Probability Mass function is the function that maps the values of a discrete random variable to their probabilities.
Suppose x1, x2, ....... are possible values of a discrete random variable X. Then p(xi) is called the probability mass function of the random variable X if,

  1. p(x) ≥ 0 for all i = 1,2,3.....
  2. $\sum_{i}p(x_{i})$ = 1.

In the simple example of the random variable X assuming the number of heads in a single toss of a coin,
X = {0, 1}
p(x) is the function that gives the probabilities of X = 0 and X = 1 in a single toss.

p(0) = p(1) = $\frac{1}{2}$ and p(0) + p(1) = $\frac{1}{2}$ + $\frac{1}{2}$ = 1


The formulas used for computing the probabilities of various discrete random variables are given below.

Binomial probability formula
In a Binomial distribution of n independent trials, the probability of exactly X successes is given by
P(X : n, p) = C(n, X). pX .qn-X where p is the probability of success in a single trial and q = 1-p.

Poisson probability formula
The probability of X occurrences in an interval of time, area, volume etc is
P(X : λ) = $\frac{e^{\lambda }\lambda ^{X}}{X!}$ where λ is the mean number of occurrence per unit.

Geometric probability formula
P(X : p) = qX . p where 0 < p < 1 and q = 1 - p
Suppose (X, Y) is a two dimensional discrete random variable and p(xi, yj) be real number associated with each (xi, yj),
i, j = 1, 2, 3...... Then p is called a joint probability Mass function of (X, Y) if
  1. p(xi,yj) ≥ 0 for all i,j
  2. $\sum_{i=1}^{\infty }\sum_{j=1}^{\infty }p(x_{i},y_{j})$ = 1

Then the marginal probability Mass functions p(xi) and q(yj) corresponding to random variables X and Y are defined as

p(xi) = $\sum_{j}p(x_{i}y_{j})$ and q(yj) = $\sum_{i}p(x_{i}y_{j})$

Let p(xi, y)j be the joint probability mass function of a two dimensional random variable (X, Y) and p(xi) and q(yj) be marginal probability mass functions corresponding to X and Y. Then the conditional probability mass function for the occurrence of
X = xi , given that Y = yj is given by

p($\frac{x_i}{y_j}$) = P(X = $\frac{x_i}{Y}$ =yj) = $\frac{p(x_{i},y_{j})}{q(y_{j})}$

and the conditional probability mass function for the occurrence of Y = y, given that X = x, is given by

p($\frac{y_j}{x_i}$) = P(Y = $\frac{y_j}{X}$ = xi) = $\frac{p(x_{i},y_{j})}{p(x_{i})}$

Solved Example

Question: The Probability Mass function of a random variable X is given below:

x
 0       1 
  2 
p(x)
 3c2  4c-10c2  5c-1
  1. Find the value of c.
  2. The largest value of X such that F(x) < $\frac{1}{2}$.

Solution:
 
∑p(x) = 1                       Property of probability Mass function
3c2 + 4c -10c2 + 5c -1 = 1
-7c2 + 9c -2 = 0

(7c-2)(c-1) = 0   ⇒ c = $\frac{2}{7}$ or c =1.

Ignoring c = 1, which makes probabilities > 1 , c = $\frac{2}{7}$

Hence the Table for Probability mass function is

x
     0       1 
  2 
p(x)
 $\frac{12}{49}$  $\frac{16}{49}$  $\frac{21}{49}$
The cumulative probabilities are

F(0) = 0 F(1) = $\frac{12}{49}$    F(2) = $\frac{28}{49}$

Hence the largest value for which F(x) < $\frac{1}{2}$ is x = 1.