It is known that the distribution of the people coming to the party is Poisson with a rate of 0.9. The DJ is going to play only if there are people. What is the probability that the DJ is going to play in front of one person exactly when it is known that no more than two people came to the party.
I'm not sure how to do it since I see it's a probability with a condition, but also I need first to calculate the Probability by the Poisson distribution? I would love any explanation.
P(2 or fewer people arrive)$=\frac{(0.9)^0 e^{-0.9}}{0!}+\frac{(0.9)^1 e^{-0.9}}{1!}+\frac{(0.9)^2 e^{-0.9}}{2!}$
P(1 person arrives)$=\frac{(0.9)^1 e^{-0.9}}{1!}$
So P(1 person arrives|no more than 2 will arive)=$\frac{\frac{(0.9)^1 e^{-0.9}}{1!}}{\frac{(0.9)^0 e^{-0.9}}{0!}+\frac{(0.9)^1 e^{-0.9}}{1!}+\frac{(0.9)^2 e^{-0.9}}{2!}}$,
since the probability of both fewer than 2 arriving and only 1 arriving is just the probability of only 1 arriving.