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Poisson Process

Please could someone help me with this question?

Customers arrive in a Poisson process at the rate of two per minute at a newspaper seller’s stand to buy the evening paper. If the seller has just 4 copies left and will not receive a supply of the next edition for 5 minutes, what is the distribution of the number of his disappointed customers?

Bascially I know how to find the probability;

Let N(t) = number of events that have occur by time t (starting at zero)

P(N(5)>4) should give me the correct probability but surely this isn't a distribution?

P(N(5)>4)= 1-P(N(5)<=4)=1-e^10{1 +10+50+1000/6 + 10000/24)

gives 1-(1933/3)e^-10

but what distribution is this?

Reply 1

Chewwy

you need to find the individual probabilities for N(5) <= 4, N(5) = 5, N(5) = 6, ....

these fully describe the distribution, yeah?

Reply 2

zcomputer5

Original post by Chewwy

you need to find the individual probabilities for N(5) <= 4, N(5) = 5, N(5) = 6, ....

these fully describe the distribution, yeah?

Ok, so computing the probabilities we find with rate 2 per minute

P(N(5)=k)=[(10)^k][e^-10]/k!

which gives

P(N(5)=0)=e^-10
P(N(5)=1)=10e^-10
P(N(5)=2)=50e^-10
P(N(5)=3)=(1000/6)e^-10
P(N(5)=4)=(10000/24)e^-10

so P(N(5)<=4)=e^-10(1 +10+50+1000/6 + 10000/24)
P(N(5)=5)=(100000/5!)e^-10
P(N(5)=6)=(1000000/6!)e^-10
P(N(5)=7)=.....

so P(N(5)>=5)=SUM from k=5 to inf [e^-10][(10^k)]/k!

Is this right? However surely I don't need to compute the stuff in blue as this people will get a paper?

(edited 13 years ago)

Reply 3

ghostwalker

Original post by zcomputer5

If D is the r.v. representing the number of disappointed customers, then:

P(D=0) = P(N(5)<=4)
P(D=1) = P(N(5) = 5)
...

P(D=n) = P(N(5) = n+4)

So, yes you do need to compute the stuff in blue.

Reply 4

zcomputer5

Original post by ghostwalker

If D is the r.v. representing the number of disappointed customers, then:

P(D=0) = P(N(5)<=4)
P(D=1) = P(N(5) = 5)
...

P(D=n) = P(N(5) = n+4)

So, yes you do need to compute the stuff in blue.