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315 LearnersLast updated on November 26, 2025


The discrete probability distribution is a simple way to show the different chances of different outcomes when you can count them one by one. It gives the different values of a random variable along with its different probabilities. A discrete probability distribution contrasts with a continuous distribution, where outcomes can take any value along a continuum, as the outcome falls anywhere on a continuum. The types of discrete probability are Binomial, Poisson, and Bernoulli distributions. If a probability distribution is said to be a discrete probability distribution, it should follow these two conditions :
Imagine you flip a fair coin 3 times, and you want to count the Number of Heads (X).
This is a perfect example because:
Step 1: List all possible outcomes
There are 8 total ways the coins can land (\(2^3 = 8\)):
Step 2: Create the Distribution Table
Now, we count the probability for each number of heads.
| Number of Heads (x) | Frequency (How many ways?) | Probability P(X=x) |
| 0 | 1 (TTT only) | 1/8 (or 0.125) |
| 1 | 3 (HTT, THT, TTH) | 3/8 (or 0.375) |
| 2 | 3 (HHT, HTH, THH) | 3/8 (or 0.375) |
| 3 | 1 (HHH only) | 1/8 (or 0.125) |
Here are the formulas for the PMF and CDF of Discrete Probability Distribution.
The PMF (Probability Mass Function) gives the probability that a discrete random variable X is exactly equal to a specific value x.
Formula:
\(f(x) = P(X = x)\)
Example:
Let X be the number of Girls in a family with 3 children.
Possible Outcomes (\(2^3 = 8\) total outcomes):
The PMFs are given as:
Total Probability:
\(\frac{1}{8} + \frac{3}{8} + \frac{3}{8} + \frac{1}{8} = 1\)
Properties of PMF:
The CDF (Cumulative Distribution Function) gives the probability that the random variable X is less than or equal to a specific value x.
Formula:
\(F(x) = P(X \le x)\)
Example:
Let a random variable X be the score obtained on a biased (weighted) 4-sided die. We want to find the probability \(P(1 < X \le 3).\)
Data Table:
|
X (Score) |
PMF P(X=x) |
CDF F(x)=P(X≤x) |
| 1 | 0.1 | 0.1 |
| 2 | 0.4 | 0.1 + 0.4 = 0.5 |
| 3 | 0.3 | 0.5 + 0.3 = 0.8 |
| 4 | 0.2 | 0.8 + 0.2 = 1.0 |
Calculation for \(P(1 < X \le 3)\):
Using the CDF property:
\(P(a < X \le b) = F(b) - F(a)\)
Therefore:
\(P(1 < X \le 3) = F(3) - F(1)\)
\(P(1 < X \le 3) = 0.8 - 0.1 = 0.7\)
(Verification using PMF): \(P(1 < X \le 3)\) means we want outcomes 2 and 3.
\(P(X=2) + P(X=3) = 0.4 + 0.3 = 0.7\)
Now let’s discuss the types of discrete probability. The types are;


The Mean (also called the Expected Value) of a discrete probability distribution is the weighted average of all possible values that the random variable can take. It represents the long-term average result if you were to repeat the experiment many times.
Formula:
The mean \mu (or E(X)) is calculated by multiplying each possible outcome x by its probability P(X=x) and then adding them all together:
\(\mu = E(X) = \sum [x \cdot P(X=x)]\)
Example:
Let a random variable X be the cash prize won in a charity raffle game.
Possible Outcomes:
Calculation Table: We multiply each outcome (x) by its probability (P(x)) to find the “weighted” value.
|
Outcome (x) |
Probability P(X=x) |
Weighted Value [x⋅P(x)] |
| 0 | 0.80 | \(0 \cdot 0.80 = \mathbf{0}\) |
| 10 | 0.15 | \(10 \cdot 0.15 = \mathbf{1.5}\) |
| 50 | 0.05 | \(50 \cdot 0.05 = \mathbf{2.5}\) |
Total Mean:
\(E(X) = 0 + 1.5 + 2.5 = 4.0\)
Interpretation:
The expected value is $4. This means that if you played this game thousands of times, you would win an average of $4 per game, even though you can never actually win exactly $4 in a single turn.
The Variance measures how “spread out” or dispersed the numbers are from the mean (expected value). A high variance means the values are widely scattered; a low variance means they are clustered closely around the average.
Formula:
There are two common ways to calculate it.
Example:
Let a random variable X be the number of daily complaints received by a service center.
Possible Outcomes:
Probabilities:
Step 1: Find the Mean (\mu)
\(\mu = (0 \cdot 0.1) + (1 \cdot 0.6) + (2 \cdot 0.3)\)
\(\mu = 0 + 0.6 + 0.6 = \mathbf{1.2}\)
\(\mu^2 = 1.2^2 = 1.44\)
Step 2: Calculate \(E(X^2)\)
We need to square each outcome first, then multiply by its probability
|
Outcome (x) |
Probability P(x) |
Squared Outcome \((x^2)\) | Weighted Square\(E(X^2) = [x^2⋅P(x)]\) |
| 0 | 0.1 | \(0^2 = 0\) | \(0 \cdot 0.1 = \mathbf{0}\) |
| 1 | 0.6 | \(1^2 = 1\) | \(1 \cdot 0.6 = \mathbf{0.6}\) |
| 2 | 0.3 | \(2^2 = 4\) | \(4 \cdot 0.3 = \mathbf{1.2}\) |
| \(\sum\) | \(0+0.6+1.2=1.8\) |
Step 3: Apply the Formula
Now we subtract the square of the mean from the sum we just found.
Variance(\(\sigma^2\))
\(\sigma^2 = E(X^2) - \mu^2\)
\(\sigma^2 = 1.8 - 1.44\)
\(\sigma^2 = \mathbf{0.36}\)
Standard deviation(\(\sigma\))
\(\sigma = \sqrt{0.36} = \mathbf{0.6}\)
Here, we will discuss how to calculate a discrete probability distribution. To find a discrete probability distribution, the probability of mass function is required. Follow these steps to find the probability;
Step 1: Identify the sample, set of all possible outcomes of an experiment.
Step 2: Then you have to find the discrete random variable (x). It is a function assigning a numerical value to each outcome.
Step 3: Identifying the possible value of X.
Step 4: Finding the probability of each outcome by using the formula;
P (X = x) = Number of favorable outcomes/Total number of possible outcomes.
Step 5: Use a table to organize the results
Now let’s learn the difference between the discrete and continuous distribution.
|
Discrete Distribution |
Continuous Distribution |
|
The probability distribution for countable values |
The probability distribution for measurable values |
|
Here, the type of variable is discrete |
Here, the type of variable is continuous |
|
The graph of discrete distribution is a bar |
The graph of continuous distribution is a curve |
Discrete probability distribution is a complex topic to get a grasp on. In this section, we will discuss some tips and tricks to master Discrete probability distribution.
Now let’s learn a few common mistakes and ways to avoid them to master discrete probability distribution.
In real-life discrete probability distribution are used to find the probability for countable values, now lets few real-world applications of it;
A factory produces light bulbs, and 5% of them are defective. If a random sample of 8 bulbs is taken, what is the probability that exactly 2 bulbs are defective?
The probability of getting 2 bulbs are 0.0515 or 5.15%.
Here, to find the probability we use the equation, P(X = k) = \(\binom{n}{k}\) pk(1 - p)n-k
According to the problem,
n = 8
k = 2
p = 0.05
Substituting the values,
P(X = 2) = \(\binom{8}{2}\) (0.05)2(1 - 0.05)8-2
P(X = 2) = \(\binom{8}{2}\) (0.05)2( 0.95)6
The value of C \(\binom{8}{2}\)= 28, (0.05)2 = 0.0025, and (0.95)6 = 0.7358
So, the value of P(X = 2) = 28 × 0.0025 × 0.7358 = 0.0515.
A basketball player has a 40% probability of making a free throw. What is the probability that their first successful shot happens on the third attempt?
The probability here is 0.144 or 14.4%.
Here the probability is of geometric distribution
So, P(X = k) = (1 - p)(k-1) × p.
Here, p = 0.4 and k = 3
Therefore, P(X = 3) = (0.6)2 × 0.4
= 0.36 × 0.4 = 0.144.
A deck has 52 cards, including 4 Aces. If 5 cards are drawn randomly, what is the probability of getting exactly 2 Aces?
Probability ≈ 0.0399 or 3.99%.
Here, the probability is calculated using \(P (X = k) = \dfrac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}}.\)
Here,
N = 52
K = 4
k = 2
n = 5
So P(X = 2) =\( \dfrac{\binom{4}{2} \binom{52 - 4}{5 - 2}}{\binom{52}{5}}.\)
P(X = 2) = 0.0399 or 3.99%.
A multiple-choice quiz has 10 questions, each with 4 answer choices. A student guesses randomly on each question. What is the probability of getting exactly 3 correct answers?
The probability is 0.250 or 25%.
Here, to find the probability we use the equation, P(X = k) = C \(\binom{n}{k}\) pk (1 - p)(n - k)
According to the problem,
n = 10
k = 3
p = ¼ = 0.25
Substituting the values,
P(X = 3) = \(\binom{10}{3}\)(0.25)2(1 - 0.25)10 - 3
P(X = 3) = \(\binom{10}{3}\) (0.25)2(0.75)7
Here, C\(\binom{10}{3}\) = 120
0.252 = 0.015625
0.757 = 0.1335
P(X = 3) ≈ 120 × 0.015625 × 0.1335 ≈ 0.250.
A factory produces screws with a 2% defect rate. What is the probability that the first defective screw appears on the 5th inspection?
The probability is 0.01845 or 1.845%.
P(X = 5) = (1 - p)(k-1) × p.
Here, p = 0.02
K = 5
P(X=5) = (1 − p)(k-1)p = (0.98)4 × 0.02
As 0.984 = 0.9224
P(X = 5) = 0.9224 × 0.02 = 0.01845.
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!






