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Chapter 5: Discrete Probability Distributions
Section 5.1: Basics of Probability Distributions
As a reminder, a variable or what will be called the random variable from now on, is
represented by the letter x and it represents a quantitative (numerical) variable that is
measured or observed in an experiment.
Also remember there are different types of quantitative variables, called discrete or
continuous. What is the difference between discrete and continuous data? Discrete data
can only take on particular values in a range. Continuous data can take on any value in a
range. Discrete data usually arises from counting while continuous data usually arises
from measuring.
Examples of each:
How tall is a plant given a new fertilizer? Continuous. This is something you measure.
How many fleas are on prairie dogs in a colony? Discrete. This is something you count.
If you have a variable, and can find a probability associated with that variable, it is called
a random variable. In many cases the random variable is what you are measuring, but
when it comes to discrete random variables, it is usually what you are counting. So for
the example of how tall is a plant given a new fertilizer, the random variable is the height
of the plant given a new fertilizer. For the example of how many fleas are on prairie dogs
in a colony, the random variable is the number of fleas on a prairie dog in a colony.
Now suppose you put all the values of the random variable together with the probability
that that random variable would occur. You could then have a distribution like before,
but now it is called a probability distribution since it involves probabilities. A
probability distribution is an assignment of probabilities to the values of the random
variable. The abbreviation of pdf is used for a probability distribution function.
For probability distributions, 0 ≤ P ( x ) ≤ 1 and P ( x )
Example #5.1.1: Probability Distribution
The 2010 U.S. Census found the chance of a household being a certain size. The
data is in table #5.1.1 ("Households by age," 2013).
Table #5.1.1: Household Size from U.S. Census of 2010
Size of
household 1 2 3 4 5 6
7 or
more
Probability 26.7% 33.6% 15.8% 13.7% 6.3% 2.4% 1.5%
Solution:
In this case, the random variable is x = number of people in a household. This is a
discrete random variable, since you are counting the number of people in a
household.
This is a probability distribution since you have the x value and the probabilities
that go with it, all of the probabilities are between zero and one, and the sum of all
of the probabilities is one.
You can give a probability distribution in table form (as in table #5.1.1) or as a graph.
The graph looks like a histogram. A probability distribution is basically a relative
frequency distribution based on a very large sample.
Example #5.1.2: Graphing a Probability Distribution
The 2010 U.S. Census found the chance of a household being a certain size. The
data is in the table ("Households by age," 2013). Draw a histogram of the
probability distribution.
Table #5.1.2: Household Size from U.S. Census of 20 10
Size of
household 1 2 3 4 5 6
7 or
more
Probability 26.7% 33.6% 15.8% 13.7% 6.3% 2.4% 1.5%
Solution:
State random variable:
x = number of people in a household
You draw a histogram, where the x values are on the horizontal axis and are the x
values of the classes (for the 7 or more category, just call it 7). The probabilities
are on the vertical axis.
Graph #5.1.1: Histogram of Household Size from U.S. Census of 2010
Notice this graph is skewed right.
Table #5.1.4: Calculating the Mean for a Discrete PDF
x 1 2 3 4 5 6 7
P(x) 0.267 0.336 0.158 0.137 0.063 0.024 0.01 5
xP ( x )
Now add up the new row and you get the answer 2.525. This is the mean or the
expected value, μ = 2.525 people. This means that you expect a household in the
U.S. to have 2.525 people in it. Now of course you can’t have half a person, but
what this tells you is that you expect a household to have either 2 or 3 people,
with a little more 3-person households than 2-person households.
b.) Find the variance
Solution:
To find the variance, again it is easier to use a table version than try to just the
formula in a line. Looking at the formula, you will notice that the first
operation that you should do is to subtract the mean from each x value. Then
you square each of these values. Then you multiply each of these answers by
the probability of each x value. Finally you add up all of these values.
Table #5.1.5: Calculating the Variance for a Discrete PDF
x 1 2 3 4 5 6 7
P(x) 0.267 0.336 0.158 0.137 0.063 0.024 0.01 5
x − μ (^) - 1.525 - 0.525 0.475 1.4 75 2.475 3.4 75 4.4 75
( x − μ)
2
2.3256 0.2756 0.2256 2.1756 6.1256 12.0756 20.
( x − μ)
2
P ( x )
Now add up the last row to find the variance, σ
2 = 2.023375 people
2
. (Note:
try not to round your numbers too much so you aren’t creating rounding error
in your answer. The numbers in the table above were rounded off because of
space limitations, but the answer was calculated using many decimal places.)
c.) Find the standard deviation
Solution:
To find the standard deviation, just take the square root of the variance,
σ = 2.023375 ≈ 1.422454 people. This means that you can expect a U.S.
household to have 2.525 people in it, with a standard deviation of 1.42 people.
d.) Use a TI-83/84 to calculate the mean and standard deviation.
Solution:
Go into the STAT menu, then the Edit menu. Type the x values into L1 and
the P(x) values into L2. Then go into the STAT menu, then the CALC menu.
Choose 1:1-Var Stats. This will put 1-Var Stats on the home screen. Now
type in L1,L2 (there is a comma between L1 and L2) and then press ENTER.
If you have the newer operating system on the TI-84, then your input will be
slightly different. You will see the output in figure #5.1.1.
Figure #5.1.1: TI-83/84 Output
The mean is 2.52 5 people and the standard deviation is 1.422 people.
e.) Using R to calculate the mean.
Solution:
The command would be weighted.mean(x, p). So for this example, the process
would look like:
x<-c(1, 2, 3, 4, 5, 6, 7)
p<-c(0.267, 0.336, 0.158, 0.137, 0.063, 0.024, 0.015)
weighted.mean(x, p)
Output:
[1] 2.
So the mean is 2.525.
To find the standard deviation, you would need to program the process into R.
So it is easier to just do it using the formula.
Example #5.1.4: Calculating the Expected Value
In the Arizona lottery called Pick 3, a player pays $1 and then picks a three-digit
number. If those three numbers are picked in that specific order the person wins
$500. What is the expected value in this game?
Similarly, if the P (the variable has a value of x or less) < 0.05, then you can consider this
an unusually low value. Another way to think of this is if the probability of getting a
value as small as x is less than 0.05, then the event x is considered unusual.
Why is it "x or more" or "x or less" instead of just "x" when you are determining if an
event is unusual? Consider this example: you and your friend go out to lunch every
day. Instead of Going Dutch (each paying for their own lunch), you decide to flip a coin,
and the loser pays for both. Your friend seems to be winning more often than you'd
expect, so you want to determine if this is unusual before you decide to change how you
pay for lunch (or accuse your friend of cheating). The process for how to calculate these
probabilities will be presented in the next section on the binomial distribution. If your
friend won 6 out of 10 lunches, the probability of that happening turns out to be about
20.5%, not unusual. The probability of winning 6 or more is about 37.7%. But what
happens if your friend won 501 out of 1,000 lunches? That doesn't seem so
unlikely! The probability of winning 501 or more lunches is about 47.8%, and that is
consistent with your hunch that this isn't so unusual. But the probability of winning
exactly 501 lunches is much less, only about 2.5%. That is why the probability of getting
exactly that value is not the right question to ask: you should ask the probability of
getting that value or more (or that value or less on the other side).
The value 0.05 will be explained later, and it is not the only value you can use.
Example #5.1.5: Is the Event Unusual
The 2010 U.S. Census found the chance of a household being a certain size. The
data is in the table ("Households by age," 2013).
Table #5.1.7: Household Size from U.S. Census of 2010
Size of
household 1 2 3 4 5 6
7 or
more
Probability 26.7% 33.6% 15.8% 13.7% 6.3% 2.4% 1.5%
Solution:
State random variable:
x = number of people in a household
a.) Is it unusual for a household to have six people in the family?
Solution:
To determine this, you need to look at probabilities. However, you cannot just
look at the probability of six people. You need to look at the probability of x
being six or more people or the probability of x being six or less people. The
P ( x ≤ 6 ) = P ( x = 1 ) + P ( x = 2 ) + P ( x = 3 ) + P ( x = 4 ) + P ( x = 5 ) + P ( x = 6 )
Since this probability is more than 5%, then six is not an unusually low value.
The
P ( x ≥ 6 ) = P ( x = 6 ) + P ( x ≥ 7 )
Since this probability is less than 5%, then six is an unusually high value. It is
unusual for a household to have six people in the family.
b.) If you did come upon many families that had six people in the family, what
would you think?
Solution:
Since it is unusual for a family to have six people in it, then you may think
that either the size of families is increasing from what it was or that you are in
a location where families are larger than in other locations.
c.) Is it unusual for a household to have four people in the family?
Solution:
To determine this, you need to look at probabilities. Again, look at the
probability of x being four or more or the probability of x being four or less.
The
P ( x ≥ 4 ) = P ( x = 4 ) + P ( x = 5 ) + P ( x = 6 ) + P ( x = 7 )
Since this probability is more than 5%, four is not an unusually high value.
The
P ( x ≤ 4 ) = P ( x = 1 ) + P ( x = 2 ) + P ( x = 3 ) + P ( x = 4 )
Since this probability is more than 5%, four is not an unusually low value.
Thus, four is not an unusual size of a family.
d.) If you did come upon a family that has four people in it, what would you
think?
Solution:
Since it is not unusual for a family to have four members, then you would not
think anything is amiss.
2.) Suppose you have an experiment where you flip a coin three times. You then
count the number of heads.
a.) State the random variable.
b.) Write the probability distribution for the number of heads.
c.) Draw a histogram for the number of heads.
d.) Find the mean number of heads.
e.) Find the variance for the number of heads.
f.) Find the standard deviation for the number of heads.
g.) Find the probability of having two or more number of heads.
h.) Is it unusual for to flip two heads?
3.) The Ohio lottery has a game called Pick 4 where a player pays $1 and picks a
four-digit number. If the four numbers come up in the order you picked, then you
win $2,500. What is your expected value?
4.) An LG Dishwasher, which costs $800, has a 20% chance of needing to be
replaced in the first 2 years of purchase. A two-year extended warrantee costs
$112.10 on a dishwasher. What is the expected value of the extended warranty
assuming it is replaced in the first 2 years?
Section 5.2: Binomial Probability Distribution
Section 5.1 introduced the concept of a probability distribution. The focus of the section
was on discrete probability distributions (pdf). To find the pdf for a situation, you
usually needed to actually conduct the experiment and collect data. Then you can
calculate the experimental probabilities. Normally you cannot calculate the theoretical
probabilities instead. However, there are certain types of experiment that allow you to
calculate the theoretical probability. One of those types is called a Binomial
Experiment.
Properties of a binomial experiment (or Bernoulli trial):
- Fixed number of trials, n , which means that the experiment is repeated a specific
number of times.
- The n trials are independent, which means that what happens on one trial does not
influence the outcomes of other trials.
There are only two outcomes, which are called a success and a failure.
The probability of a success doesn’t change from trial to trial, where p = probability of
success and q = probability of failure, q^ =^1 −^ p^.
If you know you have a binomial experiment, then you can calculate binomial
probabilities. This is important because binomial probabilities come up often in real life.
Examples of binomial experiments are:
Toss a fair coin ten times, and find the probability of getting two heads.
Question twenty people in class, and look for the probability of more than half
being women?
Shoot five arrows at a target, and find the probability of hitting it five times?
To develop the process for calculating the probabilities in a binomial experiment,
consider example #5.2.1.
Example #5.2.1: Deriving the Binomial Probability Formula
Suppose you are given a 3 question multiple-choice test. Each question has 4
responses and only one is correct. Suppose you want to find the probability that
you can just guess at the answers and get 2 questions right. (Teachers do this all
the time when they make up a multiple-choice test to see if students can still pass
without studying. In most cases the students can’t.) To help with the idea that
you are going to guess, suppose the test is in Martian.
a.) What is the random variable?
Solution:
x = number of correct answers
P ( 2 correct answers) = P ( RRW) + P ( RWR) + P ( WRR)
2 3
1
2 3
1
2 3
1
2 3
1
d.) What is the probability of getting zero right, one right, and all three right?
Solution:
You could go through the same argument that you did above and come up
with the following:
r right P(r right)
0 right
1 *
0 3
3
1 right
1 3
2
2 right
2 3
1
3 right
3 3
0
Hopefully you see the pattern that results. You can now write the general formula
for the probabilities for a Binomial experiment
First, the random variable in a binomial experiment is x = number of successes.
Be careful, a success is not always a good thing. Sometimes a success is something that
is bad, like finding a defect. A success just means you observed the outcome you wanted
to see happen.
Binomial Formula for the probability of r successes in n trials is
P ( x = r ) =
n
C
r
p
r q
n − r where n
C
r
n!
r !( n − r )!
The n
C
r
is the number of combinations of n things taking r at a time. It is read “ n choose
r ”. Some other common notations for n choose r are C n , r
, and
n
r
. n! means you are
multiplying n * ( n − 1 )* ( n − 2 ) 2 * 1. As an example, 5! = 5 * 4 * 3 * 2 * 1 = 120.
When solving problems, make sure you define your random variable and state what n , p ,
q , and r are. Without doing this, the problems are a great deal harder.
Example #5.2.2: Calculating Binomial Probabilities
When looking at a person’s eye color, it turns out that 1% of people in the world
has green eyes ("What percentage of," 2013). Consider a group of 20 people.
a.) State the random variable.
Solution:
x = number of people with green eyes
b.) Argue that this is a binomial experiment
Solution:
1.) There are 20 people, and each person is a trial, so there are a fixed number
of trials. In this case, n = 20.
2.) If you assume that each person in the group is chosen at random the eye
color of one person doesn’t affect the eye color of the next person, thus the
trials are independent.
3.) Either a person has green eyes or they do not have green eyes, so there are
only two outcomes. In this case, the success is a person has green eyes.
4.) The probability of a person having green eyes is 0.01. This is the same for
every trial since each person has the same chance of having green eyes.
p = 0.01 and q = 1 − 0.01 = 0.
Find the probability that
c.) None have green eyes.
Solution:
P ( x = 0 ) =
20
C
0
0
20 − 0 ≈ 0.
d.) Nine have green eyes.
Solution:
P ( x = 9 ) =
20
C
9
9
20 − 9 ≈ 1.50 × 10
− 13 ≈ 0.
e.) At most three have green eyes.
Solution:
At most three means that three is the highest value you will have. Find the
probability of x is less than or equal to three.
The binomial formula is cumbersome to use, so you can find the probabilities by using
technology. On the TI-83/84 calculator, the commands on the TI-83/84 calculators when
the number of trials is equal to n and the probability of a success is equal to p are
binompdf ( n , p , r ) when you want to find P ( x = r ) and binomcdf ( n , p , r ) when you want
to find P ( x ≤ r ). If you want to find P ( x ≥ r ) , then you use the property that
P ( x ≥ r ) = 1 − P ( x ≤ r − 1 ) , since x ≥ r and x < r or x ≤ r − 1 are complementary events.
Both binompdf and binomcdf commands are found in the DISTR menu. Using R, the
commands are P ( x = r ) = dbinom ( r , n , p ) and P ( x ≤ r ) = pbinom ( r , n , p ).
Example #5.2.3: Using the Binomial Command on the TI-83/
When looking at a person’s eye color, it turns out that 1% of people in the world
has green eyes ("What percentage of," 2013). Consider a group of 20 people.
a.) State the random variable.
Solution:
x = number of people with green eyes
Find the probability that
b.) None have green eyes.
Solution:
You are looking for P^ ( x^ =^0 ). Since this problem is x = 0 , you use the
binompdf command on the TI-83/84 or dbinom command on R. On the TI-
83/84, you go to the DISTR menu, select the binompdf, and then type into the
parenthesis your n , p , and r values into your calculator, making sure you use
the comma to separate the values. The command will look like
binompdf ( 20 ,.01, 0 ) and when you press ENTER you will be given the
answer. (If you have the new software on the TI-84, the screen looks a bit
different.)
Figure #5.2.1: Calculator Results for binompdf
On R, the command would look like dbinom(0, 20, 0.01)
P ( x = 0 ) = 0.8179 . Thus there is an 81.8% chance that in a group of 20
people none of them will have green eyes.
c.) Nine have green eyes.
Solution:
In this case you want to find the P ( x = 9 ). Again, you will use the binompdf
command or the dbinom command. Following the procedure above, you will
have binompdf ( 20 ,.01, 9 ) on the TI-83/84 or dbinom(9,20,0.01) on R. Your
answer is P ( x = 9 ) = 1.50 × 10
− 13
. (Remember when the calculator gives you
1.50 E − 13 and R give you 1.50 e − 13 , this is how they display scientific
notation.) The probability that out of twenty people, nine of them have green
eyes is a very small chance.
d.) At most three have green eyes.
Solution:
At most three means that three is the highest value you will have. Find the
probability of x being less than or equal to three, which is P^ ( x^ ≤^3 ). This uses
the binomcdf command on the TI-83/84 and pbinom command in R. You use
the command on the TI-83/84 of binomcdf^ ( 20 ,.01,^3 ) and the command on R
of pbinom(3,20,0.01)
Figure #5.2.2: Calculator Results for binomcdf
Your answer is 0.99996. Thus there is a really good chance that in a group of
20 people at most three will have green eyes. (Note: don’t round this to one,
since one means that the event will happen, when in reality there is a slight
chance that it won’t happen. It is best to write the answer out to enough
decimal points so it doesn’t round off to one.
2.) If you assume that each child in the group is chosen at random, then
whether a child has autism does not affect the chance that the next child
has autism. Thus the trials are independent.
3.) Either a child has autism or they do not have autism, so there are two
outcomes. In this case, the success is a child has autism.
4.) The probability of a child having autism is 1/88. This is the same for
every trial since each child has the same chance of having autism. p =
and q = 1 −
Find the probability that
c.) None have autism.
Solution:
Using the formula:
P ( x = 0 ) =
10
C
0
0 87
10 − 0
Using the TI-83/84 Calculator:
P ( x = 0 ) = binompdf ( 10 , 1 ÷ 88 , 0 ) ≈ 0.
Using R:
P ( x = 0 ) = pbinom ( 0 , 10 , 1 / 88 ) ≈ 0.
d.) Seven have autism.
Solution:
Using the formula:
P ( x = 7 ) =
10
C
7
7 87
10 − 7
Using the TI-83/84 Calculator:
P ( x = 7 ) = binompdf ( 10 , 1 ÷ 88 , 7 ) ≈ 2.84 × 10
− 12
Using R:
P ( x = 7 ) = dbinom ( 7 , 10 , 1 / 88 ) ≈ 2.84 × 10
− 12
e.) At least five have autism.
Solution:
Using the formula:
P ( x ≥ 5 ) = P ( x = 5 ) + P ( x = 6 ) + P ( x = 7 )
+ P ( x = 8 ) + P ( x = 9 ) + P ( x = 10 )
10
C
5
5 78
10 − 5
10
C
6
6 78
10 − 6
10
C
7
7 78
10 − 7
10
C
8
8 78
10 − 8
10
C
9
9 78
10 − 9
10
C
10
10 78
10 − 10
Using the TI-83/84 Calculator:
To use the calculator you need to use the complement.
P ( x ≥ 5 ) = 1 − P ( x < 5 )
= 1 − P ( x ≤ 4 )
= 1 − binomcdf ( 10 , 1 ÷ 88 , 4 )
Using R:
To use R you need to use the complement.
P ( x ≥ 5 ) = 1 − P ( x < 5 )
= 1 − P ( x ≤ 4 )
= 1 − pbinom ( 4 , 10 , 1 / 88 )
Notice, the answer is given as 0.000, since the answer is less than 0.000.
Don’t write 0, since 0 means that the event is impossible to happen. The
event of five or more is improbable, but not impossible.
