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 Post subject: simple poisson distribution questionPosted: Sun, 22 Apr 2012 02:16:09 UTC
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Joined: Wed, 15 Feb 2012 07:19:33 UTC
Posts: 31
A garage has 2 vans and 3 cars, which can be hired out for a day at a time. Requests for the hire of a van follow a Poisson distribution with a mean of 1.5 requests per day and requests for the hire of a car follow an independent Poisson distribution with a mean of 4 requests per day.

Find the probability that, on any particular day, there is at least one request for a van and at least two requests for a car, given that there are a total of 4 requests on that day. (The answer is 0.286, apparently.)

My method:

Let V be number of requests for a van per day. Then V ~ Po (1.5).
Let C be number of requests for a car per day. Then C ~ Po (4).
Let T be total number of requests per day. Then T ~ Po (5.5).

Required probability =
= 0.656.

Can someone explain where I went wrong? Thanks!

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 Post subject: Re: simple poisson distribution questionPosted: Sun, 22 Apr 2012 02:18:00 UTC
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Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 14006
Location: Austin, TX
thesocialnetwork wrote:
A garage has 2 vans and 3 cars, which can be hired out for a day at a time. Requests for the hire of a van follow a Poisson distribution with a mean of 1.5 requests per day and requests for the hire of a car follow an independent Poisson distribution with a mean of 4 requests per day.

Find the probability that, on any particular day, there is at least one request for a van and at least two requests for a car, given that there are a total of 4 requests on that day. (The answer is 0.286, apparently.)

My method:

Let V be number of requests for a van per day. Then V ~ Po (1.5).
Let C be number of requests for a car per day. Then C ~ Po (4).
Let T be total number of requests per day. Then T ~ Po (5.5).

Required probability =
= 0.656.

Can someone explain where I went wrong? Thanks!

You neglected to incorporate that necessarily 4 requests were made in the events in the numerator. You'll need to use conditional probabilities.

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 Post subject: Re: simple poisson distribution questionPosted: Sun, 22 Apr 2012 02:20:49 UTC
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Joined: Wed, 15 Feb 2012 07:19:33 UTC
Posts: 31
But I did incorporate them - aren't the two events in the numerator the only two events that could happen given the conditions in the question? There aren't any other instances where the total requests add up to four, with a different number of requests for vans/cars, are there?

P(A|B) = P(A B)/P(B)

in the above, the probabilities of the events are the probabilities of the only events that could happen given a) four requests in total b) at least one request for a van c) at least 2 request for a car

hence isn't that the 'intersect' of all three conditions?

Last edited by thesocialnetwork on Sun, 22 Apr 2012 02:27:13 UTC, edited 3 times in total.

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 Post subject: Re: simple poisson distribution questionPosted: Sun, 22 Apr 2012 02:25:00 UTC
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Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 14006
Location: Austin, TX
thesocialnetwork wrote:
But I did incorporate them - aren't the two events in the numerator the only two events that could happen given the conditions in the question? There aren't any other instances where the total requests add up to four, with a different number of requests for vans/cars, are there?

Ah, I see what's going on here. No. Why do the number of requests have to be limited by the options for the vehicles? Just because they only have 2 vans why do they only have to get 2 requests at most? Also I would still use conditional probabilities in this, because with the added information that there are 4 requests, the number of van and car requests are no longer independent.

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 Post subject: Re: simple poisson distribution questionPosted: Sun, 22 Apr 2012 02:32:31 UTC
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Joined: Wed, 15 Feb 2012 07:19:33 UTC
Posts: 31
hm in my view, i was using conditional probability. An exemplification of what I mean is given below:

probability of A occurring given that B had occurred = P(A|B) = P(A B)/P(B). In the numerator, that's "intersect" of events A and B, meaning they both happen.

P(A|B) = P(A B)/P(B)

in the above, the probabilities of the events are the probabilities of the only events that could happen given a) four requests in total b) at least one request for a van c) at least 2 request for a car

hence isn't that the 'intersect' of all three conditions? (Your point about "why should no. of requests be limited by vehicles" is taken into account in the denominator isn't it?)

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