Download Binomial Distribution

Binomial Distribution
We begin our study on the Binomial Distribution with an introduction to Bernoulli
trials. Jakob Bernoulli was a mathematician who advanced the understanding of
probability theory. Today we associate his name most often with Bernoulli trials,
a probability term describing events with exactly two possible outcomes, such as
heads and tails. In Bernoulli trials the probability remains constant from trial to
trial, and trials are independent of one another. If we add to this another condition
– a fixed number of trials - we now meet the conditions of a special type of
setting known as a “binomial setting”. If an event is part of a binomial setting then
the probability can be found using a special formula. A series of random
variables from this setting formed into a distribution is known as a “binomial
distribution
We can now determine if an event follows a binomial setting if we combine the
Bernoulli trials and add a condition that the number of trials be fixed.
A situation is binomial if:
•
Each observation (trial) has only two possible outcomes, called success
and failure
•
The probability of success, we’ll call it p, is the same for each observation
(trial).
•
Each observation (trial) is independent.
•
The number of observations (trials) is fixed.
If the event is binomial, then we can create a random variable distribution
(remember that from Module 7?) and use that distribution to solve problems.
Consider the random event with two possible outcomes, such as heads and tails,
from tossing a coin four times. Each toss is independent of every other toss and
probability is constant (i.e., the probability of Heads is the same each toss). Also,
there are a fixed number of tosses (four) which means that there is a fixed
number of trials. So, the event “tossing a coin” meets the condition of a binomial
setting.
Suppose we define X to be the number of heads in a series of 4 tosses. Let’s
make a distribution (table) of the random variable and each outcome
X
0
1
2
3
4
p(x)
Do you see why we use the numbers 0 through 4? We are interested in the
number of heads in four tosses. It is possible that there are no (0) heads out of
four, or only 1 head out of four, or only 2 heads out of four, etc. Notice also that
the random variables are filled in but the probabilities are not? Can we find those
probabilities?
Well, let’s find them using the concepts we learned from Chapter 6. We are
tossing four coins so the sample space looks like:
HHHH
HTHH
THHT
THHH
HHHT
HHTH
HTTH
TTHH
HHTT
THTH
TTHT
TTTH
HTTT
HTHT
THTT
TTTT
So, there are 16 possible combinations. How many of the sixteen have no
heads? Only one, so P(0) = 1/16 = 0.0625. How many of the sixteen have only
one head? Four meet that condition, so P(3) = 4/16 = 0.25. We can extend that
process to come up with all of the probabilities.
X
0
1
2
3
4
p(x)
0.0625
0.2500
0.375
0.2500
0.0625
Notice this is a legitimate probability distribution because 1) every probability is
between 0 and 1 and 2) the sum of the probabilities is 1.
It was also possible to find each probability by using a formula known as the
binomial probability formula. This formula is based around a way of counting
known as “combinations”. You may have studied them in previous math classes.
If not, then you may want to look it up on your own in a math book or on an
internet website. It is beyond the scope of this class to teach that material.
The formula for finding the probability of any given random variable k is
n k
P( X  k )    p (1  p) ( n k )
k
Where n represents the number of observations in the setting, k represents the
random variable of interest, p represents the probability of success and (1-p)
represents the probability of failure.
Also,
n
  is the symbol for “n choose k”. It represents the combination.
k
So, in the example above we could have found P(X= 0) using the formula as
follows:
n k
P( X  k )    p (1  p) ( nk )
k
4
  (0.5) 0 (1  0.5) ( 40 )
0
 (1)(1)(0.0625)  0.0625
HINT: you can always find
4
 
0
n
 
k
by using your calculator. For example, to get
you type in 4, then MATH, then arrow over to PRB, then arrow down to 3,
then press ENTER, then 0, then press ENTER. You should see that The answer
is 1.
We can follow the same method to get all of our probabilities through use of the
formula.
I can almost hear you say, “that’s complicated!” I tell you that I agree!
One thing that makes the binomial formula so nice to use is that the graphing
calculator has this formula programmed already. And, truthfully, the formula is
good to know and you may get a question about using it on a test (or even the
AP Test) but it is not something that will kill your score by itself if you don’t know
that much about it. Your calculator will do all of the calculating for you. The trick
is in knowing how to use it.
Let’s get all of the probabilities from the table above by using the calculator. Go
into L1 and enter {0,1,2,3,4}. Now press <2nd> <DISTR> <0:binompdf(>
Now type 4,0.5,L1)L2. Now check the values in L2. What you just did was
create probabilities for the coin flip problem from above….all in one table. Your
finished work should look like this:
We can also use this same process to find a single value.
Suppose we have a binomial situation where we have six observations (trials)
and our probability of success is 0.2. Suppose also that we are interested in
finding the probability at 3 (i.e., P(X=3)). To find the binomial probability for n=6,
p=.2, X=3, use the command binompdf(6,.2,3). The probability is .08192.
Notice that the general way to find the binomial probability P(X=k) you use the
command binompdf(n,p,k).
See that we can make a single calculation readily using binompdf. Pdf stands for
probability density function. When dealing with discrete distributions like the
binomial, a calculation of a pdf will give us a probability at exactly one random
variable.
Sometimes we need to know the sum of two or more probabilities. For example,
what if we changes our probability above from P(X=3) to P(X < 3). Notice that we
are now interested in the probability of getting less than three, rather than exactly
3. When this type of problem arises we can use a cdf with the binomial
distribution, but the calculator defines it a little differently. It sums all the
probabilities for X=0 up to X= the given value of X.
Let’s create a table using the “n=6, p=.2, X=3,” binomial information we were
given earlier.
X
p(x)
0
1
2
3
4
0.2621 0.3932 0.2458 0.0820 0.0154
5
6
0.0015 0.00006
(notice the probability values do not add to precisely 1, that is because they have
all been rounded)
To find P(X< 3) what we want to do is add P(X = 0) with P(x = 1) and P(X = 2).
We can let our calculator do this by using the binomcdf command. To do this
press 2nd, then Vars, then scroll to binomcdf, then press ENTER and type in 6,
0.2 and 2. Why “2”? Because the binomcdf command will always add up the
probabilities from the lowest random variable in the distribution through the last
one of interest. We are interested in finding P(X < 3) so “X=3” is not included in
our table