poisson distribution

the problem says: let Y denot a random variable that has a Poisson distribution with mean lambda = 2. Find the following probabilities.

a. P(Y = 4)
b. P(Y > than or = 4 | Y > than or = 2)

i got (e^-2 * 2^4)/ 4! = 0.090 for the first one

im not sure about the second one

P( Y > or equal to 4 | Y > or equal to 2) =

1 - P(Y < 4 | Y > or equal to 2) =

1 - P(Y < or equal to 3 | Y > or equal to 2) =

1 - (P( Y = 3) + P( Y = 2))=

1 - ( e^-2 lamdba^3/3! + e^-2 lamdba^2/2!)

Not quite.

\(\displaystyle \frac{Pr(x\ge\;4)}{Pr(x\ge\;2)} = \frac{1-Pr(0)-Pr(1)-Pr(2)-Pr(3)}{1-Pr(0)-Pr(1)}[\tex]\)

Thanks. I already had edited to repair that, but my PC wigged out. You can see how long it took me to fix it.

why do you divide the two probabalitys

That's the "given that" part. It limits the distribution for the other part. That's why we call it "conditional".