The number of phone calls received at a call centre in a minute is distribuited as a Poisson random variable and the average number of calls in a minute is 3.
What is the probability that in a minute no calls occurs? $$P(X=0)=e^{-3}$$
What is the probability of receiving at least two calls in two minute, since that the number of calls in a certain minute is independent from the one received in the next minute?
I solved this using the fact that the number of call in 2 minute are distributed as a Poisson random variable with parameter $2*3=6$ $$P(Y \ge2)=1-P(Y<2)=1-P(Y=1)-P(Y=0)=1-7*e^{-6}$$
is it right?
I could have solved this using two Poisson r.v. $X_1, X_2$ of parameters $3$ both. and $$P(X_1+X_2 \ge2)=1-P(X_1+X_2<2)=1-P(X_1+X_2=1)-P(X_1+X_2=0)=1-7*e^{-6}$$ the same result.