The Probability Mass Function (PMF) 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) â‰¥ 0 and
- âˆ‘_{xÏµRange(x)} P_{x }(x) = 1
Here the Range(X) is a countable set and it can be written as { x_{1}, x_{2}, x_{3}, â€¦.}. This means that the random variable X takes the value x_{1}, x_{2}, x_{3}, â€¦.
Definition
The Probability Mass function is defined on all the values of 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 probabilityÂ discrete functionÂ (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 â€œmassâ€œ indicates the probabilities that are concentrated on discrete events.
What is the difference between PMF and PDF?
The difference between PMF and PDF:
PMF | |
Solution ranges between numbers of discrete random variables | The solution is in a range of continuous random variables |
Uses discrete random variables | Uses continuous random variables |
Also, read:
Related Articles | |
Probability Density Function | Cumulative Distribution Function |
Probability Distribution Formula | Binomial Probability Formula |
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 theÂ 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. In order to find the probability of getting correct and incorrect answers, the probability mass function is used.
PMF of Poisson Distribution
Likewise binomial, PMF have 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,
X | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
P(X=x) | 0 | k | 2k | 2k | 3k | k^{2} | 2k^{2} | 7k^{2}+k |
- Determine the value of k
- Find the probability (i) P(Xâ‰¤ 6), (ii) P(3<xâ‰¤ 6 )
Solution :
(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 â‰¤ 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(Xâ‰¤ 6) = 83/100
(ii) P(3<xâ‰¤ 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<xâ‰¤ 6 ) = 33/100.
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