Question

Suppose that the number of accidents occurring on a highway per hour follows a Poisson distribution...

Suppose that the number of accidents occurring on a highway per hour follows a Poisson distribution with a mean of 1.25.

  1. What is the probability of exactly three accidents occur in hour?
  2. What is the probability of less than two accidents in ten minutes?
  3. What is the probability that the time between two successive accidents is at least ten minutes?
  4. If ten minutes have gone by without an accident, what is the probability that an accident will occur in the next twenty five minutes?

Homework Answers

Answer #1

Here number of accidents will follow a poisson distribution.

p(x) = e-x/x!

= 1.25

(a) p(X = 3) = e-1.251.253/3! = 0.0933

(b) Expected number of accident in 10 minutes = 1.25 * 10/60 = 1.25/6 = 0.2083

P(x < 2) = P(x = 0) + P(x = 1)

= e-0.20830.20830/0! + e-0.20830.20831/1! = 0.9810

(c) P(at least ten minutes between 2 accidents) =P(no accident in 10 minutes)

= e-0.20830.20830/0! = 0.8119

(d) The last 10 minutes will not impact the next 25 minutes.

so,

P(an accident will occur in next 25 minutes) = 1- P(no accidient in next 25 minutes) = 1 - e-25*1.25/60

= 1 - 0.5940 = 0.4060

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