Need help for this question

[cyc]

Let X1 and X2 form a random sample from a Poisson distribution. The Poisson distribution has a mean of 1. If Y = min(X1, X2), then P(Y=1) = ?



In the answers, it says
P(Y=1) = P[(X1=1) intersection (X2 >= 1) ] + P[ (X2=1) intersection (X1 >=2)]

The part that I dont' understand is why in the second part of the equation, X1 >= 2? isn't it supposed to be >= 1 ?

By the way, the answer is (2e-3)/e^2

Thanks

[Gandalf]

This part

In the answers, it says
P(Y=1) = P[(X1=1) intersection (X2 >= 1) ] + P[ (X2=1) intersection (X1 >=2)]

is right. I didn't try to plug in the values to get the answer in terms of e.

To express the answer as the sum of two probabilities, you don't want any overlap. In this case, you are counting the case X1 = X2 = 1 in the first expression, so you must be sure to exclude it from the second.