What is the Probability of 2 Events Occurring in a Poisson Process?

[dargar]

Homework Statement

Events X, Y, Z are all Poisson processes. Event X has a rate of 1 per unit time , event Y has a rate of 2 per unit time and event Z has a rate of 3 per unit time.

Find the probability that 2 events (of any type) occur during the interval (0, 3).

Homework Equations

Maybe this is relevant
P(N=k) = $$\frac{(\lambda t)^k e^{-\lambda t}}{k!}$$

The Attempt at a Solution

So $$\lambda_X$$ = 1, $$\lambda_Y$$ = 2 and $$\lambda_Z$$ = 4. Also k = 2 and t =3.

Is it correct to think of it as say A = X $$\cup$$ Y $$\cup$$ Z. Then the answer is:

P(N=2) = $$\frac{(7(3))^2 e^{-7(3)}}{2!}$$ where $$\lambda_A = 1 + 2 + 4 = 7.$$

 

[korican04]

Homework Statement

Events X, Y, Z are all Poisson processes. Event X has a rate of 1 per unit time , event Y has a rate of 2 per unit time and event Z has a rate of 3 per unit time.

Find the probability that 2 events (of any type) occur during the interval (0, 3).

Homework Equations

Maybe this is relevant
P(N=k) = $$\frac{(\lambda t)^k e^{-\lambda t}}{k!}$$

The Attempt at a Solution

So $$\lambda_X$$ = 1, $$\lambda_Y$$ = 2 and $$\lambda_Z$$ = 4. Also k = 2 and t =3.

Is it correct to think of it as say A = X $$\cup$$ Y $$\cup$$ Z. Then the answer is:

P(N=2) = $$\frac{(7(3))^2 e^{-7(3)}}{2!}$$ where $$\lambda_A = 1 + 2 + 4 = 7.$$

I believe that this is correct. If X Y and Z are independent then a random variable say A=X+Y+Z would have a poisson distribution with rate of $$\lambda_X$$ +$$\lambda_Y$$+$$\lambda_Z$$
Although you initially wrote $$\lambda_Z$$ =3 ,but put down 4.