post
    You in Melbourne as well?
    Sydney. It's a cloudy day, but this house has an uncanny ability to absorb and amplify heat. :P
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