The Binomial and Poisson Distributions


The Binomial and Poisson are discrete distributions -- for they are based
on counting successes rather than measuring data like a continuous distribution does.
In other words, the data values are frequencies rather than measures. For example,
if we're counting the number of customers in a Metro Store who buy the ice cream that's
on special, we would label a "buy" as a success and we'd build our probability distribution
from the frequencies recorded. Obviously, frequencies are integers since there's
no such thing as "half a buy". With a continuous distribution like the normal distribution,
our data comes from measuring, say the lengths of a school of fish or the weight of a
sample of bags of flour -- so the data can take on real number values, not just integers.

The Binomial Distribution

The Binomial Distribution counts the number of successes in a fixed number of trials.
For example, we could count the number of 5's obtained in 20 rolls of a single die,
or the number of Heads that come up in 200 coin tosses. These are both binomial events.

The word "binomial" in algebra means an expression with 2 terms. Here is means the
experiment has exactly 2 possible outcomes -- either a
success or failure -- for each trial.

These 4 characteristics identify the Binomial Distribution

If the value of P(x) cannot be found in a table, we use this formula

P(x) = n C x (p) x (1 - p)n - x,
where x is the number of successes,
n = # of trials, p = probability of a success on any given trial.

When we analyze the formula we see that:

nC x is the number of ways to get x successes in n trials.
x is the probability of x successes, and,
(1 - p)
n - x is the probability of n - x failures.
Since p = probability of a success, (1 - p) = the probability of a failure


Say we want to know the probability of getting 3 Heads in 5 coin tosses.
In this case, n = 5,
x = 3, and p = ½,
so P(3H) =
5 C 3 (½) 3 (1 - ½) 5 - 3 = 10 (1/8)(1/4) = 0.312
Let's discuss the
5 C 3 part of the formula. We want 3 heads in 5 tosses,
so we need to multiply the probabilities of 3 successes and 2 failures by the number
of ways this can happen. For instance,we could get {1H, 2T, 2H}, or {2H, 2T, 1H}
or {2H, 1H, 2T} etc. The
n C x part of the formula counts the number of ways to
get the desired outcome.

Had we wanted the probability of getting at most 3 Heads in 5 coin tosses,
we would sum the probabilities of getting 0, 1, 2 or 3 Heads in 5 tosses.
That is, we would find P(0) + P(1) + P(2) + P(3), using the correct values for x, n and p.
Had we wanted the probability of getting at least 4 Heads in 5 coin tosses --
we would find 1 - [ P(0) + P(1) + P(2) + P(3) ] or we would sum P(4) + P(5).


3 more characteristics of the Binomial Distribution

1) if n stays constant while p approaches 0.5, the shape of the Binomial
Distribution becomes more symmetric.(more like the normal curve)

2) if p stays constant while n gets very large, the shape of the Binomial
Distribution becomes more symmetric.(more like the normal curve)

3) the mean µ = np and the variance r ² = np(1 - p) (formulas 4 and 5 on list).


Since the outcomes of a coin toss forms a binomial population, we'll do some coin tossing.

With a coin toss, there are exactly 2 possible outcomes -- Heads or Tails.
Each outcome is independent of the last, since the coin has no idea
what side came up on the last toss.
The probability of a success -- we'll choose Heads (H) as a success -- is always ½.

Though we used the formula to find P(3H) above, there are of course tables
of Binomial Probabilities
. Problem is, most tables offer only a limited number
of values for p, (the probability of a success) and include values for n
between 1 and 20. So if n = 25 and p = .67 the tables are useless. Then we use
either the formula or our calculators to find the needed probability.

Let's repeat the illustration only this time we'll use a Tables of Binomial
Probabilities instead of the formula to find our answers.

Example 1

Using Binomial Probability tables, find:

a) the probability of getting exactly 3 heads in 5 coin tosses.
b) the probability of getting at most 3 heads in 5 coin tosses.

The table entries for n = 5 and p = 0.5 are:

n x P(x)
5 0 0.031
  1 0.156
  2 0.312
  3 0.312
  4 0.156
  5 0.031

a) the probability of getting exactly 3 heads in 5 tosses = 0.312

b) the probability of getting at most 3 heads in 5 tosses = .031 + .156 + 2(.312) = 0.811
We could have found{ 1 - [P(4) + P(5)]} = 1 - (.156 + .031) = 0.813. The small
difference is due to rounding to 3 decimal places.

Calculator Notes

To use the TI-83 calculator to compute these probabilities, we use either binompdf
or binomcdf from the DISTR menu (2nd + VARS). For a question like (a) above,
to find the probability of exactly x successes in n trials, we use binompdf.
For (b) part type questions, we use binomcdf, since pdf stands for probability density function and cdf stands for cumulative density function. The order of data entry is {n, p, x} for
both types of questions. So for this questions we'd enter {5, 0.5, 3} to find the values.


Example 2

If the probability is 0.60 that a car stolen in Montreal will be recovered, find the
probability that:
a) at most 3 of 10 stolen cars will be recovered.
b) at least 7 of 10 stolen cars will be recovered.

Since neither of these asks for exactly 3 or 7 cars will be recovered, we will use
the binomcdf function with n = 10, p = 0.60. For a) we set x = 3 and for b) x = 6.
Also in b) part, we'll subtract the calculator result from 1 since the question says
at least 7.

a) P(x 3) is found with binomcdf {10, 0.60, 3} = 0.05476 or 0.055.
b) P(x
¥ 7) is found with 1 - (binomcdf {10, 0.60, 6}) = 1 - 0.61722 = 0.38228

Note: we could also just add up the values we get from binompdf with x = 7, 8 , 9 and 10.
Either way we get the same answer.


The Poisson Distribution

When n is large, and p is small for a Binomial population or sample, we often approximate
the probabilities using the Poisson Distribution. The general rule of thumb is to do this
when n
¥ 100 and np < 10.

In general use however, The Poisson Distribution counts the number of successes
over a fixed interval of time or within a specified region.
For example: the number of murders recorded in Detroit in a month, or the
number of cars parked at the airport during a 48 hour period.

To apply the Poisson distribution 2 conditions must be met:
1) The number of successes that occur in any interval is independent
of those that occur in other non-overlapping intervals.

2) The probability of a success in an interval is proportional to the size of the interval.

The formula for calculating Poisson probabilities is:
where µ = the mean or expected number of successes,
e = the constant e, and x = the number of successes observed or counted.

The standard deviation of the Poisson Distr. =

Note: many statistics books use k (lambda) unstead of l for the expected mean value.
This is most often done when using the Poisson Distribution with no connection to the
Binomial Distribution.


Example 3

If a bank receives an average of 6 bad checks per day, what is the probability
that on any given day, it will receive 4 bad checks?


So, l or k = 6, and x = 4. Substituting into the formula we get:

Like the Binomial and other distributions, we don't often use the formula because we
use tables or our calculators. Like the Binomial, the TI-83 also produces Poisson
probability values. The correct order for data entry is poissonpdf {
l, x}.



1) If incompatibility is given as the reason for separating in 55% of all divorces,
find the probability that this will be the reason given for 4 of the next 6 divorces.


2) A food distributor claims that 80% of the cans of mixed nuts she sells contain
at least 3 pecans. If 8 of the cans are tested, what is the probability that:


3) It is known that 20% of all patients taking a certain drug will become drowsy within
2 minutes, what is the probability that of 14 randomly chosen patients on this drug:


4) The number of monthly breakdowns of a computer is a random variable
having the Poisson distribution with k = 1.8.
Find the probabilities that this computer will:


5) The number of emergency calls received per day by the Montreal 911 service
is a random variable having the Poisson distribution with k = 5.5.
What is the probability that on any given day, there will be only 4 emergency calls?



1) n = 6, p = 0.55 and x = 4
Using tables, we find that p = 0.55 is not included, so we use the formula or calculator.
binompdf(6, .55, 4) = 0.278


2) n = 8, p = 0.80
a) P(0) = binompdf(8, .80, 0) = 0.00. We could use tables to find this since 0.8 is listed.

b) P(1) = binompdf(8, .80, 1) = 0.001.

c) P(x > 6) = 1 - binomcdf(8, .80, 6) = 0.50332.


3) n = 14, p = 0.20
a) P(x [ 2) = binomcdf(14, .20, 2) = 0.448

b) P(x ¥ 5) = 1 - binomcdf(14, .20, 4) = 0.12984

c) P(x = 2, 3 or 4) = P(x = 2) + P(x = 3) + P(x = 4) = 0.672


4) Using Poisson Distribution with k = 1.8.
a) x = 0

b) x = 1 t


5) Using Poisson Distribution with k = 5.5.

x = 4 t


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