Summarise a probability distribution and its expectation
List possible values, check a complete probability table and use expected value to reason about repeated outcomes.
Section 8 · Summarise a probability distribution and its expectation
Exam scope and assessment notes
This lesson maps the Summarise a probability distribution and its expectation section of Cambridge IGCSE Statistics 0479 (2027 and 2028–2030 syllabus versions). This is original StudyForA teaching and self-check practice, not Cambridge material, an official mark scheme, a timed mock or an official grade. Charts and written working are not automatically assessed; human subject-teacher review remains pending. Use the syllabus version for your exam year. Cambridge offers 0479 in the June series only and does not make it available in every administrative zone. Check the current Cambridge 0479 page and confirm your centre's entry eligibility with your exams officer.
Before you begin
- Use the earlier 0479 ideas where needed; write units, working and a contextual interpretation for each result.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
A fair coin has two equally likely outcomes. What probability must each outcome have?
What you will learn
- Calculate probabilities for every outcome in a simple discrete case
- Present a probability distribution and verify that its probabilities sum to 1
- Calculate and interpret expectation
A distribution accounts for every possible value
A probability distribution table pairs each possible value x with P(X=x). Outcomes must be mutually exclusive and collectively exhaustive; the probabilities therefore sum to 1. If one probability is missing, subtract the known total from 1.
Expectation is a long-run average, not a promised result
Expected value E(X) = ΣxP(X=x). Multiply each outcome by its probability and add. It describes the average over many repetitions; a single trial can be any possible outcome and need not equal the expectation.
Pause and explain
A prize game's expected payout is $2. What does this mean?
Worked example
A game pays $0 with probability 0.5, $4 with probability 0.3 and $10 with probability 0.2. Find the expected payout.
Show the worked solution
- Check probabilities: 0.5 + 0.3 + 0.2 = 1.
- Multiply outcomes by probabilities: 0×0.5 + 4×0.3 + 10×0.2.
- Add: 0 + 1.2 + 2 = 3.2.
Answer Expected payout = $3.20 per play in the long run. This does not mean every play pays $3.20.
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.
A discrete variable X takes values 0, 1 and 3 with probabilities 0.2, 0.5 and k. Find k and E(X).
Give me a hint
First use the total probability, then calculate ΣxP(X=x).
Compare my reasoning
- k = 1 − 0.2 − 0.5 = 0.3.
- E(X) = 0×0.2 + 1×0.5 + 3×0.3.
- E(X) = 1.4.
The missing probability is k = 0.3, and the probability-weighted mean is E(X) = 1.4.
Look for these in your work
- Probabilities total one
- Expected value terms correct
- Arithmetic correct
A repair plan costs $12. There is a 0.1 probability of a $50 repair being needed and a 0.9 probability of no repair cost. Find the expected total cost and explain what it does not tell the owner.
Give me a hint
Calculate expected repair cost, then include the fixed plan cost.
Compare my reasoning
- Expected repair cost = 50×0.1 + 0×0.9 = $5.
- Expected total = $12 + $5 = $17.
- The owner will not necessarily spend exactly $17; it is a long-run average under the stated probabilities.
Expected total cost is $17. A single outcome costs either $12 or $62, so $17 is not a guaranteed bill.
Look for these in your work
- Expected variable cost correct
- Fixed cost included
- Long-run meaning explained
A distribution table lists outcomes 1, 2 and 5, but the probability for outcome 2 is omitted. The other probabilities are 0.15 and 0.35. Complete it and find the expectation.
Give me a hint
Use 1 minus the known probabilities; then apply the weighted sum.
Compare my reasoning
- Missing probability = 1 − 0.15 − 0.35 = 0.5.
- E(X)=1×0.15 + 2×0.5 + 5×0.35.
- E(X)=0.15+1+1.75=2.9.
P(X=2) is 0.5 so the distribution totals 1; weighting each outcome gives E(X) = 2.9.
Look for these in your work
- Missing probability correct
- All outcomes weighted
- Expectation correct
Common mistakes
- Forgetting an outcome when checking that probabilities total 1
- Summing outcomes without weighting by probability
- Treating an expected value as a guaranteed single outcome
What can you explain now?
A distribution has known probabilities 0.25, 0.4 and one missing value. What is the missing probability?
Compare with the explanation
0.35
The probabilities must sum to 1, so 1 − 0.25 − 0.4 = 0.35.
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Make it stick
How do you calculate expectation, and why may a single result differ from it?
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.