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# S2 Poisson vehicles in both directions, >2 arrive watch

1. I can't get to the answer in book.

Question: A petrol station has service areas on both sides of a motorway, one to serve east-bound traffic and the other for west-bound traffic. The number of east-bound vehicles arriving at the station in one minute has a Poisson distribution with mean 1.6, the two distributions being independent.

(i) (b) Find the probability that in a one-minute period more than two vehicles arrive at this petrol station.

My working is like this.

East bound vehicles per minute mean = 0.9
West bound vehicles per minute mean = 1.6

Let T = total vehicles per minute (ie in both directions)

T ~ Poisson(2.5)

P(T > 2) = 1 - P(0) - P(1) or 1 - P(1)cumulative

= 1 - 0.2873 = 0.713 (3 d.p.)

[0.2873 is Poisson mean=2.5, X = 1]

I can't see how else you would do this question???

Book answer is 0.456.

Can any one point me in the right direction?
2. (Original post by acomber)

P(T > 2) = 1 - P(0) - P(1) or 1 - P(1)cumulative

Can any one point me in the right direction?
"P(T > 2) = 1 - P(0) - P(1)"

That's not correct - as in it's incomplete on the RHS.

Sufficient?
3. (Original post by ghostwalker)
"P(T > 2) = 1 - P(0) - P(1)"

That's not correct - as in it's incomplete on the RHS.

Sufficient?
I feel so dumb! Sorry.
4. (Original post by acomber)
I feel so dumb! Sorry.
I take it from that, that you see the problem.

P(T > 2) = 1 - P(0) - P(1) - P(2)
5. (Original post by ghostwalker)
I take it from that, that you see the problem.

P(T > 2) = 1 - P(0) - P(1) - P(2)
Yes, I couldn't see for looking. Thanks for pointing out. Attention to detail!

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