Question

Let Y denote a random variable that has a Poisson distribution with mean λ = 3. (Round your answers to three decimal places.)

(a) Find P(Y = 6)

(b) Find P(Y ≥ 6)

(c) Find P(Y < 6)

(d) Find P(Y ≥ 6|Y ≥ 3).

Answer 1

Solution:

We are given λ = 3

P(Y=y) = λ^y*exp(-λ)/y!

Part a

P(Y=6) = 3^6*exp(-3)/6!

P(Y=6) = 729* 0.049787/720

P(Y=6) = 0.050409

Required probability = 0.050409

Part b

We have to find P(Y≥6)

P(Y≥6) = 1 - P(X<6) = 1 – P(X≤5)

P(X≤5) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5)

P(Y=0) = 3^0*exp(-3)/0! = 0.049787

P(Y=1) = 3^1*exp(-3)/1! = 0.149361

P(Y=2) = 3^2*exp(-3)/2! = 0.224042

P(Y=3) = 3^3*exp(-3)/3! = 0.224042

P(Y=4) = 3^4*exp(-3)/4! = 0.168031

P(Y=5) = 3^5*exp(-3)/5! = 0.100819

P(X≤5) = 0.049787 + 0.149361 + 0.224042+ 0.224042+ 0.168031+ 0.100819

P(X≤5) = 0.916082

P(Y≥6) = 1 – P(X≤5)

P(Y≥6) = 1 – 0.916082

P(Y≥6) = 0.083918

Required probability = 0.083918

Part c

P(Y<6) = P(X≤5)

P(Y=0) = 3^0*exp(-3)/0! = 0.049787

P(Y=1) = 3^1*exp(-3)/1! = 0.149361

P(Y=2) = 3^2*exp(-3)/2! = 0.224042

P(Y=3) = 3^3*exp(-3)/3! = 0.224042

P(Y=4) = 3^4*exp(-3)/4! = 0.168031

P(Y=5) = 3^5*exp(-3)/5! = 0.100819

P(X≤5) = 0.049787 + 0.149361 + 0.224042+ 0.224042+ 0.168031+ 0.100819

P(X≤5) = 0.916082

Required probability = 0.916082

Part d

P(Y≥6|Y≥3) = P(Y≥6)/(P(Y≥3)

P(Y≥6) = 0.083918

(P(Y≥3) = 1 – P(Y≤2)

P(Y=0) = 3^0*exp(-3)/0! = 0.049787

P(Y=1) = 3^1*exp(-3)/1! = 0.149361

P(Y=2) = 3^2*exp(-3)/2! = 0.224042

P(Y≤2) = 0.049787 + 0.149361 + 0.224042

P(Y≤2) =0.42319

(P(Y≥3) = 1 – P(Y≤2)

(P(Y≥3) = 1 – 0.42319

(P(Y≥3) = 0.57681

P(Y≥6|Y≥3) = P(Y≥6)/(P(Y≥3)

P(Y≥6|Y≥3) = 0.083918/0.57681

P(Y≥6|Y≥3) = 0.145486

Required probability = 0.145486