poisson process(urgent)

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圖片參考：http://img181.imageshack.us/img181/4575/31920795.png


2010-10-18 21:37:01 補充：
http://img181.imageshack.us/img181/4575/31920795.png

2010-10-21 20:41:43 補充：
Some elaborations:
The probability that there are 2 customers during the first 5 minutes can be split into 3 cases:
(1) 2 customers during the first 2 minutes; then no customer in subsequent 3 minutes
Probability = e^(-1/3) (1/3)^2)/2! * e^(-1/2) (1/2)^0/0! = 0.024144

2010-10-21 20:42:13 補充：
(2) 1 customer during the first 2 minutes; then 1 customer in subsequent 3 minutes
Probability= e^(-1/3) (1/3)^1)/1! * e^(-1/2) (1/2)^1/1! = 0.072433
(3) no customer during the first 2 minutes; then 2 customers in subsequent 3 minutes
Probability= e^(-1/3) (1/3)^0)/0! * e^(-1/2) (1/2)^2/2! = 0.054325

2010-10-21 20:42:44 補充：
The sum of these = 0.024144+0.072433+0.054325=0.150902
exactly equal to e^(-5/6) (5/6)^2/2!
Conditional probability, given 2 customers in first 5 minutes for :
2 customers in first 2 minutes = 0.024144 / 0.150902 = 0.16

2010-10-21 20:42:48 補充：
1 customer in first 2 minutes = 0.072433 / 0.150902 = 0.48
0 customer in first 2 minutes = 0.054325 / 0.150902 = 0.36
Conditional probability that at least one customer in first 2 minutes = 0.16 + 0.48 =0.64

For a 5-minute period, λ = 10/12 = 5/6

Therefore, in general, the probability that there are 2 customers arrived in 5 mins is:

λ2 x e-λ/2! = 0.1509

Then, probability that 2 customers arrived in 2 mins is: (for a 2-minute period, λ = 10/30 = 1/3)

λ2 x e-λ/2! = 0.0398

So for (a), we can use the conditional prob. to find the ans which is:

0.0398/0.1509 = 0.2638

For (b), the probability that no customers arrived in the first 2 minutes but 2 customers arrived in the first 5 minutes is:

e-1/3 x 0.1509

Hence the probability that at least one arrived in the first 2 minutes is:

0.1509 x (1 - e-1/3)/0.1509 = 0.2835