Last updated on June 18th, 2025
A random variable is a way to measure a random experiment's result by assigning numerical values to its outcome. In a probability experiment, assigning a precise number to each possible outcome helps in mathematical analysis and prediction. Random variables are classified into two categories since data can be either continuous or discrete. In this article, we will explore random variables and their properties.
A random variable is one that has an unknown value, or a function that assigns values to every experiment result. It is typically represented by letters and divided into two groups: continuous, which can have any value within a specified or continuous range, and discrete, which takes specific values. Random variables are crucial in probability and statistics, as they help in the quantification of uncertainty. Some key takeaways of a random variable are listed below:
Struggling with Math?
Get 1:1 Coaching to Boost Grades Fast !
Based on the types of values they can have, random variables are divided into two categories:
There is a finite number of possible values for a discrete random variable. In simple terms, we can say that a specific set of countable values characterizes a discrete random variable.
For a clearer insight, consider an experiment where a coin is tossed four times. The number of times the coin lands on heads is denoted as X.
Depending on how many heads appear, x can have any of the following values such as 0, 1, 2, 3, or 4. There are no other possible values for X.
The probability function associated with a discrete random variable is known as the probability mass function (PMF). If X is a discrete random variable and its PMF is P(xi).
Here, we have to keep in mind that each probability value (pi) lies between 0 and 1, i.e., 0 ≤ pi ≤ 1.
Also, the possible values of X are covered by the sum of all the probability values, which equals 1, i.e., ∑pi = 1.
Continuous random variables have an endless number of possible values and can take any value within a given range or interval. It can have an infinite number of possible values.
For better clarity, we can consider an experiment that measures the rainfall in a city over a year.
The possible result is a continuous figure since the amount can be 30 inches, 30.5 inches, 30.75 inches, or even 30.752 inches. It is clear that there are countless possible values.
Because of this, rainfall is a continuous random variable rather than a discrete variable. The corresponding probability function of continuous random variables is known as the probability density function (PDF).
If X is a continuous random variable, then the probability of X falling within an interval is P (x < X < x + dx) ≈ f(x)dx.
The key characteristics of a probability density function (PDF) are:
For each value of x, the probability density function, (fx), always remains between 0 and 1. So, the probabilities are never negative or greater than 1 (0 ≤ f(x) ≤ 1).
Additionally, over all possible values of X, ∫ f(x) dx = 1 since the total area under the curve of f(x) equals 1.
The distribution’s PDF is referred to as P(X).
In many real-life situations, we can use the random variable to predict the outcomes, analyze the data, and make well-informed decisions. Here are some of the real-life applications of the concept:
A random variable represents the numerical result of a random event. It provides a specific number for each possible outcome in a probability experiment. Understanding the concepts of random variables helps in making accurate predictions and avoiding errors during calculations. Here are some common mistakes and their helpful solutions that will enhance our mathematical and problem-solving skills.
Level Up with a Math Certification!
2X Faster Learning (Grades 1-12)
Sam has drawn a single card from a standard deck of 52. Let z be the values of the drawn card (Ace = 1, 2 to 10 = face value, Jack = 11, Queen = 12, King =13). Find P(Z = 10).
1 / 13
A deck of 52 contains four 10s. These are the only cards that satisfy Z =10. Here, the total number of possible outcomes is 52, since Sam has drawn a single card from the deck.
Next, we can apply the probability formula:
P (Z = 10) = Total number of favorable outcomes / Total number of possible outcomes
P (Z = 10) = 4 / 52
P (Z = 10) = 1 / 13
This means, there is a 1 in 13 chance that the card will be a 10.
A bus arrives every 5 to 15 minutes. Let T be the waiting time. Find P(T < 10) assuming a uniform distribution.
0.5 or 50%
For a uniform distribution between a = 5 and b = 15, the probability is calculated as:
P (a ≤ T≤ b) = b - a / Range
Here, the range is 15 - 5 = 10
P (5 ≤ T≤ 10) = 10 - 5 / 15 - 5 = 5 / 10 = 0.5
P (T < 10) = 0.5 or 50%
This means there is a 50% probability that the waiting time for the bus will be less than 10 minutes.
A fair coin is tossed 2 times. Let Y be the number of heads. Find P (Y = 1).
1 / 2 or 50%
Here the possible outcomes are 4.
So the random variable Y can take values 0, 1, or 2.
Favorable outcomes for Y =1 (exactly 1 head appears). From the possible outcomes, the favorable cases where Y = 1 are:
HT (1 head, 1 tail)
TH (1 head, 1 tail)
There are 4 total outcomes, so the probability of each outcome is:
P (Each outcome) = 1 /4
P (Y =1) = P(HT)+ P(TH)
P (Y =1) = 1 / 4 + 1/ 4 = 2 / 4 = 1 / 2
P (Y =1) = 1 / 2 or 50%
The probability of getting exactly one head when tossing a fair coin twice is 1 / 2 or 50%.
The heights of students in a school follow a normal distribution with a mean of 150 cm and a standard deviation of 10 cm. Find the probability that a randomly selected student is taller than 160 cm.
0.1587 or 15.87%
We can use the Z-Score formula:
Z = X - μ / σ
Here, X = 160 (desired height)
μ = 150 (mean)
σ = 10 (standard deviation)
Z = 160 - 150 / 10 = 10 / 10 = 1
From the z table, P (Z < 1) = 0.8413
The probability of being taller than 160 cm is:
P(X > 160) = 1 − P(X ≤ 160)
P(X > 160) = 1 − 0.8413 = 0.1587 or 15.87%
The probability that a randomly selected student is taller than 160 cm is 0.1587 or 15.87%.
A student takes a test with possible scores ( 30, 40, 50, 60), each equally likely. Find the expected test score.
The expected test score is 45.
The possible test scores: 30, 40, 50, 60
Each score has probability P(X = x)
P(X = 10) = P(X = 30) = P(X = 40) = P(X = 50) = P(X = 60) = 1 / 4
The formula for calculating the expected mean is:
E [X] = ∑ xi P (X = xi)
E [X] = (30 × 1 / 4) + (40 × 1 / 4) + (50 × 1 / 4) + (60 × 1 / 4)
E [X] = 30 / 4 + 40 / 4 + 50 / 4 + 60 / 4
E [X] = 7.5 + 10 + 12.5 + 15
E [X] = 45
Hence, the expected test score is 45.
Turn your child into a math star!
#1 Math Hack Schools Won't Teach!
Struggling with Math?
Get 1:1 Coaching to Boost Grades Fast !