Discrete Probability Distributions
The Bernoulli Distribution – any random variable denoting the
presence or absence of a certain condition in an observed phenomenon.
One of the outcomes is termed a “success” and the other a “failure”.
The Bernoulli distribution equation is
x = 0, 1
for 0  p  1
E(X) = p and V(X) = p(1-p)
The Binomial Distribution
A random variable Y possess a binomial distribution if
1. The experiment consists of a fixed number n of trials.
2. Each trial can result in one of only two possible outcomes, called
“success” and “failure”.
3. The probability of “success”, p, is constant from trial to trial.
4. The trials are independent.
5. Y is defined as the number of successes among the n trials.
The Binomial Distribution
y = 0, 1, 2, …, n
for 0  p  1
E(Y) = np and V(Y) = np(1-p)
Example: Suppose a large lot contains 10% defective fuses. Four fuses
are randomly sampled from the lot.
a. Find the probability that exactly one fuse in the sample of four is
defective.
b. Find the probability that at least one fuse in the sample of four is
defective.
Solution:
a.
b.
= 0.2916
= 0.3439
Example: In a study of lifetimes for a certain type of battery, it was
found that the probability of a lifetime X exceeding 4 hours is 0.135. If
three such batteries are in use in independently operating systems, find
the probability that
a. Only one of the batteries lasts 4 hours or more.
b. At least one battery lasts 4 hours or more.
Solution:
Let y = the number of batteries (out of three) lasting 4 hours or more.
We can reasonably assume that Y has a binomial distribution with n = 3
and p = 0.135. Hence,
a.
b.
= 0.303
= 0.647
Example: An industrial firm supplies 10 manufacturing plants with a
certain chemical. The probability that any one firm calls in an order on a
given day is 0.2, and this probability is the same for all 10 plants. Find
the probability that, on a given day, the number of plants calling in order
is
a. Exactly 3
b. At most 3
c. At least 3
Solution:
Let Y = the number of plants (out of 10) calling in orders on the day in
question. If the plants order independently, then Y can be modelled to
have a binomial distribution with n = 10 and p = 0.2.
a. The probability of exactly 3 out of 10 plants calling in order is
= 0.201
b. The probability of at most 3 out of 10 plants in orders is
= 0.107 + 0.269 + 0.302 + 0.201 = 0.879
c. The probability of at least 3 out of 10 plants calling in orders is
= 1 – (0.107 + 0.269 + 0.302) = 0.322
The Geometric and Negative Binomial Distributions
The Geometric Distribution



Is a special case of negative binomial distribution
Is interested in the random variable Y, the number of trial, on which
the first success occurs
The results of y successive trials are then as follows with F = failure
and S = success
(y- 1) trials

Last (yth) trial
Therefore, the probability of observing failures on first (y-1) trials
and success on the last (yth) trial (i.e. the probability of requiring y
trials till the first success is observed) is
Success on Trial
Number
1
2
3
4
.
.
.
y
Sequence
The Geometric Distribution
S
FS
FFS
FFFS
.
.
.
FF…FS
Probability
p
(1-p)p
(1-p)(1-p)p
(1-p)(1-p)p
.
.
.
(1-p)(1-p)…(1-p)p
Example: A recruiting firm finds that 30% of the applicants for a certain
industrial job have advanced training in computer programming.
Applicants are selected at random from the pool and are interviewed
sequentially.
a. Find the probability that the first applicant having advanced training
is found on the fifth interview.
b. Suppose the first applicant with the advanced training is offered the
position, and the applicant accepts. If each interview costs Php300,
find the expected value and variance of the total cost of
interviewing incurred before the job is filled.
Solution:
a.
b. The total cost of interviewing in C = 300Y.
and
The standard deviation of the total cost is
= Php836.66.
The Negative Binomial Distribution
Let Y denote the number on the trial on which the rth success occurs in a
sequence of independent Bernoulli trials with p denoting the common
probability of “success”. The negative binomial distribution is defined by
two parameters, r and p.
The Negative Binomial Distribution equation is
Example: In the previous example, 30% of the applicants for a certain
position have advanced training in computer programming. Suppose three
jobs requiring advanced programming training are open. Find the
probability that the third qualified applicant is found on the fifth interview,
if the applicants are interviewed sequentially and at random.
Solution:
Let Y denote the number of the trial on which the third qualified candidate
is found. Then Y can reasonably be assumed to have a negative binomial
distribution with r = 3 and p = 0.3.
The Poisson Distribution
The Poisson distribution occurs when we count the number of occurrences
of an event over a given time period or length or volume. For example
 The number of flaws in a square yard of fabric
 The number of bacterial colonies in a cubic centimetre of water
 The number of times a machine fails in the course of a workday
The Poisson distribution equation is
Example:
For a certain Manufacturing industry, the number of industrial accidents
averages three per week.
a. Find the probability that no accident will occur in a given week.
b. Find the probability that two accidents will occur in a given week.
c. Find the probability that at most four accidents will occur in a given
week.
d. Find the probability that two accidents will occur in a given day.
Solution:
With  = mean number of accidents per week = 3, hence
a. P(No accident in a given week) = p(0) =
b. P(Two accidents in agiven week) = p(2) =
c. P(at most 4 accidents in a given week)
= p(0) + p(1) + p(2) + p(3) + p(4)
=
d. Since we are interested in the number of accidents on a given day,
we need to obtain a new  on a per day basis which is  = 3/7
=0.2857. Where the 7 represents the number of days in a week.
Hence,
P(Two accidents on a given day) = p(2) =
0.031
The Hypergeometric Distribution
In general, suppose a lot consists of N items, of which k are of one type
(called successes) and N-k are of another type (called failures). Suppose
n items are sampled randomly and sequentially from the lot, with none of
the sampled items being replaced, that is sampling without replacement.
Let Y denote the total number of successes among the n sampled items.
Then the probability distribution of Y is described as the hypergeometric
distribution.
The hypergeometric distribution is defined by three parameters, n, k, and
N.
The Hypergeometric distribution equation is
Example: A personnel director selects two employees for a certain job
from a group of six employees, of which one is female and five are male.
Find the probability that the female is selected for one of the jobs.
Solution:
If the selections are made at random and if Y denotes the number of
females selected, the hypergeometric distribution would provide for the
behaviour of Y.
Hence N = 6, k = 1, and n = 2, and y = 1.