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Math Help - Poisson Distribution Help

  1. #1
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    Poisson Distribution Help

    Fire alarms go off in a building an average of 10 times per year. Find the probability of MORE than one fire alarm in the month of December.



    So far I have 10 alarms/12 months= 5/6 alarms per month

    So the probability of one alarm going off in December is 5/6 and the probability of MORE than one is 1/6?

    idk, is this correct?
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  2. #2
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    Re: Poisson Distribution Help

    Where did you use the Poisson distribution? The average (mean) number of alarms in a month is 5/6. That does NOT mean the probability of an alarm in one month is 5/6.

    The Poisson probability distribution with mean \lambda is P(x= k)= \frac{\lambda^ke^{-\lambda}}{k!}.

    Also "more than 1" means "not 0 or 1". You forgot to include the probability of 0 alarms in a month.
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  3. #3
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    Re: Poisson Distribution Help

    Alright so I did the Poisson distribution and have a probability of .5654 for "not zero" and .6378 for "not 1"

    So to get the probability of "not zero or one", do I just multiply the two probabilities together to get .3606?

    Thanks for the help btw
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  4. #4
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    Re: Poisson Distribution Help

    No because "not zero" includes many of the cases of "not 1". Instead find the probability of "is zero" and "is one" and add to get the probability of "either zero or one". Then subtract that from 1.
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  5. #5
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    Re: Poisson Distribution Help

    Ok I think I get it now. Thank you. I came up with an answer of .2054 if you'd like to check my work
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