The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

Poisson Distribution Question

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 :smile:

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 :smile:
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 :smile:


Yep.

A quick way to do it is to say that P(A=k)=1k!⋅4ke−4\displaystyle \mathbb{P}(A=k) = \dfrac{1}{k!}\cdot 4^k e^{-4} and P(B=9−k)=1(9−k)!⋅59−ke−5\mathbb{P}(B=9-k) = \dfrac{1}{(9-k)!}\cdot 5^{9-k} e^{-5}. Then the product is just P(A=k,B=9−k)=1k!(9−k)!⋅0.8k⋅59e−9\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 (5e)9∑k=090.8kk!(9−k)!\displaystyle \left( \dfrac{5}{e} \right)^9 \sum_{k=0}^9 \dfrac{0.8^k}{k! (9-k)!}
(edited 6 years ago)
Original post by RDKGames
Yep.

A quick way to do it is to say that P(A=k)=1k!⋅4ke−4\displaystyle \mathbb{P}(A=k) = \dfrac{1}{k!}\cdot 4^k e^{-4} and P(B=9−k)=1(9−k)!⋅59−ke−5\mathbb{P}(B=9-k) = \dfrac{1}{(9-k)!}\cdot 5^{9-k} e^{-5}. Then the product is just P(A=k,B=9−k)=1k!(9−k)!⋅0.8k⋅59e−9\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 (5e)9∑k=090.8kk!(9−k)!\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 :smile:
Original post by Labrador99
Thanks so much! That was really helpful :smile:


Alternatively, note that if A∼Poi(4)A \sim \mathrm{Poi} (4) and B∼Poi(5)B \sim \mathrm{Poi} (5) then due to them being independent, (A+B)∼Poi(9)(A+B) \sim \mathrm{Poi} (9) which provides the quickest way to do this question.
...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
(edited 6 years ago)
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∼Poi(4)A \sim \mathrm{Poi} (4) and B∼Poi(5)B \sim \mathrm{Poi} (5) then due to them being independent, (A+B)∼Poi(9)(A+B) \sim \mathrm{Poi} (9) which provides the quickest way to do this question.


Thank you both :smile:

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you