Chapter 5 Discrete Probability Distributions

5.1 Random Variables

Random variable

A random variable is a variable whose value is determined by the outcome of a random

experiment.

Discrete random variable

A random variable that assumes countable values is called a discrete random variable.

Example

1. The number of cars sold at a dealership during a given month.

2. The number of houses in a certain block.

3. The number of heads obtained in three tosses of a coin.

Continuous random variable

A random variable that can assume any value contained in one or more intervals is called

a continuous random variable.

Example

1. The height of a person.

2. The time taken to complete an examination.

3. The weight of a baby.

4. The price of a house.

5.2 Probability distribution of a discrete random variable

Probability distribution of a discrete random variable lists all the possible values that

the random variable can assume and their corresponding probabilities.

The ordered pairs (x, P(x)) where )()( xXPxP is called the probability distribution

of the discrete random variable X.

Two characteristics of a probability distribution:

1. 0 ≤ P(x) ≤ 1 for each value of x

2. ∑ P(x) = 1

Example 5.1: The following table lists the probability distribution of the number of

breakdowns per week for a machine based on past data.

Breakdowns per week 0 1 2 3

Probability 0.15 0.20 0.35 0.30

a) Present this probability distribution graphically.

b) Find the probability that during a given week, the number of breakdowns for

this machine is

i) exactly 2 ii) 0 to 2

iii) more than 1 iv) at most 1

Solution:

Example 5.2: According to a survey, 60% of all students at a large university suffer from

math anxiety. Two students are randomly selected from this university. Let x denote the

number of students in this sample who suffer from math anxiety. Develop the probability

distribution of x.

Solution:

Mean

The mean of a discrete random variable x is the value that is expected to occur per

repetition, on average, if an experiment is repeated a large number of times. It is denoted

by and calculated as

)(xxP

The mean of a discrete random variable x is also called its expected value and is denoted

by )(xE ; that is,

)()( xxPxE

Example 5.3: (Refer to Example 5.1)

Find the mean number of breakdowns per week for this machine.

Solution:

Standard deviation

The standard deviation of a discrete random variable x measures the spread of its

probability distribution and is computed as

22

22

)(

))(()(

xPx

xExE

Example 5.4: Loh Corporation is planning to market a new makeup product. It will earn

an annual profit of $4.5 million if this product has high sales and an annual profit of $1.2

million if the sales are medium, and it will lose $2.3 million a year if the sales are low.

The probabilities of these three scenarios are 0.32, 0.51 and 0.17 respectively.

a) Let x be the profits (in millions of dollars) earned per annum by the company

from this product. Write the probability distribution of x.

b) Calculate the mean and standard deviation of x.

Solution:

5.3 The Binomial Probability Distribution

5.3.1 Factorials and combinations

Factorial 123)...3)(2)(1(! nnnnn

where 0! = 1

Combinations (order does not matter)

The number of combinations for selecting x from n distinct elements is given by the

formula )!(!

!

xnx

n

x

nCx

n

5.3.2 The Binomial Experiment

An experiment that satisfies the following four conditions is called a binomial

experiment.

1. There are n identical trials.

2. Each trial has two and only two outcomes. These outcomes are usually called a

success and a failure. Each trial is called a Bernoulli trial.

3. P(success) = p , P(failure) = q, and 1 qp . The probabilities p and q remain

constant for each trial.

4. The trials are independent.

Example 5.5: Consider the experiment consisting of 10 tosses of a coin. Determine

whether or not it is a binomial experiment.

Solution:

1. There are a total of 10 trials (tosses) and they are all identical. All 10 tosses are

performed under identical conditions. Here n = 10.

2. Each trial has only two possible outcomes: a head and a tail. Let a head be called

a success and a tail be called a failure.

3. The probability of obtaining a head is ½ and that of a tail is ½ for any toss. The

sum of these two probabilities is 1.0. Also, these probabilities remain the same for

each toss.

4. The trials are independent. The result of any preceding toss has no bearing on the

result of any succeeding toss.

Consequently, the experiment consisting of 10 tosses is a binomial experiment.

The Binomial Random Variable The random variable X that represents the number of successes in n trials for a binomial

experiment is called a binomial random variable.

Binomial Probability Distribution The probability distribution of X in a binomial experiment is called the binomial

probability distribution or the binomial distribution and denoted by B(n, p). We will

write this as ).,(~ pnBX

Binomial Formula

For a binomial experiment, the probability of exactly x successes in n trials is given by

the binomial formula

xnxqpx

nxXP

)(

where n = total number of trials

p = probability of success

q = 1 – p = probability of failure

x = number of successes in n trials

n – x = number of failure in n trials

To find the probability of x successes in n trials for a binomial experiment, the only

parameters needed are those of n and p.

Example 5.6: Five percent of all DVD players manufactured by a large electronics

company are defective. A quality control inspector randomly selects three DVD players

from the production line. What is the probability that exactly one of these three DVD

players is defective?

Solution:

Example 5.7: In a Robert Half International survey of senior executives, 35% of the

executives said that good employees leave companies because they are unhappy with the

management (USA TODAY, February 10, 2009). Assume that this result holds true for the

current population of senior executives. Let x denote the number in a random sample of

three senior executives who hold this opinion. Write the probability of x and draw a bar

graph for this probability distribution.

Solution:

Binomial Probability Table

Apart from using the binomial distribution formula, we can also refer to a special

statistical table to obtain binomial probabilities. The statistical table enables us to obtain

the probability of r successes for a binomial variable with parameters n and p (n trials and

probability of a success equals p). We shall refer to the binomial table listed in The New

Cambridge Statistical Tables. This particular table tabulate cumulative binomial

probabilities in the form of:

P(X ≤ x) = P(X = 0) + P(X = 1) + ... + P(X = x)

i.e. the probability of obtaining at most r successes from n trials.

The following is an example of some binomial probabilities tabulated in the Cambridge

Statistical Tables:

n = 5 r = 0 1 2 3 4

p = 0.01 0.9510 0.9990 number of successes, x = 2

.02

.03

.04

.

.

.

.9039

.8587

.8154

.9962

.9915

.9852

0.9999

.9997

.9994

0.50 0.0313 0.1875 0.500 0.8125 0.9688

To facilitate the usage of this table, the following relationships are useful to know:

1. P(X = r) = P(X ≤ r) − P(X ≤ r −1)

2. P(X < r) = P(X ≤ r −1)

3. P(X ≥ r) = 1− P(X ≤ r −1)

4. P(X > r) = 1− P(X ≤ r)

Note: the right-hand side of the above equations all contain the inequality “ ≤ ”.

Example 5.8: Suppose X ∼ Bin(n = 5, p = 0.04). Find the following probability from the

binomial statistical table:

(i) P(X ≤ 2) , (ii) P(X < 2) , (iii) P(X = 2) , (iv) P(X ≥ 2) , (v) P(X > 2) .

Solution:

P(X ≤ 2) = 0.9994

The Cambridge Statistical Table tabulate binomial probabilities for p = 0.01 until p =

0.50 only. For a binomial event with probability of a success greater than 0.50, one

should consider it's complementary event. This is because, if the probability of a success,

p > 0.50, then the probability of it's complement (a failure), would be less than 0.5, i.e. (1

– p) ≤ 0.5.

Let X be the number of successes and Y be the number of failures from n binomial

trials. If the probability of a success is denoted by p and the probability of a failure is

denoted by q, then

X ∼ Bin(n, p) and Y ∼ Bin(n, q), with p + q = 1

Since the number of trials for a binomial experiment is fixed at n, therefore,

X + Y = n or X = n – Y.

Thus,

P(X = r) = P(n – Y = r)

= P(Y = n – r)

and P(X ≤ r) = P(n – Y ≤ r)

= P(Y ≥ n – r)

Example 5.9: Let X ∼ Bin(n = 10, p = 0.90). Find P(X = 5) dan P(X ≥ 8).

Shape of the Binomial Distribution

For any number of trials n:

1. The binomial probability distribution is symmetric if p = 0.50 .

Table 5.1: Probability Distribution

of x for n = 4 and p = 0.50

2. The binomial probability distribution is skewed to the right if p is less than

0.50.

Table 5.2: Probability Distribution

of x for n = 4 and p = 0.30

x P(x)

0 0.0625

1 0.2500

2 0.3750

3 0.2500

4 0.0625

x P(x)

0 0.2401

1 0.4116

2 0.2646

3 0.0756

4 0.0081

3. The binomial probability distribution is skewed to the left if p is greater than

0.50.

Table 5.3: Probability Distribution

of x for n = 4 and p = 0.80

Mean and Standard deviation of a binomial distribution

Suppose ),(~ pnBX . Then the mean and standard deviation of X are

np and npq

where pq 1 .

Suppose ),(~ pnBX and XnY . Then

)1,(~ pnBY .

Example 5.10: According to Harris Interactive Survey conducted for World Vision and

released in February 2009, 56% of teens in the United State volunteer time for charitable

causes. Assume that this result is true for the current population of U.S. teens. A sample

of 60 teens of 60 teens is selected. Let x be the number of teens in this sample who

volunteer time for charitable causes. Find the mean and standard deviation of the

probability distribution of x.

Solution:

x P(x)

0 0.0016

1 0.0256

2 0.1536

3 0.4096

4 0.4096

5.4 The Hypergeometric Distribution

The hypergeometric probability distribution is useful for determining the probability of a

number of occurrences when sampling is done without replacement. It counts the

number of successes (x) in n selections from a population of N elements, k of which are

successes and (N-k) of which are failures.

Hypergeometric Probability Density Function

Consider a statistical experiment where a sample of n observations are to be taken

without replacement from a population of size N. The population contains k items that are

labeled 'success' and N - k items that are labeled 'failure'. If a random variable X assumes

the value equal to the number of successes in the sample of size n, then X has a

hypergeometric distribution with parameters N, n and k. The random variable X is said

to be hypergeometrically distributed with parameters N, n and k, and has the following

probability density function:

nN

xnkN

xk

C

CCxXP

, x = 0, 1, 2, , k

Example 5.11: Dawn Corporation has 12 employees who hold managerial positions. Of

them, 7 are female and 5 are male. The company is planning to send 3 of these 12

managers to a conference. If 3 managers are randomly selected out of 12,

a. find the probability that all 3 of them are female.

b. find the probability that at most 3 of them is a female.

Solution:

5.5 The Poisson Probability Distribution

The Poisson distribution is a discrete probability distribution that applies to occurrences

of some events over a specified interval. The random variable X is the number of

occurrences of the events in an interval. The interval can be time, distance, area, volume

or some similar unit.

Properties of Poisson Process

1. The number of outcomes occurring in an interval is independent of the number

that occurs in any other interval.

2. The probability that a single outcome will occur during a very short interval is

proportional to the length of the interval and does not depend on the number of

outcomes occurring outside this interval.

3. The probability that more than one outcome will occur in such a short interval is

negligible.

The following are examples of discrete random variable for which the Poisson

probability distribution can be applied.

1. The number of telemarketing phone calls received by a household during a given

day.

2. The number of mistakes typed on a given page.

3. The number of customers entering a grocery store during a one-hour interval.

Poisson Probability Distribution Formula

The probability of x occurrences in an interval is

,2,1,0!

)(

xx

exP

x

where is the mean number of occurrences in that interval .

The Poisson distribution is denoted by Poisson( ) or P( ).

Example 5.12: A washing machine in a laundry shop breaks down an average of three

times per month. Find the probability that during the next month this machine will have

a) exactly two breakdowns

b) at most one breakdown

Solution:

The Poisson probability Table

Like the binomial probabilities, the Poisson probabilities can also be obtained from a

special statistical table. The statistical table gives the probability of obtaining r outcomes

for a Poisson variable with parameter μ . We shall refer to the Poisson table listed in The

New Cambridge Statistical Tables. This particular table tabulate cumulative Poisson

probabilities in the form of:

P(X ≤ r) = P(X = 0) + P(X = 1) + ... + P(X = r)

i.e. the probability of obtaining at most r outcomes.

The following is an example of some Poisson probabilities tabulated in the Cambridge

Statistical Tables:

r = 0 1 2 3

0.00 1.000

.02 0.9802 0.9998

3.00 0.0498 0.1991 0.42332 0.6472

Example 5.13: Let X ~ Poisson (3). Find (i) P(X 2), (ii) P(X = 2), (iii) P(X 2)

Solution:

P(X ≤ 2) = 0.4232

Mean and Variance of the Poisson probability distribution

Suppose )(Poisson~ X , then 2 .

Poisson approximation of the Binomial Distribution

A binomial distribution with a sufficiently large number of trials, n, and a sufficiently

small probability of success, p, can be approximated with a Poisson distribution. We will

consider n to be sufficiently large when n ≥ 100 and p to be sufficiently small when p <

0.1, such that np < 10.

Let X ∼ Bin (n, p) with n ≥ 100 and p < 0.1, such that np < 10. A Poisson

appproximation of X results in:

X ~ Poisson (μ = np )

Example 5.14: Cynthia’s Mail Order Company provides free examination of its products

for 7 days. If not completely satisfied, a customer can return the product within that

period and get a full refund. According to past records of a company, an average of 2 of

every 10 products sold by this company are returned for a refund. Find the probability

that exactly 6 of the 40 products sold by this company on a given day will be returned for

a refund by using (i) binomial distribution, (ii) Poisson approximation.

Solution