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    Hi

    I have a question on the Poisson distribution that I am a bit stuck on. Any help on the method of such a question would be great

    A take away restaurant has 2 incoming phone lines, A and B. The number of incoming calls in a 15 minute period on both lines are independent Poisson variates with means 4 and 5 respectively. Find the probability that in a 15 minute period:
    a) More than 3 calls come in on line A
    b) Less than 5 calls come in on line B
    c) Exactly 9 calls come in to the restaurant.

    I have done parts a and b without any bother, but am not sure about the method for part c. I was thinking I may have to find the probabilities of each way of making a total of 9 calls (i.e. 0 and 9, 1 and 8, etc.), and multiply them (i.e. P(a=0)*P(B=9), before adding all of them together???

    Thanks in advance
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    (Original post by Labrador99)
    I was thinking I may have to find the probabilities of each way of making a total of 9 calls (i.e. 0 and 9, 1 and 8, etc.), and multiply them (i.e. P(a=0)*P(B=9), before adding all of them together???

    Thanks in advance
    Yep.

    A quick way to do it is to say that \displaystyle \mathbb{P}(A=k) = \dfrac{1}{k!}\cdot 4^k e^{-4} and \mathbb{P}(B=9-k) = \dfrac{1}{(9-k)!}\cdot 5^{9-k} e^{-5}. Then the product is just \mathbb{P}(A=k, B=9-k) = \dfrac{1}{k!(9-k)!}\cdot 0.8^k \cdot 5^9 e^{-9} so the overall probability for the question is just \displaystyle \left( \dfrac{5}{e} \right)^9 \sum_{k=0}^9 \dfrac{0.8^k}{k! (9-k)!}
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    (Original post by RDKGames)
    Yep.

    A quick way to do it is to say that \displaystyle \mathbb{P}(A=k) = \dfrac{1}{k!}\cdot 4^k e^{-4} and \mathbb{P}(B=9-k) = \dfrac{1}{(9-k)!}\cdot 5^{9-k} e^{-5}. Then the product is just \mathbb{P}(A=k, B=9-k) = \dfrac{1}{k!(9-k)!}\cdot 0.8^k \cdot 5^9 e^{-9} so the overall probability for the question is just \displaystyle \left( \dfrac{5}{e} \right)^9 \sum_{k=0}^9 \dfrac{0.8^k}{k! (9-k)!}
    Thanks so much! That was really helpful
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    (Original post by Labrador99)
    Thanks so much! That was really helpful
    Alternatively, note that if A \sim \mathrm{Poi} (4) and B \sim \mathrm{Poi} (5) then due to them being independent, (A+B) \sim \mathrm{Poi} (9) which provides the quickest way to do this question.
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    ...or, the even quicker method is to form a new variable (A + B) with Poisson distribution and lambda = 9. You are allowed to do this as the question states that A and B are independent.

    Overlapped with RDKGames
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    (Original post by old_engineer)
    ...or, the even quicker method is to form a new variable (A + B) with Poisson distribution and lambda = 9. You are allowed to do this as the question states that A and B are independent.

    Overlapped with RDKGames
    (Original post by RDKGames)
    Alternatively, note that if A \sim \mathrm{Poi} (4) and B \sim \mathrm{Poi} (5) then due to them being independent, (A+B) \sim \mathrm{Poi} (9) which provides the quickest way to do this question.
    Thank you both
 
 
 
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