Binomial Distributions

Binomial Distributions
Unlike normal distributions which are continuous, the binomial distributions is discrete. Discrete
distributions provide nice bridges linking frequency histograms with probability, on the way to
understanding probability with continuous distributions.
The most basic element of a binomial distribution is a Bernoulli experiment.
A Bernoulli experiment is a simple experiment with only two possible outcomes. This could be
something like flipping a coin (heads or tails). It could also be something like rolling a die and
getting a three. Here, our experimental outcome is three or everything else (1,2,4,5,6).
Almost any phenomenon can be reduced to a binomial outcome. We denote these outcomes
success and failure, although there is no implied normative meaning. So, for instance, taking a
treatment for cancer either leads to an improvement or not. If we are interested in the
likelihood that a treatment does not improve a cancer patients condition, success is failure to
treat the disease.
Success is normally noted as p; failure as q.
Since there are only two outcomes, we know that the sum of the two probabilities must equal 1,
or p + q = 1. Furthermore, q = 1-p and p = 1-q. The binomial distribution is known as a one
parameter distribution. This is because if you know one outcome, you implicitly know the
other.
A binomial experiment is a number of independent Bernoulli experiments.
Specifically, a binomial experiment has the following properties.
• The experiment consists of n identical trials.
• Each trial results on one of two possible outcomes.
• The probability of success on a single trial, p, is invariant across trials, and q = 1 - p.
• The trials are independent.
• We are interested in X, the number of successes in n trials.
The number of successes, r, represents a binomial variable. What is interesting about this
variable is that we can associate a probability with each outcome, r, and derive a probability
distribution.

Let's look at an easy example. If we toss a fair coin, defining getting a head on a toss as success,
we know that p = 0.5 and q = 0.5. We propose to flip this coin four times; this is our binomial
experiment. Each flip of the coin, or trial, constitutes a Bernoulli experiment.
How do we enumerate success? We know that in four flips of the coin that there are five possible
values of r; 0, 1, 2, 3, and 4. The binomial distribution associates a probability with each of
these outcomes. So, for instance, what is the probability that r equals 4 [p(r=4)]? This is
asking what is the probability of p(H and H and H and H). This is equal to (q)(q)(q)(q) = p 4 =
0.5 4 = 0.0625. Thus, there is a 6.25% chance of tossing a coin four times and observing a head
on each toss.
What about getting one head and three tails [p(r=1)]? How many ways are there to get one head
and three tails? We can list them as follows:

H T T T
T H T T
T T H T
T T T H
There are 4 ways to get 1 success
and they have equal
probability
p(HTTT) = (p)(q)(q)(q) = pq 3 = (0.5)(0.5 3 ) = 0.0625
p(THTT) = (q)(p)(q)(q) = pq 3 = (0.5)(0.5 3 ) = 0.0625
p(TTHT) = (q)(q)(p)(q) = pq 3 = (0.5)(0.5 3 ) = 0.0625
p(TTTH) = (q)(q)(q)(p) = pq 3 = (0.5)(0.5 3 ) = 0.0625
Summing these independent probabilities we get 0.0625+0.0625+0.0625+0.0625 = 0.25. Thus,
there is a 25% chance that in four flips, we observe a single head.

We can also use this method to calculate all the other probabilities for the values of r. But, there
is a shortcut. What we are actually asking is, given 4 trials, how many ways can we distribute r
successes across these trials. This then becomes a simple combination problem. The formula
for combinations is:
⎛n⎞
⎜ ⎟ =
⎝ r ⎠
n!
r!( n − r)!
So, given all the possible values of r, we get:
r
0
1
2
3
4
Combinations
⎛4⎞
⎜ ⎟ =
⎝0
⎠
0!
⎛4⎞
⎜ ⎟ =
⎝1⎠
1!
⎛4⎞
⎜ ⎟ =
⎝2
⎠ 2!
⎛4⎞
⎜ ⎟ =
⎝3⎠
3!
⎛4⎞
⎜ ⎟ =
⎝4
⎠
4!
4!
= 1
0 !
( 4 − )
4!
= 4
1 !
( 4 − )
4!
= 6
2 !
( 4 − )
4!
= 4
3 !
( 4 − )
4!
= 1
4 !
( 4 − )
We can also use a general formula to calculate the individual probabilities associated with any one
outcome of r. This formula is:
p r q n-r
Putting both this formula and that for combinations together, we get:
⎛n⎞
⎜ ⎟ p
⎝ r ⎠
r
q
n−r

⎛n⎞
⎜ ⎟ p
⎝ r ⎠
r
q
n−r
So, the full binomial probability distribution for this problem is given as:
p(r) Binomial Probability
⎛4⎞
⎟⎠
⎝0
0
4
p(0) ⎜ ( 0.5 )( 0.5 )
⎛4⎞
⎟⎠
⎝1
1
3
p(1) ⎜ ( 0.5 )( 0.5 )
⎛4⎞
⎟⎠
⎝2
2
2
p(2) ⎜ ( 0.5 )( 0.5 )
⎛4⎞
⎟⎠
⎝3
p(3) ⎜ ( 0.5 )( )
3 0. 5 1
⎛4⎞
⎟⎠
⎝4
4
0
p(4) ⎜ ( 0.5 )( 0.5 )
0.0625
0.2500
0.3750
0.2500
0.0626
0.50
Binomial Distribution
(n=3, p=0.33, q=0.67)
Probability of Successe
0.40
0.30
0.20
0.10
0.00
0 1 2 3 4
Number of Successes