Some standard, off the shelf probability distributions
Recall every random variable X has a probability distribution (probability
function). The probability distribution, expected value and variance of the
random variable describe the “theoretical behavior” of a random variable.
There are a number of important discrete probability distributions which allow us
to describe many common experiments.
The discrete uniform distribution;
Example: Let X be an integer in the range 1..n chosen at random.
Probability function:
P(x) = 1/n for x=1,2,3,4…n
E(X) =  = (1/n)(1) + (1/n)2+…(1/n)n
= (1/n)[ 1+2+3..+n]
= 1(/n) (n) (n+1)/2
(n+1)/2
E(x2) = (1/n)(12) + (1/n)22+…(1/n)n2
= 1/n( n (n+1) (2n+1)/6)
= (n+1)(2n+1)/6
Var(X) = 2 = E(x2) – [E(x)]2
= (n+1)(2n+1)/6 - (n+1)2/4 =(n2-1)/12
Example:
Roll a die: X is number of spots
X has a uniform discrete distribution
Range(X) = {1,2,3,4,5,6}
P(x) = 1/6 for x = 1..6
= E(X) = (6+1)/2 = 3.5
2 = Var(X) = (62 – 1)/12 = 2.92
 = 1.71
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Example:
Pick a whole number at random from 1 to 100
X is the number
Range(X) = {1,2,..100}
P(x) = 1/100for x = 1..100
= E(X) = (100+1)/2 = 55.5
2 = Var(X) = (1002 – 1)/12 = 833.25
 = 28.87
Example:
Let X be an integer in the range 10..20.
P(x) = 1/11 for x = 10,11,12,13,14,..20
To find E(X):
Let Y be a number in the range 1..11.
Thus, E(X) = (1+11)/2 = 6.
Now, X = Y + 9.
E(X) = E(Y+9) = E(Y) + 9 = 6+9 = 15
Var(X) = Var(Y+9) = Var(Y) = (112 – 1)/12 = 10
Note: the Discrete Uniform Distribution is really a family of distributions.
If X is n integer chosen at random in the range 1..6 then X has a uniform discrete
distribution with parameter 6.
If Y is a random integer in the range 1..564 then Y has a uniform discrete
distribution with parameter 564.
Thus there is not one discrete uniform distribution but infinitely many. However
all are similar in the sense that the pmf is p(x) = 1/n for x = 1...n
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The Binomial Distribution:
A Binomial Experiment:
Consider an experiment with the following properties:
1. The experiment consists of n independent trials.
For example : Flip a coin 45 times, Roll Dice 37 times
2. Each trial has two possible outcomes: Success(S) or Failure(F)
Example: Flip a coin 45 times. Each of the 45 trials has two possible
outcomes : Heads(S) or Tails(F)
Roll dice 37 times. Each of the 37 trials has two possible outcomes
Doubles(S) or not Doubles(F)
3. The trials are independent. The outcome of one trial does not affect
the outcome of another trial
4. The probability of success on each trial is the same (p) [The probability
of failure is q = 1-p.]
For example: Flip a coin 45 times. If Heads is success then the
probability of success on each trial is ½ . This does not change from
trial to trial.
Roll dice 37 times. If Doubles denotes success then the probability of
success on each trial is 1/6 and the probability of failure on each trial is
5/6. There is no change in probability from trial to trial.
An experiment with properties (1-4) is called a binomial experiment.
Consider a binomial experiment with n trials and probability of success on each
trial p. Let X be the number of successes in n trials. X is a random variable.
Range(X) = {0,1,2,3,4..n}. What is the probability function of X? What is E(X)?
What is Var(X)?
Example:
Roll dice 4 times. Let X be the number of times doubles appears.
X is a random variable. Range(X) = {0,1,2,3,4}
This is a binomial experiment:
1. There are 4 independent trials
2. On each trial--- Success: Rolling Doubles; Failure: Not Rolling Doubles
3. The trials are independent
4. On each trial; P(Success) = p = 1/6; P(5ailure) = q = 5/6
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Our first problem is to find the probability function for X;
x
0
1
2
3
4
P(x)
?
?
?
?
?
Here is a picture of the whole experiment:
Now lets find P(X = 2) ( the probability that the number of successes/doubles is
2).
This happens for the following sequences
Sequence
probability
DDD’D’
DD’DD’
DD’D’D
D’DDD’
D’DD’D
D’D’DD
(1/6)2(5/6)2
(1/6)2(5/6)2
(1/6)2(5/6)2
(1/6)2(5/6)2
(1/6)2(5/6)2
(1/6)2(5/6)2
Roll number where success
occurs
{1,2}
{1,3}
{1,4}
{2,3}
{2,4}
{3,4}
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Notice there are 6 branches – one for every group of 2 successes i.e C(4,2) = 6
Thus p(2) = (1/6)2(5/6)2+(1/6)2(5/6)2+(1/6)2(5/6)2+(1/6)2(5/6)2+(1/6)2(5/6)2+(1/6)2(5/6)2
=
6(1/6)2(5/6)2
Similarly, to find p(3) :
Sequence
probability
Rolls where success occurs
3
DDDD’
(
(1/6) (5/6)
{1,2,3}
DDD’D
(1/6)3(5/6)
{1,2,4}
3
DD’DD
(1/6) (5/6)
{1,,3,4}
D’DDD
(1/6)3(5/6)
{2,3, 4}
Notice that there are four brabches – on for every group of 3 successes: C(4,3) =
4
P(3) = 4* (1/6)3(5/6)
We can use the tree diagram to see that
p(0) = (5/6)4 -------------------------same as (1/6)0(5/6)4
p(1) = 4* (1/6) (5/6)3
p(2) = 6* (1/6)2(5/6)2
p(3) = 4* (1/6)3(5/6)
p(4) = (1/6)4--------------------same as (1/6)4(5/6)0
i.e.
x
0
1
2
3
4
P(x)
1*(1/6)0(5/6)4
4* (1/6) (5/6)3
6* (1/6)2(5/6)2
4* (1/6)3(5/6)
1*(1/6)4(5/6)0
=.482
=.386
=.116
=.015
=.001
ToTAL : 1
Or as we have seen in the example
x
0
1
2
3
4
P(x)
C(4,0)*(1/6)0(5/6)4
C(4,1)* (1/6) (5/6)3
C(4,2)* (1/6)2(5/6)2
C(4,3)* (1/6)3(5/6)
C(4,4)*(1/6)4(5/6)0
=.482
=.386
=.116
=.015
=.001
ToTAL :
1
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In general, given a binomial experiment with n trials where
p = probability of success
q = 1-p = probability of failure
and
X is the number of successes (in n trials) then the probability function is
p( r) = P(X = r) = C(n,r)prqn-r
Also, μ= E(X) = np; σ2 = npq, and σ=
2
Like the discrete uniform distribution, the binomial distribution is really a family of
distributions—one member for eah pair (n,p). The random variable X in the
above example had a
binom(4,1/6)
distribution.
binom(4,.1666) μ= .664 σ=.745
binom(20,.1666) μ= 3.33 σ=1.66
binom(4,.5) μ=2, σ=1
binom(20,.5) μ= 10 σ=2.23
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Example:
Suppose that 10% of all widgets produced by a machine are defective.
If 5 widgets are chosen at random:
a. what is the probability that exactly 2 are defective
Let X = the number of defective widgets…so “defective is success”
The distribution of X is binom(5,.1).
n=5
p = .1
q = 1-p
P(x = 2) = C(5,2)(.1)2(.9)3 = (5*4)/(2*1) (.1)2(.9)3 = 10*(.1)2(.9)3 = .0729
b . Find the probability that at least 4 are defective:
P(x > 4) = p(X=4) + p(X=5)=
C(5,4) (.1)4(.9)1 + C(5,5) (.1)5(.9)0 = 5*.00009 + 1*.00001 = .00046
c. Find the probability that there is at least one defective
Remember: the opposite of “at least one” is “none”
P(X > = 1) = 1 – P(X = 0) = 1- C(5,0)(.1)0(.9)5 = 1 - .59049 = .40951
d. Find the expected number of defective widgets, the variance, and standard
deviation
μ = E(X) = np = 5*.1 = .5; σ2 = npq = 5*.1*.9 = .45 ; σ= .671
Example
In a large population, 20 people are selected at random. and asked if they watch
Spongebob Squarepants. In fact, 30% of the population watches Spongebob.
Suppose that X is the number who watch Spongebob. Find
a. the probability that X is at most 10.
The distribution of X is binom(10,.4)
n = 20
p = .3
q = .7
Note : μ = np = 20*.3 = 6; σ2 =npq = 20*.3*.7 = 4.2
Using the cumulative table:
P(X < 10 ) = .983
b. the probability that X is exactly 10:
P(x = 10) = P(X < 10) – P(X < 9) = .983 - .952 = .031
c. the probability that at least one person watches Spongebob
P(at least one) = 1 – P(X = 0) = 1 - .001 = .999
d. the probability that between 5 and 12 (inclusive) watch Spongebob
P(5 < X < 12) = P(X < 12) – P(X < 4) = .999 - .238 = .761
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Poisson Distribution and Random events
Any process with the following characteristics is called a Poisson process.
1. The occurrences of some event in time or space are independent
2. The probability of a single occurrence of an event in an interval of time or
space is proportional to the length of the interval of time or space
3. In an “infinitely small” interval of time or space the probability of more than
one occurrence of an event is ~ 0. As an interval of time or space
approaches 0, the probability of an occurrence of an event approaches 0.
Consider a Poisson process where the average number of occurrences of some
event during an interval is λ. If X be the number occurrences of the event in
some interval of time or space, then X is said to have a Poisson Distribution with
parameter λ X is a random variable.
Example:
1. Suppose on the average a Keebler Chocolate Chip Cookie has 16 chips
per cookie. If X is the number of chips in a cookie selected at random,
then X has a Poisson distribution with parameter λ = 16. Range(X) =
{1,2,3,4,….}
2. Suppone, on the average, 6 cars cross an intersection per minute. Let X
be the number of cars that cross in any minute. Then X has a Poisson
Distribution with λ = 6.
3. Suppose that the average number of calls coming to a switchboard is
16/hour. Let X be the number that come in any half hour period.
Range(X) = {0,1,2,3,4,5,6,……..}. X has a Poisson distribution with λ = 8.
Suppose that X is random variable and the distribution of X is Poisson with
parameter λ. Then
1. Range(X) = {1,2,3,4,5…..}
2. the probability function for X is
p( r ) = P(X = r ) = λre-λ/r!
3. μ = E(X) = λ
4. σ2 = λ; σ =

As before, the Poison distribution is really a family of distributions. There is a
different distribution for each value of λ
8
λ=3
λ=6
λ = 13
λ = 25
Poisson distribution
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Example:
On the Average a Keebler Chocolate Chip Cookie contains 16 chips. Find
the probability that a cookie selected at random has exactly 18 chips.
Let X be the number of chips in a cookie selected at random. Assume that the
diustribution of X is Poisson with λ = 16.
Find P(X = 18)
P(X = 18) = λ18e-λ/18! = 1618e-16/18! ~ .083
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Example; Using the Cumulative table
Suppose that on the average there are 6 traffic accidents at a certain
intersection per year. In a given year,
a. what is the probability that 3 or fewer accidents occur?
Let X be the number of accidents that occur in a year. Assume that X has
a Poisson distribution with λ = 6.
P(X < 3 ) = .151
b. What is the probability that at least one accident occurs in a year?
P(X > 1) = 1 - P(X = 0) = 1-.002 = .998
c. What is the probability that 4 or more accidents occur?
P(X > 4) = 1 – P(X < 3 ) = 1 - .151 = .849
Relationship between binomial and Poisson distributions:
In a binomial experiment if
n is large and
p is small
then
binom(n,p) ~poisson(λ) where λ = np
In other words, the poisson distribution can be used to approximate the binomial
distribution
Example:
Suppose X is binom(400, .005)
n = 400,
p = .005
q = .995
Use the binomial probability mass function to find P(X = 1):
P(X = 1) = C(400,1)*.0051.995399 = 0.27066943301711285515159151027602
Now suppose Y is Poisson with λ = np = 400*.005 = 2
P(Y = 1) = λ1e-λ/1! = 21e-2/1! = 2e-2 = 0.27067056647322538378799898994497
The approximation is generally good if:
n > 100
p < .01
np < 20
which was the case in the example above.
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