The Probability Mass Function (PMF) also called as 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 falls on the condition :
Here the Range(X) is a countable set and it can be written as { x_{1}, x_{2}, x_{3}, ….}. This means that the random variable X takes the value x_{1}, x_{2}, x_{3}, ….
The probability Mass function is defined on all the values of \(\mathbb{R}\), where it takes all the argument 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 a positive value.
The probability mass function is also called a discrete probability function where it produces distinct outcomes. This is the reason why probability mass function is used in computer programming and statistical modeling. 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.
Some of the probability mass function examples that use binomial and Poisson distribution are as follows :
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. In order to find the probability of getting correct and incorrect answers, the probability mass function is used.
The probability mass function example is given below :
Let X be a random variable, and P(X=x) is the PMF given by,
X |
0 |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
P(X=x) |
0 |
k |
2k |
2k |
3k |
k^{2} |
2k^{2} |
7k^{2}+k |
(1) We know that \(\sum x_{i}=1\)
Therefore,
0 + k + 2k + 2k + 3k + k^{2} + 2k^{2} + 7k^{2}+ k = 1
9k + 10k^{2} = 1
10k^{2} + 9k – 1 = 0
10k^{2} + 10k – 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 probability value should not be zero
The value of k is 1/10
(2) (i) P(X ≤ 6) = 1 – P( x > 6)
= 1 – ( 7k^{2}+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 + k^{2 }+ 2k^{2}
= (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.