Question

Assume that the number of telephone calls arriving at a switchboard by time t (minutes) is described by Poison process {N(t)}. On average, one call comes in every 10 minutes.

Given that exactly one call arrived in 0 < t ≤ 15, calculate the conditional probability that it occurred in 10 < t ≤ 15.

Answer 1

rate = 0.1 call/min

P(k call in x min) = e^(-0.1*x) * (0.1*x)^k / (k!)

P(1 call in 10<t<=15 | 1 call in 0<t<=15)

= P(1 call in 10<t<=15 and only 1 call in 0<t<=15) / P(1 call in 0<t<=15)

= P(1 call in 10<t<=15)*P(0 call in 0<t<=10) / P(1 call in 0<t<=15)

= [e^(-0.1*5) * (0.1*5)^1 / (1!)]*[e^(-0.1*10) * (0.1*10)^0 / (0!)] / [e^(-0.1*15) * (0.1*15)^1 / (1!)]

= 0.3333

P(1 call in 10<t<=15 | 1 call in 0<t<=15) = 0.3333

(please UPVOTE)