Chapter 15 Binomial Distribution
Properties
– Two possible outcomes (success and failure)
– A fixed number of experiments (trials)
– The probability of success, denoted by p, is
the same on every trial
– The trials are independent
• Example: Suppose 75% of all drivers wear
their seatbelts. Find the probability that
four drivers might be belted among five
cars waiting traffic light?
Chapter 15 Binomial Distribution
• We look for the number of k successes in
n trial. Here k is less than or equal to n.
• Let X = number of successes in n trials.
• P(X=x) = C(n,x)*p^x*q^{n-x}
• Here p = probability of success, q = 1-p =
probability of failure.
Chapter 15 Binomial Distribution
• Example: Suppose 20 people come to the
blood drive. What is the probability that
there are 2 or 3 universal donors?
Solution:
• P(X=2) = C(20,2)(.06)^{2}(.94)^{18}=
0.2246
• P(X=3) = C(20,3)(.06)^{3}(.94)^{17}=
0.0860
• Answer: 0.3106
Chapter 15 Binomial Distribution
• To compute the C(n,x) number from the
TI-83 do the following
•
•
•
•
•
1. Type your number
2. Press MATH
3. Select Prob
4. Press nCr
5. Type your second number
Examples
2. A certain tennis player makes a
successful first serve 70% of the time.
Assume that each serve is independent
of the others. If she serves 6 times,
what’s the probability she gets
a) all 6 serves in
b) exactly 4 serves in?
c) at least four serves in?
d) no more than 4 serves in?
Examples
2 binomial model: p = 0.7, q = .3, n = 6
a) C(6,6)*.7^6*.3^0 = 0.118
b) C(6,4)*.7^4*.3^2 = 15*.7^4*.3^2 = .324
c) C(6,4)*.7^4*.3^2 = 0.324
C(6,5)*.7^5*.3^1 = 0.303
C(6,6)*.7^6*.3^0 = 0.118
Answer: 0.745
d) 1 – (.303+.118) = .575
Example
A statistics test contains 5 multiple choice
questions, each of which has four choices.
Suppose a student guesses the answer to
each question. Find the probability
distribution of X, the number of questions
the student answers correctly.
Example
X=k
0
1
2
3
4
5
P(X=k)
.2373
.3955
.2636
.0878
.0146
.0009
Concluding Remarks on
1. Binomial as an Approximation to
Hypergeometric Distribution
Example: A city has 1000 residents of whom 450
are male. 200 are to be selected at random
without replacement.
Clearly this is a hypergeometric distribution
problem with parameters:
N = 1000, n = 450, M = 200.
Binomial Approximation of Hypergeometric
Distribution
• Compute the probabilities for the following
values of m:
• m = 90, 91, 92, …, 110.
Example: A city has 10 residents of
whom 4 are male. 5 are to be selected
at random without replacement.
This is a hypergeometric distribution
problem with parameters: N = 10, n =
4, M = 5.
Compare the hypergeometirc and
binomial distributions.
Concluding Remarks
2. How much variation is typical?
Example:
Blood is drawn and a blood count is
performed by measuring two things:
 # of white blood cells
 proportion of the different types of white blood
cells.
These measurements are compared to typical
measurements to determine whether there is
cause for concern.
Concluding Remarks
Example (neutrophil)
Suppose that the typical proportion of\
neutrophils is 0.6. In a blood count it was
found that 45 out of 100 white blood cells
were neutrophils. Is there enough evidence
to be concerned? In other words the number
45 typical or not?
Concluding Remarks
We can model this by the binomial
distribution.
Let X = # of neutrophils in 100 white blood
cells drawn from a person whose
proportion of neutrophils is normal (i.e.
0.6).
We would like to study the probabilty
distribution
Concluding Remarks
Question: Is it surprising to find a blood
count different from 60?
Let us find the likelihood of getting 60
neutrophils in a 100 blood count (if we
assume that the proportion of neutrophils in
a normal person is 0.6)
Concluding Remarks
• This can be modeled by the binomila
distribution.
P( X = 60) = C(100,60)(.6^60)*(.4^(40)
=0.081
• Hence it is not surprising to find number
neutrophils to be other than 60
• How surprising is it to find 45 neutrophils?
Concluding Remarks
• Better question will be: How likel is it to
find the number of neutrophils to be less
than 45?
• We compute P(X <= 45) = .0017.
• Hence it is very unlikely that this happens
under normal situations. The differnece
between 60 and 45 is probably significant!
• Cause to be concerned is probably
justified!
Concluding Remarks
• When we compute P(X < = m), we are
finding what is called “a cumulative
distribution function”.
• Example: Let X be binomial with
parameters n = 5 and p = 0.4. Find the
cumulative distribution function of X.
• Note: m = 0, 1, 2, 3, 4, 5.
Concluding Remarks
m
0
1
2
3
4
5
P(X=m)
0.07776
0.2592
0.3456
0.2304
0.0768
0.01024
P(X<=m)
0.07776
0.33696
0.68256
0.91296
0.98976
1
Concluding Remarks
3. Is selection random? (Historical case:
Discrimination against Mexican-Americans-Read
the course pack (page 102)
Problem
Suppose in a county (Hidalgo County, TX)
79.1% of the population is of Hispanic origin.
From 1962-1972, 870 were summoned to
serve on a grand jury. Of these 339 had
Spanish surnames. How likely is this if the jury
selection was random?
Concluding Remarks
• Let X = # of Hispanics who served on a
grand jury.
• Compute P(X < = 339)
n = 870, p = .791, m = 0, 1, 2, 3, 4, …, 339.
P(X < = 339) = 4.18*10^(-148) ~ 0. This will
be the probability if the jury selection was
random. Exremely unlikely!