The Probability Mass Function (PMF) is also called a probability function or frequency function which characterizes the distribution of a discrete random variable. Let X be a discrete random variable of a function, then the probability mass function of a random variable X is given by

P_x (x) = P( X=x ), For all x belongs to the range of X

It is noted that the probability function should fall on the condition :

• P_x (x) ≥ 0 and
• ∑_xϵRange(x) P_x (x) = 1

Here the Range(X) is a countable set and it can be written as { x₁, x₂, x₃, ….}. This means that the random variable X takes the value x₁, x₂, x₃, ….

These can also be stated as explained below.

The probability mass function P(X = x) = f(x) of a discrete random variable is a function that satisfies the following properties:

• P(X = x) = f(x) > 0; if x ∈ Range of x that supports
• \(\sum_{x\epsilon Range\ of x}f(x)=1\)
• \(P(X\epsilon A)=\sum_{x\epsilon A}f(x)\)

Definition

The Probability Mass function is defined on all the values of R, where it takes all the arguments of any real number. It doesn’t belong to the value of X when the argument value equals to zero and when the argument belongs to x, the value of PMF should be positive.

The probability mass function is usually the primary component of defining a discrete probability distribution, but it differs from the probability density function (PDF) where it produces distinct outcomes. This is the reason why probability mass function is used in computer programming and statistical modelling. In other words, probability mass function is a function that relates discrete events to the probabilities associated with those events occurring. The word “mass“ indicates the probabilities that are concentrated on discrete events.

What is the difference between PMF and PDF?

The difference between PMF and PDF:

Also, read:

Applications of Probability Mass Functions

• Probability mass function plays an important role in statistics. It defines the probabilities for the given discrete random variable. It integrates the variable for the given random number which is equal to the probability for the random variable.
• It is used to calculate the mean and variance of the discrete distribution.
• It is used in binomial and Poisson distribution to find the probability value where it uses discrete values.

Some of the probability mass function examples that use binomial and Poisson distribution are as follows :

PMF of Binomial Distribution

In the case of the binomial distribution, the PMF has certain applications, such as:

• To find the number of successful sales calls
• To find the number of defective products in the production run
• Finding the number of head/tails in coin flipping
• Calculating the number of male and female employees in a company
• Finding the vote counts for two different candidates in an election

Consider an example that an exam contains 10 multiple choice questions with four possible choices for each question in which the only one is the correct answer. To find the probability of getting correct and incorrect answers, the probability mass function is used.

PMF of Poisson Distribution

Likewise binomial, PMF has its applications for Poisson distribution also.

• To find the monthly demands for a particular product
• Calculating the hourly number of customers arriving for a bank
• Finding the hourly number of accesses to a particular web server
• Finding the number of typos in a book

Examples with PMF Table

The probability mass function example is given below :

Question : Let X be a random variable, and P(X=x) is the PMF given by,

1. Determine the value of k
2. Find the probability (i) P(X≤ 6), (ii) P(3<x≤ 6 )

Solution :

(1) We know that;
∑P(x_i)=1
Therefore,
0 + k + 2k + 2k + 3k + k² + 2k² + 7k²+ k = 1
9k + 10k² = 1
10k² + 9k – 1 = 0
10k² + 10k – k -1 = 0
10k(k + 1) -1(k + 1) = 0
(10k – 1) ( k + 1 ) = 0
So, 10k – 1 = 0 and k + 1 = 0
Therefore, k = 1/10 and k = -1
k=-1 is not possible because the probability value ranges from 0 to 1.
Hence, the value of k is 1/10.

(2) (i) P(X ≤ 6) = 1 – P( x > 6)

= 1 – ( 7k²+k )

= 1 – (7(1/10)² + ( 1/ 10) )

= 1 – (7/100 + 1/10)

= 1 – ( 17/100)

= ( 100 – 17)/100

= 83/100

Therefore , P(X≤ 6) = 83/100

(ii) P(3<x≤ 6 ) = P( x =4) + P ( x = 5 ) + P ( X = 6)

= 3k + k² + 2k²

= (3/10) + (1/10)² + 2 (1/10)²

= 3/10 + 1/100 + 2/100

= 3/10 + 3/100

= ( 30+3)/100

= 33/100

P(3<x≤ 6 ) = 33/100.

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