Thread: Poisson Distribution, Independent Events

1. Suppose that X and Y are independent Poisson random variables with parameters 1 and 2 respectively, Find:

a) P(X = 1 and Y = 2);
b) P((X + Y) / 2) >= 1);
c) P(X = 1 | (X +Y)/2 = 2)[/quote]I found a) by multiplying the individual probabilities that X=1 and Y=2 because X and Y are independant


How can I use the independance of X and Y to calculate b) and c)?


2). Radioactive substances emit alpha particles/ The number of such particles reaching a counter over a given time period follows the Poisson distribuiton. Suppose two substances emit alpha particles independantly of each other. The first substance gives out alpha particles according to the Poission(3.87) distribution while the second substance emits alpha particles which reach the counter according to the Poisson(5.41) distribution. Find the chance that the counter is hit by at most 4 particles[/quote]Again I need to know how to use the independence of each substances alpha particle emission frequencies to calculate this probability


I was thinking along the lines of:

X: denotes number of alpha particles from substance 1 reaching the counter
Y: denotes number of alpha particles from substance 2 reaching the counter

P(X + Y <= 4) = P(X <= 2) + P(Y <=2)

Thanks.

Originally Posted by Hasan1

1. Suppose that X and Y are independent Poisson random variables with parameters 1 and 2 respectively, Find:

a) P(X = 1 and Y = 2);
b) P((X + Y) / 2) >= 1);
c) P(X = 1 | (X +Y)/2 = 2)

I found a) by multiplying the individual probabilities that X=1 and Y=2 because X and Y are independant


How can I use the independance of X and Y to calculate b) and c)?

[snip][/quote]
b) Calculate 1 - P( (X + Y) / 2) < 1).

c) Apply Bayes Theorem.

Yes but how do I actually calculate 1 - P((X+Y)/2 < 1) ?
The answer for b is 0.8008, but I don't know how they get that.

I tried breaking that into

1 - P(X+Y < 2)
1 - P(X < 1) + P(Y < 1) (which gave me the wrong answer)
or 1 - P(X < 2) + P(Y < 2) ( which also gave me the wrong answer)

since X ~ Poisson(1) shouldn't P(X<1) be (e^-1)(1^0)/(0!) and simiilarly
since Y~ Possion(2) -> P(Y<1) (e^-2)(2^0)/(0!) ?

Originally Posted by Hasan1

Yes but how do I actually calculate 1 - P((X+Y)/2 < 1) ?
The answer for b is 0.8008, but I don't know how they get that.

[snip]

What pairs (X, Y) satisfy (X+Y)/2 < 1 ....?

Those would be:

(X,Y) = (0,0)
(X,Y) = (0,1)
(X,Y) = (1,0)

So does that make

P((X+Y)/2< 1)

equivalent to:

P(X=0 and Y=0) + P(X=1 and Y=0) + P(X=0 and Y=1) ?

Originally Posted by Hasan1

Those would be:

(X,Y) = (0,0)
(X,Y) = (0,1)
(X,Y) = (1,0)

So does that make

P((X+Y)/2< 1)

equivalent to:

P(X=0 and Y=0) + P(X=1 and Y=0) + P(X=0 and Y=1) ?

Yes.