Build a discrete probability distribution from outcomes
Define the numerical variable, group elementary outcomes by its value and keep every probability in the table.
Section 5.4 · Discrete random variables
Exam scope and assessment notes
Probability & Statistics 1 (Paper 5, AS & A Level) · Mathematics 9709, 2026–2027, section 5.4. A probability table needs non-negative probabilities summing to one. E(X) is a probability-weighted mean; Var(X) = E(X²) − [E(X)]², with squared units. A binomial model B(n,p) counts successes in a fixed non-negative integer number n of independent two-outcome trials with a constant success probability p. A geometric model Geo(p) counts the trial of the first success, starting at 1, with independent trials and a constant 0 < p ≤ 1. Formula proofs are not required. The expectation of the geometric distribution is required; its variance is not required by section 5.4. Normal approximations belong to section 5.5 and are not used here. Exact model probabilities do not establish real-world independence or constant rates without evidence. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- Enumerate equally likely elementary outcomes.
- Use fractions and unordered selection counts.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
How many heads occur in outcome HT?
What you will learn
- Define a discrete random variable and its possible values.
- Construct a complete probability table from a specified experiment.
A variable assigns a number to each outcome
A random variable X assigns a numerical value to each elementary outcome. Discrete means its possible values can be listed as a finite or countably infinite set; they need not be consecutive integers. For two independent fair coin tosses, define X as the number of heads. HH gives 2, HT and TH give 1, and TT gives 0.
The distribution table has X values 0, 1, 2 with probabilities 1/4, 1/2, 1/4. Each elementary toss string is equally likely, but the three X values are not. Check that every possible value is represented and the probabilities sum to one. Probabilities attach to values, not to the number of columns in a table.
Use the experiment’s sampling mechanism
For one fair die let Y = 1 if the result is greater than 4 and Y = 0 otherwise. Then P(Y = 0) = 4/6 = 2/3 and P(Y = 1) = 2/6 = 1/3. A variable can deliberately summarise several outcomes into one value, and its definition determines which outcomes are grouped.
Draw two balls uniformly without replacement from three red and two blue, and let X count red balls. Ten unordered pairs are equiprobable. X = 0 has one blue-blue pair, X = 1 has six mixed pairs and X = 2 has three red-red pairs. The table probabilities are 1/10, 6/10, 3/10. This table is valid although the draws are dependent; a binomial assumption is unnecessary.
Pause and explain
For two fair independent coins, what is P(X = 1 head)?
Worked example
Two independent fair coins are tossed. Build the distribution of X, the number of heads.
Show the worked solution
- List TT, TH, HT and HH, each with probability 1/4.
- Their X values are 0, 1, 1 and 2.
- Combine equal X values to obtain probabilities 1/4, 1/2, 1/4 for 0, 1, 2.
Answer X: 0, 1, 2; probabilities: 1/4, 1/2, 1/4.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Build the distribution of X, the number of heads in two independent fair coin tosses.
Give me a hint
Group the four equally likely strings by their head counts.
Compare my reasoning
- TT gives 0 heads; HT and TH each give 1; HH gives 2.
- The counts are 1, 2 and 1 out of four elementary outcomes.
- Write probabilities 1/4, 1/2, 1/4 and check their sum is one.
X: 0, 1, 2; P(X): 1/4, 1/2, 1/4.
Look for these in your work
- Define the head-count variable.
- Combine both one-head outcomes.
Roll one fair die. Define Y = 1 for a result greater than 4 and Y = 0 otherwise. Build its distribution.
Give me a hint
The indicator groups four die outcomes as zero and two as one.
Compare my reasoning
- Results 1–4 give Y = 0.
- Results 5–6 give Y = 1.
- The probabilities are 4/6 = 2/3 and 2/6 = 1/3.
Y: 0, 1; probabilities: 2/3, 1/3.
Look for these in your work
- Use the specified indicator definition.
- Include the whole sample space exactly once.
Draw two balls uniformly without replacement from three red and two blue. Let X count red balls. Build the distribution using unordered pairs.
Give me a hint
Count all ten pairs, then split them by zero, one or two red balls.
Compare my reasoning
- X = 0 has 2C2 = 1 pair; X = 1 has 3 × 2 = 6 pairs.
- X = 2 has 3C2 = 3 pairs.
- Divide by 5C2 = 10 to obtain 1/10, 6/10, 3/10.
X: 0, 1, 2; probabilities: 1/10, 6/10, 3/10.
Look for these in your work
- Preserve the without-replacement model.
- Check the three disjoint counts sum to ten.
Common mistakes
- Define the variable rather than only naming the experiment.
- Variable values need not be equiprobable.
- Use the stated replacement mechanism when grouping outcomes.
What can you explain now?
Can a discrete variable take values −1, 0 and 2?
Compare with the explanation
Yes; discreteness does not require consecutive positive integers.
The possible values form a list, and their probabilities define the distribution.
After trying it yourself, choose your next review. This is your self-assessment.
Your review choice appears on Today. Sign in to sync it across devices.
Make it stick
Define X, list its possible values, combine their contributing outcomes and check the complete probability table.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.