Answer to Question #323823 in Statistics and Probability for Bless

Question

(a) The random variable Y has a Poisson distribution and is such that P(Y = 0) = P(Y = 1). What is P(Y 2 = 1)?

(b) Cars arrive at a toll both according to a Poisson process with mean 80 cars per hour. If the attendant makes a one-minute phone call, what is the probability that at least 1 car arrives during the call?

Answer 1

a) Poisson distribution is described by the formula $P(y)=\lambda^y e^{-\lambda}/y!$ ,

where $\lambda$ is distribution's mean

$P(0)=P(1)\Lrarr \lambda^0e^{-\lambda}/0!=\lambda^1e^{-\lambda}/1!\Lrarr\lambda=1\\ P(0)=P(1)=1/e,\space P(2)=1/(2!\cdot e)=1/(2e)$

b) Mean for 1-minute interval is 80 / 60 = 4 / 3.

Probability that neither car will arrive:

$P(0)=((4/3)^0e^{-4/3})/0!=e^{-4/3}$

Probability that at least one car will arrive is equal to $1-P(0)=1-e^{-4/3}\approx0.7364=73.64\%$