Question

The traffic incidents in Melbourne and Sydney follow a Poisson process with the rate
of 5 and 6 incidents per hour, respectively.
(a) Find the probability that no traffic accidents will occur in Melbourne in the next
30 minutes.
(b) Find the expected time (in minutes) until 10 new incidents occur in Sydney.
(c) Find the expected time (in minutes) until 10 new incidents in total occur in
Melbourne and Sydney. Hint: For this and the next question, you can assume
that traffic incidents in the two cities are independent.
(d) Assuming exactly 5 traffic incidents in total will occur in Melbourne and Sydney
in the next 10 minutes, find the probability that not more than 2 traffic incidents
will occur in Melbourne during the same period of time. Hint: Use the formula
of conditional probability.

Answer 1

a) For melbourne, there are on an average 5 incidents per hour that is equivalent to 2.5 incidents in 30 minutes. Therefore the probability of no incidents in the next 30 minutes is computed here as:

= e-2.5 = 0.0821

Therefore 0.0821 is the required probability here.

b) Expected minutes until 10 new incidents occur in sydney is computed here as:

= 10/average number of incidents in an hour in sydney

= 10/6 hours

that is (10/6)*60 = 100 minutes

Therefore 100 minutes is the expected waiting time here.

c) For melbourne and sydney in total, the mean number of incidents is computed as: 6 + 5 = 11 incidents. Therefore the expected waiting time for a total of 10 new incidents is computed here as:

= 10/11 hours

that is (10/11)*60 = 54.54 minutes

d) Given that a total of 5 incidents occur in a 10 minute period in sydney and melbourne combined, probability that not more than 2 incidents occur in Melbourne is computed here as:

= [ P(M = 0)P(S = 5) + P(M = 1)P(S = 4) + P(M = 2)P(S = 3) ] / P(M + S = 5)

Therefore, the probability now is computed here as:

Therefore 0.5848 is the required probability here.