# S2 Poisson - many ways to get 4 from period of 5 units

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Thread starter 6 years ago
#1
The number of wombats that are killed on a particular stretch of road in Australia in any one day can be modelled by a Poisson (0.42) random variable.
i) Calculate the probability that exactly two wombats are killed on a given day on this stretch of road.
ii) Find the probability that exactly four wombats are killed over a 5-day period on this stretch of road.

Part i) is easy enough.

P(X = 2) = e^-0.42 x 0.42^2/2! = 0.0580 - I can do that part ok.

But part ii) seems to have so many possible permutations.

You have 5 days.

So you could have 4 on day one (I can calculate that probability) and no wombats killed on other days. Or nothing day 1, 4 wombats killed day 2, then no more wombats killed for remaining 3 days. I think that is nCr = 5C4 = 5. Yes?

But that is only 4 and nones. What about eg 3 on day 1, then 1 next day, then nothing. so all those combinations. Is there a way to work out how many of those there are?

There must be a way to do this. can someone help please.

By the way I found this link:

http://www.thestudentroom.co.uk/show....php?t=1240369

which says:

ii) extend the model from 1 day to 5 days...multiply the mean by 5 and use this in the Possum formula.

How does that work?

E(X) for 5 day is 5x0.42=2.1

But I don't understand.

By the way, the answer is supposed to be 0.099. Board is MEI.
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6 years ago
#2
(Original post by acomber)
ii) Find the probability that exactly four wombats are killed over a 5-day period on this stretch of road.

You have 5 days.

So you could have 4 on day one (I can calculate that probability) and no wombats killed on other days. Or nothing day 1, 4 wombats killed day 2, then no more wombats killed for remaining 3 days. I think that is nCr = 5C4 = 5. Yes?

But that is only 4 and nones. What about eg 3 on day 1, then 1 next day, then nothing. so all those combinations. Is there a way to work out how many of those there are?

There must be a way to do this. can someone help please.
This way does work, eventually. I prefer the following modification: it's P(four on the first day) P(none in three days) + P(three on the first day) P(one in three days) + … + P(none on the first day) P(three in three days).

P(n in m days) is clearly what we need. We can easily find P(n in 1 day) by just the standard Poisson formulae. How about P(n in 2 days)? It's just P(n in 1 day) P(0 in 1 day) + P(n-1 in 1 day) P(1 in 1 day) + …

In this way, you can build up a formula for P(n in m days) in terms of smaller numbers of days.

By the way I found this link:

http://www.thestudentroom.co.uk/show....php?t=1240369

which says:

ii) extend the model from 1 day to 5 days...multiply the mean by 5 and use this in the Possum formula.
This is the fastest way to do it. It hinges on the fact that the Poisson distribution is scale-invariant: if X is Poisson with parameter m describing events which happen in time intervals t, then the number of events happening in time interval kt follows a Poisson distribution with parameter km.
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Thread starter 6 years ago
#3
(Original post by Smaug123)
This way does work, eventually. I prefer the following modification: it's P(four on the first day) P(none in three days) + P(three on the first day) P(one in three days) + … + P(none on the first day) P(three in three days).

P(n in m days) is clearly what we need. We can easily find P(n in 1 day) by just the standard Poisson formulae. How about P(n in 2 days)? It's just P(n in 1 day) P(0 in 1 day) + P(n-1 in 1 day) P(1 in 1 day) + …

In this way, you can build up a formula for P(n in m days) in terms of smaller numbers of days.

This is the fastest way to do it. It hinges on the fact that the Poisson distribution is scale-invariant: if X is Poisson with parameter m describing events which happen in time intervals t, then the number of events happening in time interval kt follows a Poisson distribution with parameter km.
Yes I see the short cut method is much faster. It is fairly intuitive that if the mean death rate is 0.42 per day then over a 5 day period it would be 5 x 0.42 = 2.1. Then use X ~ Poisson(2.1). Thanks.
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