Probability of something happening?

Question

Automobiles arrive at the Elkhart exit of the Indiana Toll Road at the rate of two per minute. The distributions of the arrivals approximates a Poisson distribution.

A) What is the probability that no automobiles arrive in a particular minute?

B) What is the probability that at least one automobile arrives during a particular minute?

Answer 1

Use the Poisson distribution.

Mean λ = 2

The probability of k occurrences with mean λ is:

p(k | λ) = e^(-λ)*(λ^k)/k!

For no occurrences, we want the probability for k=0 with λ=2.

p(0 | 2) = e^(-2)*(2^0)/0! = e^(-2)*1/1 = e^(-2) = 1/e^2 = 0.135

The probability that there is at least 1 occurrence is:

p(1+ | 2) = 1 - p(0 | 2) = 1 - 1/e^2 = 0.865

Answer 2

λ = 2 (2 per minute, hope you understand this)

for A, it's asking P(x = 0). just plug it into the pmf,

P(x,2) = [(e^-λ)(λ^x)]/(x!)

plug in and win! remember that 0! (0 factorial) = 1.

for B, it's asking P(x >= 0). but remember, P(x>=0) = [1 - P(x=0)], since the probabilities add up to 1. so, just subtract your part A answer from 1.

Answer 3

Given X ~ Poisson ( lambda = 2)
P(X=x) = [(lambda^x)(e^-lambda)]x!

P(X=0 | lambda = 2) = (2^0)(e^-2) / 0! = (1*e^-2)/1 = e-2

P( X is at least 1) = 1 - P(X=0) = 1-e^-2