Random variables are fundamental concepts in probability and statistics, playing a crucial role in data analysis and decision-making processes. Understanding random variables is essential for students as it lays the groundwork for advanced topics such as statistical inference, regression analysis and machine learning.

There are two main types of random variables: discrete and continuous. Discrete random variables take on a countable number of distinct values, while continuous random variables take on an infinite number of possible values within a given range.

This article aims to provide practice problems on random variables, enhancing students’ comprehension and application skills.

What are Random Variables?

A random variable is a numerical outcome of a random phenomenon. It is a function that assigns a real number to each possible outcome in a sample space. Random variables can be classified into two main types:

• Discrete Random Variables: These take on a countable number of distinct values. Examples include the number of heads in a series of coin tosses or the number of students in a class.

• Continuous Random Variables: These can take on any value within a given range or interval. Examples include the height of students in a class or the time it takes to run a marathon.

Random variables are fundamental in probability and statistics as they allow us to quantify and analyze the outcomes of random events. They form the basis for statistical inference, modelling real-world phenomena, and making predictions based on data.

Random Variables Formulas

Sample Space (S): All possible outcomes of the experiment.

Probability Distribution: Describes the probability of each value the random variable can take.

• Discrete Case: Often expressed as a probability mass function (PMF) which assigns a probability to each specific value.
• Continuous Case: Often expressed as a probability density function (PDF) which describes the probability over a range of values.

Expected Value (Mean): The average value you expect to get if you were to repeat the experiment many times.

• Discrete Case: E(X) = Σ(x × P(X = x)) for all possible values x.
• Continuous Case: E(X) = ∫ x × f(x) dx (integral over the range of X).

Variance: Measures how spread out the values of the random variable are from the expected value.

Expected Value (Mean) E(X)

• Discrete: E(X) = ∑[x⋅P(x)]
• Continuous: E(X) = ∫_-∞^∞x⋅f(x) dx

Variance Var(X)

• Discrete: Var(X) = ∑[(x-μ)² × P(x)] or Var(X) = E(X²)-(E(X))²
• Continuous: Var(X) = ∫_-∞^∞​ (x-μ)² × f(x)dx or Var(X) = E(X²)-(E(X))²

Probability Density Function (pdf) and Cumulative Distribution Function (CDFa random variable):

• FX​(x) = P(X≤x)
• fX​(x) = d/dx ​FX​(x) (for continuous random variables)

Standard Deviation σ

• σ = √Var(X)

Random Variables Practice Problems with solution

Problem 1: Let X be a random variable with the following probability mass function (pmf): P(X = 1) = 0.2, P(X = 2) = 0.5, and P(X = 3) = 0.3. Find the expected value E(X).

Solution:

E(X) = 1×0.2 + 2×0.5 + 3×0.3

= 0.2 + 1 + 0.9 = 2.1

Problem 2: A fair six-sided die is rolled. Let X be the outcome of the roll. Find the variance Var(X).

Solution:

E(X) = 1 + 2 + 3 + 4 + 5 + 6/6

= 3.5

E(X²) = 1² + 2² + 3² + 4² + 5² + 6²

= 15.667

Var(X) = E(X²) – (E(X))²

= 15.667 -3.5²

= 2.9167

Problem 3: Let Z be a random variable with the following probability mass function (pmf): P(Z = -1) = 0.3, P(Z = 0) = 0.4, and P(Z = 1) = 0.3. Find the expected value E(Z)and the variance Var(Z).

Solution:

E(Z) = (-1)×0.3 + 0×0.4 + 1×0.3

= 0

E(Z²) = (-1)²×0.3 + 0²×0.4 + 1²×0.3

= 0.6

Var(Z) = E(Z²)-(E(Z))² = 0.6-0

= 0.6

Problem 4: A random variable W follows a uniform distribution on the interval [2, 4]. Find the expected value E(W) and the variance Var(W).

Solution:

E(W) = 2 + 4/2

= 3

Var(W) = (4-2)² /12 = 4/12

= 1/3

Problem 5: Let X be a Bernoulli random variable with parameter p = 0.4. Find the expected value E(X) and the variance Var(X).

Solution:

E(X) = p = 0.4

Var(X) = p(1-p) = 0.4⋅(0.6)

= 0.24

Problem 6: Suppose X and Y are independent random variables with X following a normal distribution with mean 2 and variance 1, and Y following a normal distribution with mean 3 and variance 4. Find the distribution of Z = X + Y.

Solution:

Z∼N(μX + μY,σX² + σY²)

= N(2 + 3,1 + 4)

= N(5,5)

Problem 7: A random variable X follows a geometric distribution with parameter p = 0.3. Find the expected value E(X).

Solution:

E(X) = 1/p

= 1/0.3

= 10/3≈3.33

Problem 8: If Z is a standard normal random variable, find P(Z > 1.96).

Solution:

Using standard normal distribution table:

P(Z > 1.96)

= 1 – P(Z ≤ 1.96)

= 1-0.975 = 0.025

Problem 9: Let X be a random variable with the following probability mass function (pmf): P(X = 0) = 0.4, P(X = 1) = 0.3, and P(X = 2) = 0.3. Find the expected value E(X) and the variance Var(X).

Solution:

E(X) = 0⋅0.4 + 1⋅0.3 + 2⋅0.3

= 0.9

E(X²) = 0^2×0.4 + 1^2×0.3 + 2^2×0.3 = 0 + 0.3 + 1.2

= 1.5

Var(X) = E(X²)-(E(X))² = 1.5-0.9² = 1.5-0.81

= 0.69

Problem 10: If Z is a standard normal random variable, find P(-2 ≤ Z ≤ 2).

Solution:

Using standard normal distribution table

P(-2 ≤ Z ≤ 2 )

= P(Z ≤ 2) – P(Z ≤ -2)

= 0.9772 – 0.0228

= 0.9544

Random Variables Practice Problems

Q1. Let X be a discrete random variable with the following pmf: P(X = 1) = 0.1, P(X = 2) = 0.3, P(X = 3) = 0.4, and P(X = 4) = 0.2. Find the expected value E(X).

Q2. A random variable Y follows a uniform distribution on the interval [1, 5]. Find the probability that Y is greater than 3.

Q3. If Z is an exponential random variable with rate parameter λ = 2, find the probability that Z is less than 1.

Q4. Let X be a normal random variable with mean μ = 4 and variance σ² = 9. Find P(1 ≤ X ≤ 7).

Q5. A fair coin is flipped 10 times. Let X be the number of heads obtained. Find the probability that X equals 5.

Q6. Let Y be a continuous random variable with the pdf fY(y) = 1/2sin⁡(y) for 0 ≤ y ≤ π. Find the expected value E(Y).

Q7. A random variable Z follows a Poisson distribution with parameter λ = 6. Find the probability that Z equals 4.

Q8. Let X be a Bernoulli random variable with parameter p = 0.7. Find the expected value E(X) and the variance Var(X).

Q9. A random variable X follows a binomial distribution with parameters n = 8 and p = 0.4. Find the probability that X is at least 6.

Q10. If Z is a standard normal random variable, find P(Z ≤ -1.28).

Also Check:

• What is Probability Distribution
• Expected Value

Random Variables Practice Problems – FAQs

What is Concept of Probability and Random Variables?

In probability a random variable is a real valued function whose domain is the sample space of the random experiment. It means that each outcome of a random experiment is associated with a single real number, and the single real number may vary with the different outcomes of a random experiment.

Can Random Variables only have One Value?

A random can have a single value but it is not necessary that it only has one value. For example, the square root of a number can have two values, positive and negative. It is possible that the probability of a value of a random variable is zero. This would mean that the event cannot occur.

What is the Range of Random Variables?

Range of a random variable X, shown by Range(X) or RX, is the set of possible values for X. In the above example, Range(X) = RX = {0,1,2,3,4,5}. The range of a random variable X, shown by Range(X) or RX, is the set of possible values of X.

What is the General Function of a Random Variable?

A random variable is a rule that assigns a numeric value to every possible outcome in a sample space. Random variables may be discrete or continuous in nature. A random variable is discrete if it assumes only discrete values within a specified interval.

How do we Solve the Probability of Random Variables?

Suppose a variable X can take the values 1, 2, 3, or 4. The probability that X is equal to 2 or 3 is the sum of the two probabilities: 

P(X = 2 or X = 3)

= P(X = 2) + P(X = 3)

= 0.3 + 0.4 = 0.7

Similarly, the probability that X is greater than 1 is equal to 1 – P(X = 1) = 1 – 0.1 = 0.9, by the complement rule.