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:
The number of successes in two disjoint (non-overlapping) regional intervals is independent.
The specified region could take many forms. For instance, it could be a length, an area, a volume, a period of time, etc.
The probability of a success during a small regional interval is proportional to the entire length of the regional interval.
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
Define the number of success as \(X\).
Suppose successes occur with intensity (same as: at a frequency or rate) of \(\lambda\) per unit time or space.
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.
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.
What is the probability of 3 accidents in any month?
Plot the distribution of accidents from 0 to 20.
what is the probability of 3 accidents or less in any month?
Results
What is the probability of 3 accidents in any month? We let \(X\) be the number of possible accidents per month at this intersection. Out data is \(X = x = 3\) and \(\lambda = 6\). We can then compute \[
\begin{array}{c}
P(X = 3) = \frac{\lambda^{x}e^{-x}}{x!} \\ \\
= \frac{6^{3}e^{-3}}{3!} \\ \\
= \frac{(216)(0.002478)}{3x2x1} \\ \\
= \frac{0.5354105}{6} \\ \\
= 0.08923508
\end{array}
\] that is, 8.9%.
Plot the distribution of accidents from 0 to 20.
What is the probability of 3 accidents or less in one month?
Add up the probabilities for observing 0, 1, 2, and 3 accidents in one month. This is the same as saying that these are the probabilities that 0, or 1, or 2, or 3 accidents (union) occur.
Sum up 0.002478752, 0.014872513, 0.044617539, and 0.089235078 to get 15.1%.
Short exercises
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%
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.
Find the probability that no calls come in a given 1 minute period. Answer: 4.978%
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 \(P(X_1 + X_2 \geq 2)\).
Let’s organize the answer this way \[
P(X_1 + X_2 \geq 2) = 1-P(X_1 + X_2 < 2)
\] In this way we only need to look at no calls (0 minutes) and 1 minute calls, or, for each possible sequence of 0 and 1 minute calls each sequence happening together (intersection) and independently. We have