Imagine this

  • Your corporate website get 2 hits per minute
  • There have been 3 pandemics in the last 100 years
  • Your satellite TV fails to record a show at least 4 times a month
  • Your most important client has 6 meltdowns a month

What do you want to know?

  • How probable any and / or all of the events will happen!

Why?

  • So you can plan your life a bit better – allocate scarce time and money more efficiently, at the very least

Enter M. Poisson

Key assumptions

The Poisson random variable satisfies the following conditions:

  1. The number of successes in two disjoint (non-overlapping) regional intervals is independent.

  2. The specified region could take many forms. For instance, it could be a length, an area, a volume, a period of time, etc.

  3. The probability of a success during a small regional interval is proportional to the entire length of the regional interval.

  4. The average number of successes is known (labeled \(\lambda\)).

Apart from disjoint time intervals, the Poisson random variable thus applies to disjoint regions of any kind of space, such as geographical and market.

Applications

  • The number of deaths by horse kicking in the Prussian army (first application)
  • Birth defects and genetic mutations
  • Rare diseases
  • Car accidents
  • Traffic flow and ideal vehicle gap distance
  • Number of typing errors on a page
  • Hairs found in McDonald’s hamburgers
  • Extinction of an endangered animal in Africa
  • Failure of a machine in one month
  • Frequency of cyber attacks in a year
  • The number of loan defaults in a month

Binomial or Poisson?

  • Use the binomial when

you need to find the probability of the number of successes in so many trials.

  • Use the Poisson distribution when

you need to find the probability of a number of successes per unit of time or space.

Let’s calculate

  1. Define the number of success as \(X\).

  2. Suppose successes occur with intensity (same as: at a frequency or rate) of \(\lambda\) per unit time or space.

  3. Then, the probability of a specified number of successes \(X = x\) per unit time (or space) is \[ P(X = x) = \frac{\lambda^{x}e^{-x}}{x!} \] where \(e = 2.71828\) is the base of the natural logarithm.

  4. The mean of the distribution is \(\lambda\) and the standard deviation is \(\sqrt{\lambda}\).

Try this

A utility company is concerned about the ability of its vehicle operators to drive safely in its service region. It decides to sample a population of accidents in the 10461 zip code. One particular intersection comes to light. Analysts estimate that 6 accidents (“vehicular contacts”) per month occur at the intersection of Crosby and Westchester Avenues in the 10461 zip code in the Bronx.

  1. What is the probability of 3 accidents in any month?
  2. Plot the distribution of accidents from 0 to 20.
  3. what is the probability of 3 accidents or less in any month?

Results

  1. Plot the distribution of accidents from 0 to 20.
  1. What is the probability of 3 accidents or less in one month?

Short exercises

  1. Customers pass through the doors of a retail outlet at an average rate of 300 per hour.
  • Find the probability that none passes in a given minute.
  • What is the expected number passing in two minutes?
  • Find the probability that this expected number actually pass through in a given two-minute period.

Answers : 0.0006, 10, 12.5%

  1. A company makes windmill rotors. The probability a rotor is defective is 0.01. What is the probability that a sample of 300 rotors will contain exactly 5 defective rotors? [Use both the Poisson and the Binomial distributions to arrive at an answer.]

Answer : a little over 10%

Not so short exercises

The number of calls into your call center averages 3 per minute.

  1. Find the probability that no calls come in a given 1 minute period. Answer: 4.978%
  2. Assume that the number of calls arriving in two different minutes are independent. Find the probability that at least two calls will arrive in a given two minute period. Answer: 98.26%

Answers

Let \(X_1\) and \(X_2\) be the number of calls coming in the first and second minutes respectively. We want \(P(X_1 + X_2 \geq 2)\).