Question

06 Let {xW;t 203 follo ay Let 2X(4) t zo} follows the poisson process with average arrival rate of 5 people per 1/2 hour. Find the probability of lo arrivals in the interval of 10 minutes to 20 minutes Find the probability that any arrival has to wait for more thon 15 minutes D> P(x(1) = 10 / X(20) = 15), d) PCX (20) = 15 / XCI) = 10) e> P(x(20) = 10 / PX(19) - 8, X(18) = 6, X(17) - 4

Answer 1

We would be looking at the first 4 parts here:

We are given the average rate of arrival here as 5 arrivals in 30 minutes ( that is half an hour)

a) From interval of 10 minutes to 20 minutes, that is a 10 minute interval, the number of arrivals here could be modelled as:

X ~ Poisson(5/3)

Therefore the probability here is computed as:

P(X = 10) = 10) (5/3) 10 10 /3

Therefore 0 is the approx. probability here.

b) Probability that an arrival has to wait for more than 15 minutes is computed as the probability of no arrival in 15 minutes. The average rate of arrival in a 15 minute period is 5/2 = 2.5

Therefore the probability here is computed as:

=e-2.5 = 0.0821

Therefore 0.0821 is the required probability here.

c) The probability here is computed using Bayes theorem as:

P(X(10) = 10 X (20) = 15) - P(X 10 = 10)P(X10 = 15 – 10) P(X20 = 15)

P(X(10) = 10 X (20) = 15) - P(X10 = 10)P(X10 = 5) P(X20 = 15)

(5/3)10 -5/3 -5/3 P(X(10) = 10X (20) = 15) = – 10 51 15) - = 0.000078 (10/3) -10/3

Therefore 0.000078 is the required probability here.

d) Here, the required probability is:

P(X(20) = 15 | X(10) = 10)

which means given that there are 10 arrivals in first 10 minutes, probability of 15 arrivals in 20 minutes is computed as the probability that there would be 5 arrivals in the next 10 minutes. Which is computed as:

P(X10 = 5) = 6/35 5) -e-5/3 = 0.002222

Therefore 0.002222 is the required probability here.