Discrete random variable questions in Cambridge A Level Mathematics (9709) Probability and Statistics 1 follow a very regular pattern. You find the probability of each value, often after finding a constant k, set out the probability distribution table, and then calculate \mathrm{E}(X) and \mathrm{Var}(X). The marks are spread across those steps, so a careful answer collects most of them even if one value goes wrong. This post shows how recent mark schemes divide the marks, works through one question in full, and gives you three real questions to try.
The three things you need
A probability distribution table. List every value x that X can take, with \mathrm{P}(X = x) underneath. The probabilities must add up to 1.
The expectation.
$$\mathrm{E}(X) = \sum x\,\mathrm{P}(X = x)$$The variance.
$$\mathrm{Var}(X) = \sum x^2\,\mathrm{P}(X = x) - \left[\mathrm{E}(X)\right]^2$$That is all the theory. The skill is in finding the probabilities correctly and setting out the arithmetic so each mark is visible.
How the marks are divided
The table is marked in stages. In Paper 51, May/June 2026, a three-mark table was marked B1 for a table with the correct x values and at least one correct probability, B1 for three more correct probabilities, and B1 for all six. Paper 52, May/June 2024 used the same shape. So a table with the right values and a few correct probabilities still scores. The schemes also say probabilities need not be in the table if they are clearly linked to their values, but a table is the clearest way to do that.
Finding k can be its own mark. In Paper 52, May/June 2023, the first B1 was for k = \frac{1}{14}, from 3k + 3k + 8k = 1. In Paper 51, Oct/Nov 2024 it was also a B1, for k = \frac{1}{28}.
Variance has two method marks. In Paper 52, May/June 2023 and Paper 51, May/June 2026, \mathrm{Var}(X) was M1 for the \mathrm{E}(X) calculation, M1 for "appropriate variance formula using their (\mathrm{E}(X))^2 value", and A1 for the answer. "Their" matters: if your \mathrm{E}(X) is wrong but you use it correctly in the variance formula, the second method mark can still be earned. Write both sums out, not just the results.
"Show that" needs the working. Paper 52, May/June 2024 asked you to show that \mathrm{P}(X = 7) = 0.2 for one mark. The scheme's working identifies 7 as $5 + $2, then gives \frac{1}{5} \times \frac{2}{4} \times 2. The \times 2 is there because the two coins can be picked in either order. Since 0.2 is given in the question, the mark is for the working that reaches it.
The codes M, A and B are explained in how Cambridge mark schemes work.
Where the probabilities come from
Questions get their probabilities in one of two ways, and each needs a different first step.
From a formula with a constant. The question gives \mathrm{P}(X = x) in terms of x and k. Substitute every value of x, add the results, set the total to 1 and solve for k. Then substitute k back. Watch for values where the formula gives the same probability, such as x = -2 and x = 2 when the formula uses x^2.
From a situation. Dice, spinners, coins or counters are described, and you work out the probabilities yourself. Here the risk is missing outcomes or counting some twice. Two methods keep you safe:
- A sample space grid for two independent events, such as two spinners. Each cell is one equally likely outcome, so a probability is the number of cells for that value divided by the total number of cells.
- A systematic list for selections without replacement, such as choosing two coins from a purse. List each pair once, then multiply by 2 where the two items can come out in either order.
Whichever method you use, the probabilities in your table must add up to 1. Check that before you calculate \mathrm{E}(X), because every later mark depends on the table being right.
Checking your answers
Three quick checks catch most errors. The probabilities add up to 1. \mathrm{E}(X) lies between the smallest and the largest value of X, because it is a weighted average of them. \mathrm{Var}(X) is positive. None of these proves the answer is right, but failing any of them proves something is wrong.
Worked walk-through: Paper 52, May/June 2023, Question 1
The random variable X takes the values -2, 2 and 3, and
$$\mathrm{P}(X = x) = k(x^2 - 1)$$The question is worth 6 marks.
(a) Draw up the probability distribution table, giving the probabilities as numerical fractions, 3 marks.
Substitute each value:
- x = -2: k(4 - 1) = 3k
- x = 2: k(4 - 1) = 3k
- x = 3: k(9 - 1) = 8k
The probabilities add to 1, so 14k = 1 and k = \frac{1}{14}. That is the first B1.
| x | -2 | 2 | 3 |
|---|---|---|---|
| \mathrm{P}(X = x) | \frac{3}{14} | \frac{3}{14} | \frac{8}{14} |
The second B1 is for the table with the correct values of x and at least one correct probability, and the third for the rest. Note "numerical fractions": 3k is not acceptable in the final table. The scheme also allows follow-through, so correct probabilities from a wrong k can still earn the last mark.
(b) Find \mathrm{E}(X) and \mathrm{Var}(X), 3 marks.
$$\mathrm{E}(X) = -2 \times \tfrac{3}{14} + 2 \times \tfrac{3}{14} + 3 \times \tfrac{8}{14} = \tfrac{24}{14} = \tfrac{12}{7}$$Showing this sum is the first M1. For the variance, first find \sum x^2\,\mathrm{P}(X = x):
$$4 \times \tfrac{3}{14} + 4 \times \tfrac{3}{14} + 9 \times \tfrac{8}{14} = \tfrac{96}{14} = \tfrac{48}{7}$$Then subtract the square of \mathrm{E}(X), which is the second M1:
$$\mathrm{Var}(X) = \tfrac{48}{7} - \left(\tfrac{12}{7}\right)^2 = \tfrac{336}{49} - \tfrac{144}{49} = \tfrac{192}{49}$$The final A1 needs both \mathrm{E}(X) = \frac{12}{7} and \mathrm{Var}(X) = \frac{192}{49}, which the scheme also accepts as 1.71 and 3.92.
Try it yourself and see which of the six marks your answer earns.
Common ways marks are lost
- Probabilities that do not add to 1. Always check the total before you calculate anything else. It catches most errors in finding k or counting outcomes.
- Forgetting both orders. When two objects are chosen, a $5 coin then a $2 coin, and a $2 coin then a $5 coin, are different outcomes.
- Squaring the wrong thing. The first sum uses x^2, not x, and you subtract [\mathrm{E}(X)]^2, not \mathrm{E}(X).
- Rounding too early. Keep fractions until the end. Rounded probabilities can make the variance wrong at the third significant figure.
- Leaving k in the table. If the question asks for numerical fractions, give numbers.
- A negative variance. It is impossible, so it is a sign to recheck the working.
Two more real questions to try
Paper 52, May/June 2024, Question 5 is about coins: one $5 coin, two $2 coins and two $1 coins, with two chosen at random and X their total value. You show that \mathrm{P}(X = 7) = 0.2, draw up the table for five values, then find \mathrm{Var}(X). The scheme's answer is 3.24. The work is in listing the pairs carefully.
Paper 51, May/June 2026, Question 1 adds the scores on two fair four-sided spinners, one numbered 1, 2, 3, 3 and the other -2, -1, -1, 1. A sample space grid of all 16 outcomes is the safest way to build the table. Then find \mathrm{Var}(X), which the scheme gives as \frac{15}{8}.
You do not need an account to try these: as a guest, each question part can be marked twice for free. Write your working by hand or type it, and the marking shows which B, M and A marks you earned.
Practise more discrete random variable questions
The free 9709 discrete random variables topical page lists 52 structured questions on this topic from 22 papers, from 2023 to 2026, each linked to its paper. For full papers, open the A Level Mathematics past papers and choose a Probability and Statistics 1 paper.
Frequently asked questions
- How do I find k in a probability distribution question?
- Write each probability in terms of k, add them, and set the total equal to 1, because the probabilities of all the values of X must add up to 1. Solve for k, then substitute it back to get each probability as a number.
- What is the formula for Var(X) for a discrete random variable?
- Var(X) is the sum of x squared times p, minus the square of E(X). Find E(X) first as the sum of x times p. Recent 9709 mark schemes give a method mark for an appropriate variance formula using your own value of E(X), so the method can score even if E(X) was wrong.
- Should I give probabilities as fractions or decimals?
- Follow the question. Several recent 9709 questions say giving the probabilities as numerical fractions, and then fractions in terms of k or rounded decimals do not match what was asked. Where it does not say, exact fractions are still the safest way to avoid rounding errors in E(X) and Var(X).
- Can Var(X) be negative?
- No. Variance measures spread and is never negative. If you get a negative value, you have probably forgotten to square the x values in the first sum, or subtracted E(X) instead of its square.