Discrete and binomial distributions
Use probability weights and recognise the assumptions behind a binomial model.
Probability & Statistics 1 (Paper 5) · Mathematics 9709, 2026–2027. This short revision overview checks weighted means and binomial assumptions; it does not cover the full section 5.4 pathway, including variance and geometric distributions. Written answers are self-checked, not officially marked.
Before you begin
- Multiply probabilities for independent events and use a complement.
- Multiply a value by its probability, including when the value is negative.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
A table lists all possible values of X as 0, 2 and 5, with probabilities 0.3, 0.4 and 0.2. What prevents it from being a valid complete probability distribution?
What you will learn
- Calculate an expected value.
- Use a binomial model for repeated trials.
Expected value is a weighted mean
For a discrete random variable, list each possible value and its probability. Probabilities must be non-negative and sum to one. The expected value is the sum of value multiplied by probability; it need not be a possible outcome.
An expectation describes a long-run average under repeated use of the model. It does not predict the exact result of the next trial.
Check the binomial assumptions
A binomial variable counts successes in a fixed number n of independent trials, each with the same success probability p and two classified outcomes. When sampling without replacement from a small population, constant p and independence usually fail.
Exactly zero successes means every trial fails, giving (1−p)ⁿ. At least one success is the complement, 1−(1−p)ⁿ.
Sources for this lesson
Pause and explain
A bag contains 6 red and 4 blue counters. Two counters are drawn at random without replacement, and X counts the red counters. Which conclusion is correct?
Worked example
X takes values 0,1,3 with probabilities 0.2,0.5,0.3. Find E(X).
Show the worked solution
- Check the probabilities: 0.2+0.5+0.3=1.
- Multiply values by probabilities: 0,0.5,0.9.
- Add the weighted values.
Answer 1.4
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.
X takes values −1, 1 and 4 with probabilities 0.2, 0.5 and 0.3. Find E(X) and explain whether it must be an outcome.
Give me a hint
Check the probability total, then keep the negative sign in the weighted sum.
Compare my reasoning
- The probabilities are non-negative and total 1.
- Weight every value: E(X) = (−1)(0.2) + (1)(0.5) + (4)(0.3) = −0.2 + 0.5 + 1.2.
- E(X) = 1.5. It is a model average, so it need not be one of the possible outcomes −1, 1 and 4.
E(X) = 1.5; it is not a possible outcome, and that is allowed.
Look for these in your work
- Check that the table is valid.
- Include the negative weighted term.
- Interpret expectation as an average rather than a prediction of the next result.
Five independent trials each have success probability 0.20. Find the probability of at least one success.
Give me a hint
The complement contains just one event: no successes.
Compare my reasoning
- The number of successes is B(5, 0.20): five fixed independent trials, with the same success probability.
- No successes means all five trials fail, so P(X = 0) = (0.80)^5 = 0.32768.
- Subtract from 1: P(X ≥ 1) = 1 − 0.32768 = 0.67232.
0.67232.
Look for these in your work
- Use the no-success event as the complement.
- Use the failure probability 0.80 in all five factors.
- Keep the exact probability under the stated model.
A bag has 3 red and 2 blue counters. Compare three random draws with replacement and thorough mixing after each draw against three draws without replacement. In each case find the probability of no red counters, and decide whether the red count is binomial.
Give me a hint
Track how many blue counters remain; do not assume both sampling rules give independent trials.
Compare my reasoning
- With replacement and mixing, every draw has red probability 3/5 independently. The red count is B(3, 3/5), and no red has probability (2/5)^3 = 8/125.
- Without replacement, no red in three draws would require three blue counters, but the bag contains only two. Its probability is 0.
- Without replacement the conditional probabilities change, so an exact binomial model is not justified. Both counts are still discrete.
With replacement: 8/125 = 0.064, binomial. Without replacement: 0, not binomial.
Look for these in your work
- State the sampling mechanism before choosing a model.
- Use the available number of blue counters to recognise an impossible event.
- Separate discrete values from binomial assumptions.
Common mistakes
- Do not average the possible values without probability weights.
- The expected value does not have to be an integer.
What can you explain now?
For four independent trials with success probability 0.25, find the probability of no successes.
Compare with the explanation
0.31640625
All four trials must fail, so the probability is 0.75⁴.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
From memory, explain why probability weights matter for E(X), then name the assumptions you must check before using a binomial formula. Revisit the section 5.4 pathway for variance, geometric models and further probability problems.
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.