Discrete Random Variables and Probability Functions
Key Terms
- A discrete random variable (DRV) X takes a countable set of values {x1, x2, …}
- A probability function assigns P(X = x) to each value; must satisfy:
- (i) P(X = x) ≥ 0 for all x
- (ii) ∑ P(X = x) = 1 (probabilities sum to 1)
- A probability distribution table lists all values x and their probabilities P(X = x)
- P(X ≤ k)
- = sum of P(X = x) for all x ≤ k (cumulative probability)
- P(a ≤ X ≤ b)
- = sum of P(X = x) for all x from a to b inclusive
- To find k in a formula: set ∑ P(X = x) = 1 and solve for k
• P(X = x) ≥ 0 for every value x
• ∑all x P(X = x) = 1
Useful probability rules:
P(X > k) = 1 − P(X ≤ k)
P(X ≥ k) = 1 − P(X ≤ k − 1)
P(a ≤ X ≤ b) = P(X ≤ b) − P(X ≤ a − 1)
Apply ∑ P(X = x) = 1:
k(1) + k(2) + k(3) + k(4) = 1
k(1 + 2 + 3 + 4) = 1
10k = 1
k = 1/10
Check: P(X=1) = 0.1, P(X=2) = 0.2, P(X=3) = 0.3, P(X=4) = 0.4. Sum = 1 ✓
| x | 1 | 2 | 3 | 4 | 5 |
| P(X=x) | 0.15 | 0.20 | 0.30 | 0.25 | 0.10 |
(a) P(X > 2) = P(X=3) + P(X=4) + P(X=5) = 0.30 + 0.25 + 0.10 = 0.65
Or: P(X > 2) = 1 − P(X ≤ 2) = 1 − (0.15 + 0.20) = 1 − 0.35 = 0.65 ✓
(b) P(1 ≤ X ≤ 3) = P(X=1) + P(X=2) + P(X=3) = 0.15 + 0.20 + 0.30 = 0.65
What Is a Discrete Random Variable?
In everyday language, we often say things like “the number of heads in 3 coin tosses” or “the number of defective items in a batch of 10.” These are examples of random variables — quantities whose value is determined by a random process. A discrete random variable (DRV) is one that can only take specific, separated values — usually whole numbers. You can count the possible outcomes, even if there are infinitely many.
Contrast this with continuous random variables (like height or temperature), which can take any value in an interval. For this topic, we focus exclusively on discrete random variables.
We conventionally use capital letters like X, Y, Z for random variables, and lower case letters like x, y, z for specific values they might take. So P(X = 3) means “the probability that the random variable X takes the value 3.”
Probability Functions and Distribution Tables
A probability function (also called a probability mass function) assigns a probability to every possible value of a DRV. It can be expressed as a formula or as a table. For a valid probability function, two conditions must hold:
Condition 1: Every probability is non-negative: P(X = x) ≥ 0 for all x. A negative probability is meaningless.
Condition 2: All probabilities sum to exactly 1: ∑ P(X = x) = 1. This reflects certainty — some outcome must occur.
A probability distribution table lists all possible values of X in one row and the corresponding probabilities in another row. It provides a complete description of the random variable’s behaviour. From a distribution table, you can calculate any probability involving X.
Why These Conditions?
The condition P(X = x) ≥ 0 is obvious — probability cannot be negative. The condition ∑ P(X = x) = 1 comes from the fact that the sample space is exhaustive: one of the listed outcomes must occur. If the probabilities summed to more than 1, we would be claiming more certainty than is possible; if they summed to less than 1, something would be missing.
In practice, these two conditions are a quick diagnostic test. If either fails, the function is not a valid probability function.
Finding an Unknown Constant k
A common problem type gives you a probability function containing an unknown constant k (e.g., P(X = x) = kx for x = 1, 2, 3, 4). The strategy is always: apply the condition ∑ P(X = x) = 1 to generate an equation, then solve for k. This is often straightforward algebra.
After finding k, always check that all probabilities are non-negative. If k turns out to be negative, something has gone wrong — re-examine your equation.
Calculating Probabilities from a Distribution
Once you have the probability distribution (either from a formula or a table), you can calculate any required probability.
P(X > k): Add up P(X = x) for all x greater than k. Alternatively, use the complement: P(X > k) = 1 − P(X ≤ k).
P(X ≥ k): Similar, but include x = k: P(X ≥ k) = 1 − P(X ≤ k − 1).
P(a ≤ X ≤ b): Add P(X = x) for x = a, a+1, …, b.
The complement rule P(X > k) = 1 − P(X ≤ k) is especially useful when it is easier to compute the cumulative probability P(X ≤ k) than to add up many individual terms.
Connection to Relative Frequency
In real-world statistics, we often estimate a probability function from data by computing relative frequencies. If we roll a die 600 times and observe 2 exactly 95 times, we estimate P(X = 2) ≈ 95/600 ≈ 0.158. With a large enough sample, the relative frequency table approximates the true probability distribution. This connection reinforces that probability is a mathematical model for long-run relative frequency.
Mastery Practice
- Fluency Show that the following is a valid probability function for X:
x 0 1 2 3 P(X=x) 0.1 0.4 0.3 0.2 - Fluency X has probability function P(X = x) = k(x + 1) for x = 0, 1, 2, 3. Find k.
- Fluency For the distribution below, find the missing probability p and then find P(X ≥ 3).
x 1 2 3 4 P(X=x) 0.25 0.30 p 0.15 - Fluency Using the distribution from Q3, find P(X > 2) and P(1 ≤ X ≤ 3).
- Understanding Which of the following is a valid probability function for X with values {1, 2, 3}?
(A) P(X = x) = x/4 (B) P(X = x) = (4 − x)/9 (C) P(X = x) = (x − 1)/3
Justify your answer. - Understanding A probability function is given by P(X = x) = c/x for x = 1, 2, 3, 4, 5, 6. Find c. Then construct the probability distribution table and calculate P(3 ≤ X ≤ 5).
- Understanding A bag contains 3 red, 2 blue, and 1 green marble. One marble is drawn. Let X = 1 if red, X = 2 if blue, X = 3 if green. Construct the probability distribution table for X and find P(X ≠ 2).
- Understanding X has distribution:
Find (a) P(X < 1) (b) P(−1 ≤ X ≤ 1) (c) P(X ≥ 0).x −1 0 1 2 P(X=x) 0.20 0.35 0.30 0.15 - Problem Solving X has probability function P(X = x) = k(2x − 1) for x = 1, 2, 3, 4. Find k, state the probability distribution, identify the mode, and find P(X is even).
- Problem Solving Two fair coins are tossed simultaneously. Let X be the number of heads. (a) Construct the probability distribution table for X. (b) Find P(X ≥ 1). (c) Explain why P(X = 0) + P(X = 1) + P(X = 2) = 1 must hold.