poisson distribution

[mark]

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!)

 

[tkhunny]

Not quite.

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

 

[pka]

Not quite.

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

 

[tkhunny]

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

 

[mark]

why do you divide the two probabalitys

 

[tkhunny]

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