[Maths Class Notes] on Probability Mass Function Pdf for Exam

PMF or probability mass function is a simple concept in mathematics. It is a part of statistics. When you are learning about pmf you will find it very interesting and informative. It’s an informative and useful concept. Probability Mass Function is otherwise referred to as Probability Function or frequency function. PMF characterizes the distribution of a discrete variable which is unplanned or random. Example of a discrete random variable:

Let Y be the random variable of a function, and this is its probability mass function:

Py (y) = P (Y-y), for all y belongs to the range of Y.

Here are two conditions on which the probability function should fall upon:

• P y (y) ≥ 0

• ∑ yϵRange (y) P y (y) = 1

Probability Mass Function Definition

The definition of Probability Mass Function is that it’s all the values of R, where it takes into argument any real number. There are two times when the cost doesn’t belong to Y. First, when the case is equal to zero. The second time is when the value is negative, the value of the probability function is always positive.

Another name of PMF is the Probability Discrete Function (PDF). It’s given because when you are drawing the variable, it produces distinct outcomes or results. Two places where the discrete probability function is used is computer programming and statistical modelling.

 Probability Mass Function

The simple meaning of Probability Mass Function is the function relating to the probability of those events taking place or occurring. The word ”mass” is used to denote the expectations of discrete events.

Finding The Probability Mass Function 

It’s effortless to find the PMF for a variable. Given below are the steps that you need to follow to find the PMF of a variable:

Step 1: Start solving the question by fulfilling the first condition of the PMF. ( mentioned above)

Step 2: Take all the values of P ( X- x) and add it up. There will be a whole number ( 0, 1, 2), numbers with variables ( 1y, 2y 3y) and numbers which are squared ( 2 y2, 3 y2 ).

Step 3: Start using simultaneous equations to solve the sum.

Step 4: As you start using simultaneous equations, you will get two answers in the end. 

Step 5: You need to check which of the answers fulfils these two conditions:

(i) The value of the variable is never negative. 

(ii) The amount of the variable does not equal zero.

Step 6: The answer to the question is the one that follows both the conditions which are mentioned above. 

Probability Mass Function Applications

There are many places where the probability mass function is used and applied. Here are some of the places where there’s an application of PMF:

• One of the sections where PMF is used is statistics. It plays a vital and essential role in the study of statistics. Probability Function shows the various probabilities of the discrete variable data.

• PMF combines the variable for the random number that is identical or equal to the expectation for the random variable.

• Many people use PMF to calculate two main concepts in statistics- mean and discrete distribution.

• Another place where PMF is binomial and Poisson distribution is to find the value of the variables which are distinct and random. 

There are mainly two differences between PDF and PMF. Here are the two dissimilarities between them:

Distinction Between PDF and PMF:

Probability Mass Function Solved Example

Here is a probability mass function example which will help you get a better understanding of the concept of how to find probability mass function. 

Solved Example 1:

Let X be a random variable, and P (X=x) is the PMF given below;

1. Find the value of k.

2. Find the pmf probability of 

(i) P (X ≤ 6 )

Solution:

Given: ∑P(xi)=1

1. So, 

0+k+2k+2k+3k+ k2+ 2 k2+k  = 1 

9k 10 k2 = 1  

9k 10 k2 – 1 = 0

10k 2 + 10k – k – 1 = 0

10 k ( k + 1) – 1 (k – 1) = 0

(10k – 1 ) ( k +1 ) = 0

Therefore, 

10k – 1 = 0 and k + 1 = 0

Hence, 

k= 1/ 10, and k = -1 

K= -1 is not the desired answer because the pmf probability value lies between 0 and 1 ( it’s one of the conditions which is mentioned  above) 

Conclusion: The value of k is 1/10.

2.  (i)  P (X ≤ 6 )

So, 

P (X ≤ 6 ) = 1 – P ( x > 6)

= 1 – 7 k2 + k

= 1 – ( 7 ( 1/10) 2 + (1 / 10)) 

= 1 – ( 7/ 100 + 1 / 10) 

=  1 – ( 17/ 100)

= ( 100 – 17 )/ 100

= 83/ 100

Conclusion : P (X ≤ 6 ) = 83/ 100.