Download 1 The Binomial Distribution

1
The Binomial Distribution
The Binomial Distribution is a probability experiment that meets the following
criteria.
1. Each trial has only 2 possible outcomes that are considered success or
failure.
2. There must be a fixed number of trials. For example, flip a coin 8 times,
take a multiple choice test that has 20 questions, or randomly choose 10
people from the population.
3. The outcomes are independent. This means that past events do not affect
the probability of future events. These are events with replacement or
when the population is assumed to be sufficiently large, like the population
of the United States.
4. The probability of a success must remain the same for each trial. This
might be best explained by example. Consider a multiple choice test
that has 20 questions. If each question has four possible answers, then
the probability of guessing the correct answer is one-fourth; however, if
some of the questions have five possible answers instead of four, then the
probability of guessing correctly is changed to one-fifth. In this second
scenario, the probability of success did not remain the same for each trial.
Following is an example problem that will be used to illustrate how to calculate
problems regarding the binomial distribution. About one in 10 people in the
United States are left-handed (L or lefties). Randomly draw 5 people from the
population. Construct a probability distribution and use it to find the following
probabilities.
1. three people are left-handed,
2. at most three people are left-handed,
3. at least three people are left-handed.
Constructing the probability distribution will allow us to see the sample space
and to better understand what the binomial distribution is describing. The
sample space for drawing five lefties is that it is possible that zero lefties are
picked from the crowd, one of the five people could be lefties, two of the five
could be lefties, three of the five could be lefties, four of the five could be lefties,
and finally, all of the people selected could be lefties. The sample space has six
different outcomes. Let x be the probability of selecting a desired number of
lefties, then part of the probability distribution can be constructed as follows.
1
x
0
1
2
3
4
5
P (x)
For the sake of illustration, begin by finding P (2), the probability that 2 of
the five people selected are lefties. One approach to the problem is to write an
example of picking two lefties, which also means there will be 3 righties in the
outcome. Let lefties be denoted L and righties be R. Then one example of how
this could happen is
L
L
R
R
R.
This is not the only way this could happen, but it is an example that quickly
comes to mind. From here, applying what has been learned about counting
and probability will uncover the answer. The probability of selecting a person’s
handedness is independent of the people already selected. The probability of
1
and the probability of a rightie is the complement, or
selecting a leftie is 10
1
9
1 − 10 = 10 . This gives the following solution.
L
L
R
R
R
1
10
1
10
9
10
9
10
9
10
From what has been learned, multiply the probabilities to find the answers;
however, this would not be the correct because because this event can happen
more ways than this one example. It can happen in other ways, for example,
LRLRR, or RRLRL, and there are other arrangements. Instead of trying to
find all other possibilities, notice the LLRRR is a word, or arrangement of
letters. The number of words that can found is 5!/(2!3!). Multiply the above
probabilities by the number of ways of selecting 2 lefties, or
P (2) =
5!
2!3! ·
L
L
1
10 ·
1
10 ·
R
9
10 ·
R
9
10 ·
R
9
10
This process can be continued to find all probabilities of the sample space.1
1
P (0) =
5!
·
5!
R
R
R
R
R
9
·
10
9
·
10
9
·
10
9
·
10
9
10
L
R
R
R
R
P (1) =
5!
·
1!4!
1
·
10
9
·
10
9
·
10
9
·
10
9
10
L
L
L
R
R
P (3) =
5!
·
3!2!
1
·
10
1
·
10
1
·
10
9
·
10
9
10
2
The example of P (2) has more to offer. The problem has been stated in terms
of selecting a leftie as a success. Consider the following analysis of P (2).
5!
1 1 9 9 9
·
·
·
·
·
3!2! 10 10 10 10 10
2 3
5!
9
1
=
·
·
(5 − 2)!2!
10
10
2 5−2
1
9
=5 C2 ·
10
10
P (2) =
The problem was that lefties are success. The equation above translates as, “5
choose 2 successes times the probability of a success raised to the number of
successes times the probability of failures raised to the number of failures.” In
general, any individual outcome of a binomial distribution can be found using
this technique.
The formal equation for a binomial experiment with x successes and n trials is
P (x) =n Cx px q n−x ,
where p is the probability of success, and q = 1 − p is the probability of failure.
Now the original probability distribution can be completed.
0
1
2
3
4
5
P (x)
P (0)
P (1)
P (2)
P (3)
P (4)
P (5)
P (x)
59049
100000
32805
100000
7290
100000
810
100000
45
100000
1
100000
x
Notice that the sum of the probabilities is one, so the probability distribution
seems correct.
Return to the initial problem and the three question to address.
1. Find the probability that exactly three people are left-handed. This probability can be taken directly from the probability distribution or can be
P (4) =
P (5) =
5!
·
4!1!
5!
·
5!
L
L
L
L
R
1
·
10
1
·
10
1
·
10
1
·
10
9
10
L
L
L
L
L
1
·
10
1
·
10
1
·
10
1
·
10
1
10
3
calculated directly as
P (3) =5 C3 ·
1
10
3 9
10
5−3
.
Further, the TI-84 has the capability of solving this as the function:
binompdf(5,1/10,3), where the syntax means that 5 is the number of
trials, 1/10 is the probability of a success, and 3 is the number of successes. Access binompdf by pressing 2nd , vars to access “distr”, then
scroll down to binompdf and press enter to select it. On the older model
calculators, type binompdf(5,1/10,3), and on the new models answer
the indicated questions then select “calculate”, then enter .
2. Find the probability that at most three people are left-handed. This
means the probability of zero lefties, or the probability of 1 leftie, or
the probability of 2 lefties, or the probability of 3 lefties: P (0) + P (1) +
P (2)+P (3). Since the probability distribution has been constructed, these
probabilities can be added directly as
32805
7290
81
99954
59049
+
+
+
=
= 0.99954
100000 100000 100000 100000
100000
The concept of adding from zero success to a certain number of successes
is often helpful, so the TI-84 has the function binomcdf. The binomcdf
function always begins adding from the probability of zero successes to,
and including, the last denoted success. Try the calculation again using
binomcdf. Access binomcdf by pressing 2nd , vars to access the “distr”,
then scroll down to binomcdf and press enter to select it. On the older
model calculators, type binomcdf(5,1/10,3). On the newer model TI84s, answer the indicated questions, then select “calculate”, then enter .
Notice that the syntax of binompdf and binomcdf are the same, but the
functions are different.
3. Find the probability that at least three people are left-handed. At least
three means the probability of three lefties, or the probability of four
lefties, or the probability of five lefties, i.e., P (3) + P (4) + P (5). Add each
probability directly from the probability distribution,
45
1
856
810
+
+
=
= 0.0856.
100000 100000 100000
100000
Another approach is to utilize binomcdf. To see this recall that the sum
4
of the probabilities of a probability distribution is one. Then,
P (0) + P (1) + P (2) + P (3) + P (4) + P (5) = 1
−P (0) − P (1) − P (2))
= − P (0) − P (1) − P (2))
P (3) + P (4) + P (5) = 1 − (P (0) + P (1) + P (2))
|
{z
}
|
{z
}
P (at least 3) = 1 − (binomcdf(5,1/10,2))
P (at least 3) = 0.0856
5