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224 LearnersLast updated on November 24, 2025

The probability distribution is a statistical function that calculates the relative likelihood of all possible experimental outcomes. It explains how the values of a random variable are distributed. Probability distribution helps in predicting weather conditions or test results.
A probability distribution is like the perfect or theoretical version of a frequency distribution. A frequency distribution shows how many times each value appears in a real dataset or sample.
In a sample, the number of times a value appears depends on its likelihood of occurrence. This likelihood is called probability, a number between 0 and 1, where 0 indicates that the event can never happen and 1 indicates that the event is guaranteed.
The closer the probability is to 1, the more often that value will appear in a sample. More precisely, the probability of a value is the relative frequency you would get if the sample were infinitely large. Since we can’t collect infinite data in the real world, probability distributions serve as theoretical models.
They represent the frequency distribution for the entire population. Probability distributions help describe real-world situations, such as coin toss outcomes or the weights of chicken eggs. They are also crucial in hypothesis testing, where they are used to calculate p-values.
Let’s see an example:
If a coin is tossed three times, what will be the probability distribution of the number of heads?
Solution:
The possible numbers of heads that can appear are:
(0, 1, 2, 3)
There are a total of 23 = 8 possible outcomes when a coin is tossed three times.
To get zero heads, the only possible outcome is TTT.
Thus, the probability of getting zero heads = 1/8.
To get one head, the possible outcomes are HTT, THT, TTH.
Thus, the probability of getting one head = 3/8.
To get two heads, the possible outcomes are HHT, HTH, THH.
Thus, the probability of getting two heads = 3/8.
To get three heads, the only possible outcome is HHH.
Thus, the probability of getting three heads = 1/8.
The probability distribution is as follows:
| x (Number of Heads) | 0 | 1 | 2 | 3 |
| P(x) | \(\frac{1}{8}\) | \(\frac{3}{8}\) | \(\frac{3}{8}\) | \(\frac{1}{8}\) |
The two basic forms of probability distributions that describe distinct types of random variables are discrete and continuous probability distributions. It is important to understand their key differences, as this may help in data analysis.
Discrete Probability is utilized when a random variable can take specific, countable values. A discrete random variable may take a countable number of unique values, such as 0, 1, 2, 3...
Discrete probability distributions are used primarily in instances where only a countable number of outcomes can be listed.
For example: the number of customers visiting a company in an hour, or the probability of getting tails in five coin tosses.
On the other hand, continuous probability distributions are used to describe continuous random variables, where an infinite set of numbers within a certain range can be applied. These numbers are uncountable, as each interval contains endless possibilities. For example: The exact time required to accomplish a task.
Probability distribution represents how the probabilities of a random variable can be distributed, which helps in predicting the possible outcomes.
The normal distribution, also known as bell curve, is one of the most commonly used probability distributions.
The type of probability distribution a dataset belongs to is based on its data generation process, such as the probability density function (PDF) and the probability mass function (PMF).


A probability distribution can be shown in a table using a random variable and its possible outcomes.
A random variable X is a rule that assigns a numerical value to each outcome in the sample space of a random experiment.
The probability distribution of X, written as P(X), lists:
It can be represented as:
| X (Values of Random Variable) | X1 | X2 | X3 | ... | Xn |
| P(X) (Probabilities) | P1 | P2 | P3 | ... | Pn |
Here:
Each probability Pi is greater than zero all the probabilities must add up to 1
P1 + P2 + P3 + … + Pn = 1
Probability distributions are of two types. Let’s learn how they differ:
Every random variable has a probability distribution that specifies the likelihood of each possible value. Random variables can be:
A discrete random variable uses a probability mass function (PMF) to show the probability of each possible value.
A continuous random variable is described by a probability density function (PDF) that specifies how probabilities are distributed over an interval of values.
Even if two random variables share the same probability distribution, they may still differ in how they relate to other variables, or whether they are independent.
When we generate outcomes according to a random variable’s probability distribution, the resulting observed values are called random variates.
Probability distributions help describe how likely different outcomes are in a dataset. Understanding their patterns and practicing applications makes mastering them easier.
Use familiar objects like coins, dice, cards, marbles, or candies. Which kids can understand the probability better when they can see the outcomes.
Create a small table showing outcomes and their probabilities. Visual aids help the children quickly understand how the probability distributions work.
Bar graphs, histograms, line plots, and simple charts make the concept easier to grasp. Visuals help learners see patterns, not just memorize formulas.
Simple tools like spreadsheets, apps, or online simulators can show patterns instantly.
Seeing instant graphs builds better intuition.
Students might make mistakes when solving probability distribution problems. These mistakes can be avoided with a proper understanding of the concepts and solutions. Let’s look at some common errors along with their solutions:
We now understand the significance of probability distributions in mathematics. Let’s explore how they play a pivotal role in other fields beyond math:
A fair coin is tossed 6 times. What is the probability of getting exactly 5 heads?
The probability of obtaining 5 heads in 6 tosses is 0.093 which is equal to 9.375%.
We have n = 6; probability of obtaining heads in just one toss: p = 0.5
Here, we apply the binomial probability formula:
\(P(X = x) = \frac{n!}{x!(n - x)!}\, p^{x}\, \times q^{\,n - x}\)
Here, p (X = x) is the probability of obtaining 5 heads in 6 trials; x =number of trials in total, and q = (1 – P) is the probability of getting tails.
Substituting the given values:
\(P(5) = \frac{6!}{5!(6-5)!}\,(0.5)^{5}\,\times (0.5)^{6-5}\)
Now, we calculate the binomial coefficient:
\( \frac{6!}{5! \times 1!} = \frac{6 \times 5!}{5! \times 1!} = \frac{6}{1!} = \frac{6}{1} = 6\)
Calculating the probability:
\(P(5) = 6 \times (0.5)^{5} \times (0.5)^{1}\)
\(= 6 \times \frac{1}{32} \times \frac{1}{2}\)
\(= 6 \times \frac{1}{64} = \frac{6}{64} = 0.093\)
A student attempts 20 multiple-choice questions, each with five options. What is the probability of answering exactly 8 correctly if they guess randomly?
The likelihood of 8 right answers if they guess randomly is 0.00222 which is equal to 2.22%.
Using the binomial probability formula:
\(P(X = x) = \frac{n!}{x!(n - x)!} \times p^{x} \times q^{\,n - x}\)
Where,
n = 20
x = 8
p = \(\frac{1}{5}\)
q = 1 – p = \(\frac{4}{5}\)
Calculating the binomial coefficient:
\(\frac{20!}{8!\times 12!}\)
We now expand the first 12 terms of 20! To cancel with 12!:
\(\frac{20 \times 19 \times 18 \times 17 \times 16 \times 15 \times 14 \times 13}{8!}\)
Calculating the probability:
\(\left( \frac{1}{5} \right)^{8} = \frac{1}{390625}\)
= 0.00000256
\(\left( \frac{4}{5} \right)^{12} = \frac{16777216}{244140625} \approx 0.0687\)
Combining each part by multiplying them:
\(P(8) = 12,650 \times 0.00000256 \times 0.0687 \approx 0.002224\)
\(= 12,650 × 0.000000176 = 0.032384\)
Now,
\(0.032384 × 0.0687 = 0.00222 \)
Therefore, the probability is 0.00222.
A basketball player has a 40% chance of making a free throw. If they attempt 10 free throws, what is the probability that they make exactly 4?
The probability of making 4 free throws is 0.2508 which is equal to 25.08%.
We use the formula for binomial probability:
\(P(X = x) = \frac{n!}{x!(n - x)!} \times p^{x} \times q^{\,n-x}\)
We have:
n = 10
x = 4
p = 0.4
q = 1 – p = 0.6
First, we compute the binomial coefficient:
\(\frac{n!}{x!(n - x)!} = \frac{10!}{4!(10-4)!} = \frac{10!}{4!6!}\)
Here, we expand the factorials of 10:
\(= \frac{10 \times 9 \times 8 \times 7}{8!}
= \frac{5040}{24}
= 210\)
Now, compute the probability:
\(p^4 = (0.4)^4 = 0.0256\)
\(q^6 = (0.6)^6 = 0.046656\)
Combining each part by multiplying them:
\(P(4) = 210 × 0.0256 × 0.046656\)
\(= 210 × 0.001194\)
\(= 0.2508\)
A factory produces 10 light bulbs, and each has a 3% chance of being defective. What is the probability that exactly 2 bulbs are defective?
The probability of 2 defective bulbs is 0.03195 or 3.195%.
n = 10
x = 2
p = 0.03
\(q = 1 – 0.03 = 0.97\)
First, we compute the binomial coefficient:
\(\frac{n!}{x!(n-x)!}
= \frac{10!}{2!(10-2)!}
= \frac{10!}{2!8!}\)
Expand the factorials:
\(\frac{10!}{2!}
= \frac{10 \times 9}{2!}
= \frac{90}{2}
= 45\)
Now, we compute the probability:
\(P(2) = (0.03)^2 = 0.0009\)
\(q^8 = (0.97)^8 = 0.7890\)
Combining each part by multiplying them:
\(P(2) = 45 × 0.0009 × 0.7890\)
\(= 45 × 0.0007101\)
\(= 0.03195\)
In a class of 26 students, each has an 80% chance of passing a test. What is the probability that exactly 20 students pass?
The probability of 20 students passing the test is 0.169 which is equal to 16.9%.
n = 26
x = 20
p = 0.80
\(q = 1 – 0.80 = 0.20\)
First, we compute the binomial coefficient:
\(\frac{n!}{x!(n - x)!}
= \frac{26!}{20!(26-20)!}
= \frac{26!}{20!6!}\)
Expand the factorials:
\(= \frac{26 \times 25 \times 24 \times 23 \times 22 \times 21}{6!}\)
\(= \frac{165,242,320}{720}\)
\(= 229574\)
Computing the probability:
\(P^{20} = (0.80)^{20} = 0.0115\)
\(q^6 = (0.20)^6 = 0.000064\)
Combining each part by multiplying them:
\(P(20) = 229574 × 0.0115 × 0.000064\)
\(= 229574 × 0.000000736\)
\(= 0.1690\)
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!






