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Last updated on October 7, 2025
A probability mass function is a function used specifically to predict the probability of a discrete random variable. It is used in real-life situations to calculate discrete probabilities, such as predicting business losses. In this topic, you’ll learn how to use the probability mass function to determine the likelihood of any occurrence.
Probability mass function (PMF) indicates the probability assigned to each possible value of a discrete random variable. For example, when we flip a coin twice, the PMF gives the probability of each possible outcome: heads or tails.
The formula to calculate the probability mass function is:
f(x) = P(X = x)
Where,
f(x) or P(X = x) is the probability of a discrete random variable X
X is discrete random variable
x is the possible value that the random variable takes
There are unique characteristics of probability mass function (PMF) that you might not know. We’ll explore a few key facts here:
Understanding the properties of the probability mass function enables us to identify the function easily. Here are a few properties that will help:
The probability mass function for a discrete variable X can be mathematically expressed as: f(x) = P(X = x). There are various other formulas to determine PMF for different distribution, as listed below:
The binomial distribution represents the number of possible outcomes, the likelihood of success, and the likelihood of failure.
The formula we use for the binomial distribution is:
Binomial distribution: P(X = x) = nCx × px × (1 − p)n − x
Here:
n: Number of outcomes
p: Probability of success
(1 − p): The probability of failure
The Poisson distribution represents the average and the quantity of independent events that occurred during a certain period.
The formula we use for the Poisson distribution is:
Poisson distribution: P(X = x) = (λx × e−λ) / x!
Here, the mean represented is the symbol λ.
We graphically represent the probability mass function in different forms, such as tables or graphs. Example: If a coin is flipped 4 times and x is the random variable that indicates the quantity of tails, then the probability for the mass function of the event is:
Number of Tails (X) | Outcomes | P (X = x) |
0 | {HHHH} | 1/16 |
1 | {HHHT, HHTH, HTHH, THHH} | 1/4 |
2 | {HHTT, HTHT, HTTH, THHT, THTH, TTHH} | 3/8 |
3 | {TTTH, TTHT, THTT, HTTT} | 1/4 |
4 | {TTTT} | 1/16 |
Let’s now plot this in a probability mass function graph:
We use probability mass function to understand the probability of discrete events in different real-life situations. Let’s look at some:
The probability mass function is of paramount importance in math. However, students might find it difficult to grasp the concept, leading to many mistakes. Here, we list a few common mistakes along with some tricks to avoid them.
Given a probability mass function: f(x) = bx^3, for x = 4, 5, 6.
b = 1/405 which satisfies the condition, and the probability adds up to 1.
∑x ϵ S f(x) = 1
We now substitute the given function:
\(\sum_{x=4}^{6} bx^3 = 1\)
b × (43 + 53 + 63) = 1
b × (64 + 125 + 216) = 1
b × (405) = 1
b = 1/405
If X denotes the number of heads, find the PMF of X when two coins are tossed fairly.
P(X = 0) = 1/4
P(X = 1) = 1/2
P(X = 2) = 1/4
Out of 4 outcomes (HH, HT, TH, TT), the counts of heads are 0, 1, or 2.
If two dice are rolled fairly, find P(X = 7). Let X be the sum of the numbers obtained.
P(X = 7) = 6/36 = 1/6
Outcomes giving sum 7: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)
Here, we have 6 outcomes. Total outcomes = 6 × 6 = 36
So, P(X = 7) = 6/36 = 1/6.
If X is the number of heads, find P(X = 2) when a fair coin is tossed thrice.
P(X = 2) = \(\binom{3}{2}\) (0.5)2 (0.5)1 = 0.375
Substituting the values in the formula, we get:
P(X = 2) = \(\binom{3}{2} \)(0.5)2(0.5)1
Computing step by step, we get:
\(\binom{3}{2} \) = 3
(0.5)2 = 0.25
(0.5)1 = 0.5
So, 3 × 0.25 × 0.5 = 0.375
Therefore, P(X = 2) = 0.375.
There are 10 bulbs in a box and 2 are defective. Find the PMF of X if one bulb is selected randomly. Let X = 1 if the bulb is defective and X = 0 if it works.
P(X = 1) = 2/10 = 0.2
P(X = 0) = 8/10 = 0.8
It’s a Bernoulli distribution with success probability p = 0.2.
Jaipreet Kour Wazir is a data wizard with over 5 years of expertise in simplifying complex data concepts. From crunching numbers to crafting insightful visualizations, she turns raw data into compelling stories. Her journey from analytics to education ref
: She compares datasets to puzzle games—the more you play with them, the clearer the picture becomes!