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Quantitative Aptitude · Theoretical Distributions

Random Variables and Probability Distributions for CA Foundation

Updated 1 October 2026 · Fact-checked

A random variable assigns a number to each outcome of a random experiment. A probability distribution lists its values with their probabilities. To solve questions, check that probabilities sum to 1, then compute E(X) = Σ x·p(x) and Var(X) = E(X²) − [E(X)]².

Understand Random Variables and Probability Distributions

A random variable is a rule that turns each outcome of a random experiment into a number. If you toss two coins, the outcomes are HH, HT, TH, TT. Let X be the number of heads. Then X takes the values 0, 1 or 2. X is the random variable.

A random variable is discrete if it takes countable values, such as 0, 1, 2, 3. Examples are the number of defective items in a lot or the number of calls received. It is continuous if it can take any value in an interval, such as height, weight or time. Rule of thumb: counting gives discrete, measuring gives continuous.

For a discrete variable, the probability mass function (PMF) gives p(x) = P(X = x). It must satisfy two conditions: p(x) ≥ 0 for every x, and Σ p(x) = 1. For a continuous variable, the probability density function (PDF) f(x) is used. Here f(x) ≥ 0 and the total area under the curve is 1. Probability is the area under the curve between two points, so P(X = a) = 0 for any single point a.

A theoretical distribution is a probability distribution described by a formula, built from assumptions about the experiment, rather than from observed data. Binomial, Poisson and Normal are the main ones. They let you find probabilities without running the experiment.

Two numbers summarise a distribution. Expectation E(X) is the long-run average value, the mean. Variance Var(X) measures how spread out the values are around the mean. Standard deviation is the square root of the variance.

Key formulas to remember

Conditions for a discrete distribution
p(x) ≥ 0 for all x, and Σ p(x) = 1
Use this to find an unknown constant in a given table.
Conditions for a continuous distribution
f(x) ≥ 0, and total area under f(x) = 1
For a continuous variable, P(X = a) = 0, so P(a ≤ X ≤ b) is the same whether endpoints are included or not.
Expectation (discrete)
E(X) = Σ x·p(x)
Multiply each value by its probability and add.
Expectation of X²
E(X²) = Σ x²·p(x)
Square the value x, not the probability.
Variance
Var(X) = E(X²) − [E(X)]²
Always non-negative. If you get a negative value, recheck your arithmetic.
Standard deviation
SD(X) = √Var(X)
Same units as X.
Linear change
E(aX + b) = a·E(X) + b; Var(aX + b) = a²·Var(X)
Adding a constant b does not change variance.
Cumulative probability
P(X ≤ x) = Σ p(t) for all t ≤ x
For discrete X, P(X > x) = 1 − P(X ≤ x).

How to solve Random Variables and Probability Distributions questions

Use this order for almost any question on a discrete distribution table or on expectation and variance.

  1. 1Decide whether X is discrete or continuous. Counting means discrete, measuring means continuous.
  2. 2If a constant is unknown, use Σ p(x) = 1 (discrete) or total area = 1 (continuous) to find it.
  3. 3Check every probability is between 0 and 1 after finding the constant. Reject values that give a negative probability.
  4. 4For a probability of a range, add the p(x) values in that range. Use 1 − P(...) when the complement is shorter.
  5. 5Compute E(X) = Σ x·p(x).
  6. 6Compute E(X²) = Σ x²·p(x).
  7. 7Compute Var(X) = E(X²) − [E(X)]² and take the square root if SD is asked.
  8. 8Match your answer with the options and check the units and the form asked (variance or SD).

Quickest way: Table-column method with complement check

When to use it: Use it for any MCQ that gives a probability table and asks for a constant, a probability, a mean or a variance.

  1. Write the x values in a row and the p(x) values below. Add a third row for x·p(x).
  2. Solve for the unknown constant first. Do this before anything else.
  3. Find E(X) by adding the third row. Do the mental check that E(X) lies between the smallest and largest x.
  4. Only compute E(X²) if variance or SD is asked. Use x²·p(x) directly.
  5. For P(X ≥ k) type questions, compare the number of terms. Add the shorter side and use 1 minus it if needed.
  6. If two options look close, recheck by estimating the mean. Skip the question if it needs more than about two minutes, as wrong answers cost 0.25.

Common mistakes in Random Variables and Probability Distributions

  • Forgetting to check that Σ p(x) = 1 before using a table.

    Students jump straight to E(X).

    Fix: Add the probabilities first. If one is unknown, solve for it. This often finds the answer on its own.

  • Computing E(X²) as [E(X)]².

    The two look similar and students assume squaring is the same as taking the mean of squares.

    Fix: Square each x first, multiply by p(x), then add. Only then subtract [E(X)]² to get variance.

  • Squaring the probabilities instead of the values in E(X²).

    Confusion about which column to square.

    Fix: Remember that x² means the value squared. The probability p(x) is never squared.

  • Treating P(X = a) as non-zero for a continuous variable.

    Students carry over the discrete idea.

    Fix: For a continuous variable the probability at a single point is 0. Only intervals have probability.

  • Applying the shift rule to variance as Var(X + b) = Var(X) + b.

    Students copy the rule for expectation.

    Fix: Adding a constant shifts the mean but not the spread. Var(aX + b) = a²·Var(X).

  • Giving variance when SD is asked, or the other way round.

    Rushing and not reading the last line.

    Fix: Underline what is asked. Take the square root only for SD.

Worked examples

Example 1

A random variable X has the distribution: x = 0, 1, 2, 3 with p(x) = 0.1, k, 0.3, 0.2. What is the value of E(X)? Options: (a) 1.4 (b) 1.6 (c) 1.8 (d) 2.0

Show the solution
  1. Use Σ p(x) = 1: 0.1 + k + 0.3 + 0.2 = 1, so k = 0.4.
  2. E(X) = 0(0.1) + 1(0.4) + 2(0.3) + 3(0.2).
  3. = 0 + 0.4 + 0.6 + 0.6 = 1.6.
  4. Check: 1.6 lies between the smallest value 0 and the largest value 3, as it must.

Answer: (b) 1.6

Example 2

A random variable X takes values 1, 2, 3 with probabilities 0.2, 0.5, 0.3 respectively. What is Var(X)? Options: (a) 0.49 (b) 0.61 (c) 0.72 (d) 2.1

Show the solution
  1. E(X) = 1(0.2) + 2(0.5) + 3(0.3) = 0.2 + 1.0 + 0.9 = 2.1.
  2. E(X²) = 1(0.2) + 4(0.5) + 9(0.3) = 0.2 + 2.0 + 2.7 = 4.9.
  3. [E(X)]² = 2.1² = 4.41.
  4. Var(X) = 4.9 − 4.41 = 0.49.

Answer: (a) 0.49. Option (d) 2.1 is the mean, a common trap.

Example 3

If E(X) = 5 and Var(X) = 4, what is Var(3X + 2)? Options: (a) 12 (b) 14 (c) 36 (d) 38

Show the solution
  1. Use Var(aX + b) = a²·Var(X).
  2. Here a = 3, b = 2, so a² = 9.
  3. Var(3X + 2) = 9 × 4 = 36.
  4. The constant 2 does not change the variance.

Answer: (c) 36

Exam tips

  • Questions often give a table with one unknown. Solve Σ p(x) = 1 first, because later parts depend on it.
  • Learn Var(aX + b) = a²Var(X) cold. It gives a quick answer without any table work.
  • Check whether the question asks for variance or standard deviation before you pick an option.
  • Expect conceptual MCQs: which variable is discrete, why P(X = a) = 0 for continuous X, or what a theoretical distribution is. Revise the definitions.
  • Wrong answers cost 0.25 marks. Skip a long table question if you are unsure and come back at the end.

Practice questions from Theoretical Distributions

Random Variables and Probability Distributions: frequently asked questions

What is the difference between a discrete and a continuous random variable?

A discrete random variable takes countable values, such as the number of defective pieces. A continuous random variable takes any value in an interval, such as weight or time. Discrete uses a PMF, and continuous uses a PDF.

What is a theoretical distribution in statistics?

It is a probability distribution defined by a formula based on assumptions about how the experiment works. It is not built from observed data. Binomial, Poisson and Normal are the standard examples in CA Foundation.

How do I find the expectation and variance of a random variable?

Find E(X) by adding x·p(x) over all values. Then find E(X²) by adding x²·p(x). Variance is E(X²) − [E(X)]².

Can the probability density function be greater than 1?

Yes, f(x) can exceed 1 at some points, because it is a density and not a probability. What must equal 1 is the total area under the curve. The PMF of a discrete variable can never exceed 1.