Question

Let ??₁(??) and ??₂(??) be two independent Poisson processes with rates ??₁ = 1 and ??₂ = 2, respectively. Find the probability that the second arrival in ??₁(??) occurs before the third arrival in ??₂(??). Hint: One way to solve this problem is to think of ??₁(??) and ??₂(??) as two processes obtained from splitting a Poisson process.

Answer 1

Given the average number of times an event occurs over certain time period, related Poisson distribution gives a discrete probability distribution of a number ot events occurring in a given time period.

Here we are considering 2 different independent processes having two separate rates of event occurrances.

??1 = 1 Units - Lets assume the unit is Hertz, meaning number of times per second, symbol Hz

??2 = 2 Hz

Now, the formula for Poisson probability is:

P(N; ??) = (e^-??) (??ⁿ) / n!

Where, n = actual number of events occurred

Substituting n = 1 and ?? = ??1 = 1 in the formula for ??1(??)

we get P(1; 1) = 0.368 or 36.8%

whereas, for P2 substituting n = 2 and ?? = ??2 = 2 for ??2(??)

P(2;2) = 0.2707 or 27.07%

and so now we can calculate absolute probability of second arrival in N1 happenning before third arrival in P2

that is P1 - P2 =  9.73 %