Cambridge AS & A Level Mathematics 9709/52 is the May/June 2026 Paper 5 (Probability & Statistics 1), variant 2. It lasts 1 hour 15 minutes and is worth 50 marks, and it counts towards AS Level (with Paper 1) and towards both full A Level routes. It had seven compulsory questions, answered on the question paper. The biggest was Question 7 (9 marks) on the normal distribution, and Questions 2, 4 and 6 were worth 8 marks each. This guide goes through the paper question by question with the final answers and what the official mark scheme gave marks for. The paper and its mark scheme are free on the Tutopiya Learning Portal.

For an overview of all six 9709 papers from this series, see A Level Maths 9709 May/June 2026 past papers: what came up and how it was marked.

9709/52 at a glance

Question	Topic	Marks
1	Selections: more men than women in a team; dividing 13 people into three groups	6
2	Probability distribution for heads on three biased coins; first and second successes	8
3	Counting 3-digit numbers greater than 200	4
4	Binomial probability; normal approximation with a continuity correction	8
5	Pens taken from a box: joint, total and conditional probability	7
6	Back-to-back stem-and-leaf diagram, interquartile range, a change to the mean	8
7	Normal distribution: μ and σ from two percentages; an expected number	9
	Total	50

Question-by-question walkthrough

The mark scheme uses three main types of mark. M marks are for a valid method applied to this problem, A marks are for accurate answers and need the M mark before them, and B marks are for a correct result or statement and stand alone. A starred mark (B1*) has a later DB or DM mark depending on it, FT (follow through) lets a mark follow from your own earlier value, and SC is a special-case mark, often a single B1 for a correct answer when the method marks were not earned. The cover asks for non-exact answers to 3 significant figures, and there are no marks for unsupported calculator answers.

Question 1: selections (6 marks)

A club has 6 men and 9 women; a team of 5 is chosen at random.

• (a) More men than women (4 marks): the cases are 5M 0W (6 ways), 4M 1W (15 × 9 = 135) and 3M 2W (20 × 36 = 720), a total of 861 out of ¹⁵C₅ = 3003. Answer: 861/3003 = 41/143 = 0.287. M1 for a product ⁶Cₓ × ⁹Cᵧ with x + y = 5, M1 for adding the three correct cases (or 1 minus the other three), M1 for ¹⁵C₅ as the denominator, then A1. Multiplying probabilities such as 6/15 × 5/14 × … with the right arrangements factor was an accepted alternative.
• (b) Dividing 13 people into groups of 6, 4 and 3 (2 marks): ¹³C₆ × ⁷C₄ × ³C₃ = 1716 × 35 = 60 060, or 13! ÷ (6! 4! 3!). M1 for either form, then A1 CAO (correct answer only). A correct answer with no method shown could still earn SC B1.

Question 2: three biased coins (8 marks)

Each coin shows a head with probability ⅓; X is the number of heads when all three are thrown.

• (a) Probability distribution (3 marks): P(X = 0) = 8/27, P(X = 1) = 12/27 = 4/9, P(X = 2) = 6/27 = 2/9, P(X = 3) = 1/27. B1 for the table with the right x-values and one correct probability, B1 for three correct, B1 for all four. Decimals had to be correct to at least 3 significant figures.
• (b) At least two heads for the first time on the 7th throw (2 marks): P(at least two heads) = 6/27 + 1/27 = 7/27, so the probability is (20/27)⁶ × 7/27 = 0.0428. M1 for (1 − p)⁶p with your p and no other terms, then A1.
• (c) At least two heads for the second time on the 6th or 7th throw (3 marks): on the 6th throw there is one earlier success among the first 5 throws: 5 × (7/27)² × (20/27)⁴. On the 7th, one earlier success among the first 6: 6 × (7/27)² × (20/27)⁵. Total 0.191. M1 for either term, M1 for the sum of both forms, then A1.

Question 3: 3-digit numbers (4 marks)

Digits from {1, 2, 3, 4, 5, 6, 7}, and the number must be greater than 200, so the first digit cannot be 1.

• (a) Repeats allowed, divisible by 5 (1 mark): 6 choices for the first digit, 7 for the middle and 1 for the last (5): 6 × 7 × 1 = 42. B1.
• (b) No repeats, even (3 marks): if the first digit is even (2, 4 or 6), there are 2 even digits left for the end and 5 digits for the middle: 3 × 2 × 5 = 30. If the first digit is odd (3, 5 or 7), there are 3 even digits for the end: 3 × 3 × 5 = 45. Total 75. B1 for one correct case, M1 for adding the right cases, then A1. Splitting by the last digit, or by the odd/even pattern of all three digits, earned the same marks.

Question 4: puzzles, binomial and normal (8 marks)

Haz completes the puzzle in under 10 minutes with probability 0.6 each day.

• (a) At least 7 successes in 9 days (3 marks): ⁹C₇(0.6)⁷(0.4)² + ⁹C₈(0.6)⁸(0.4) + (0.6)⁹ = 0.1612 + 0.0605 + 0.0101 = 0.232. M1 for one binomial term of the right form, A1 for the full correct expression, B1 for the answer (above 0.2315, up to 0.232). Working out 1 − P(0 to 6) was also accepted.
• (b) Fewer than 66 successes in 120 days (5 marks): mean 120 × 0.6 = 72, variance 120 × 0.6 × 0.4 = 28.8. “Fewer than 66” is 65 or fewer, so use 65.5: z = (65.5 − 72) ÷ √28.8 = −1.211, and P(Z < −1.211) = 1 − 0.8871 = 0.113. B1 for both 72 and 28.8, M1 for standardising with your mean and standard deviation (using the variance by mistake still earned this M1), M1 for a continuity correction (65.5 or 66.5), M1 for the correct area, then A1. Calling 28.8 the standard deviation lost the B1.

Question 5: pens from a box (7 marks)

The box has 3 red, 5 blue and 2 green pens. Lui keeps one, a green pen is added, then Mira keeps one, so Mira always picks from 10 pens.

• (a) Both blue (1 mark): 5/10 × 4/10 = 1/5. B1.
• (b) Mira’s pen is red (3 marks): 3/10 × 2/10 + 5/10 × 3/10 + 2/10 × 3/10 = (6 + 15 + 6)/100 = 27/100 = 0.27. M1 for a sum of products of the right structure, M1 for the right first-pick probabilities paired with the cases, then A1.
• (c) Lui’s pen is red given Mira’s pen is blue or green (3 marks): P(Lui red and Mira blue or green) = 3/10 × 8/10 = 0.24, and P(Mira blue or green) = 1 − 0.27 = 0.73, so the answer is 24/73 = 0.329. B1* for the numerator, B1* for the denominator (1 minus your part (b) was accepted), then DB1 for the answer.

Question 6: rugby players’ heights (8 marks)

Fifteen heights for each of two teams, the Hippos and the Zebras.

• (a) Back-to-back stem-and-leaf diagram (4 marks): stems 16, 17, 18 and 19. B1 for the stem, B1 for the Hippos on the left with leaves in order reading outwards (right to left) and lined up, B1 for the Zebras on the right in order and lined up, B1 for a single key naming both teams and stating cm, such as 2|17|9 meaning 172 cm for the Hippos and 179 cm for the Zebras. Commas between leaves, or leaves that are not single digits, lost marks.
• (b) Interquartile range for the Hippos (2 marks): the 4th value is 172 and the 12th is 190, so the IQR is 190 − 172 = 18 cm. M1 for the quartiles (the scheme accepted an upper quartile from 188 to 191 and a lower quartile from 168 to 176), then A1.
• (c) The new player’s height (2 marks): a mean rise of 0.5 cm over 15 players means the total rises by 7.5 cm, so the new player is 184 + 7.5 = 191.5 cm (192 accepted). M1 for 15 × ½ (or the new and old totals, 2541 + x and 2725 over 15), then A1.

Question 7: leaf lengths and the normal distribution (9 marks)

• (a) μ and σ (5 marks): 20% below 3.60 gives (3.60 − μ)/σ = −0.842. 70% above 5.40 means 30% below, so (5.40 − μ)/σ = −0.524. Subtracting, 1.8 = 0.318σ, so σ = 5.66 and μ = 8.37. B1 for each z-value (0.841 to 0.843 and 0.523 to 0.525, with sign), M1 for one standardised equation using a z-value (not a probability such as 0.20 or 0.70), M1 for solving the pair, A1 for both answers. The signs had to be consistent; a negative σ scored A0.
• (b) Expected number of 600 leaves within 0.6 cm of the mean (4 marks): for N(4.63, 0.55²), P(4.03 < X < 5.23) = P(−1.091 < Z < 1.091) = 2 × 0.8623 − 1 = 0.7246. Then 0.7246 × 600 = 434.76, so about 434 leaves (435 also accepted). M1 for standardising with 0.55, M1 for the correct area, A1 for 0.725, then B1FT for one whole-number answer from your probability. The last mark needed a single whole number without “about” or “approximately”; writing that 434.76 is about 434, so 434 leaves, was fine.

Where the marks are on 9709/52

• Permutations and combinations carried 10 of the 50 marks: Question 1 (6) and Question 3 (4).
• The normal distribution carried 9 marks (Question 7).
• Discrete random variables and the geometric and binomial distributions carried 8 marks (Question 2).
• The binomial distribution and its normal approximation carried 8 marks (Question 4).
• Representing and summarising data carried 8 marks (Question 6).
• Conditional probability carried 7 marks (Question 5).

In all, 14 of the 50 marks used the normal distribution in some form (Questions 4(b) and 7).

How to self-mark 9709/52 with the mark scheme

1. Check that your working is written down. The cover says unsupported calculator answers get no marks. Several parts (1(b), 2(b), 2(c), 6(b), 7(a), 7(b)) gave only a single special-case B1 for a correct answer without the method.
2. Award M marks for structure. In Question 1(a) and Question 5(b), a sum of products of the right form earns the first M1 even with wrong numbers. In Question 2(c), the first M1 needed a term with its arrangement factor (5 or 6); the second only needed both terms in the right form.
3. Count the method marks in the normal approximation separately. Question 4(b) had three M marks: standardising, using a continuity correction, and choosing the correct tail. Missing the correction costs one mark, not the part.
4. Check the signs in Question 7(a). Both z-values are negative here because both lengths are below the mean; inconsistent signs lost the A mark.
5. Follow the stem-and-leaf rules exactly. Leaves in order and lined up, no commas, the Hippos on the left, and one key with both teams and the units.
6. Check accuracy. Probabilities to 3 significant figures (0.287, 0.0428, 0.191, 0.232, 0.113, 0.329), with exact fractions accepted where the scheme shows them. Question 1(b) was correct answer only, and Question 7(b) needed one whole number.
7. Log each lost mark as a missing case, a missing arrangement factor, a continuity-correction slip, a wrong tail, or an accuracy slip.

Sit 9709/52 as a timed mock

The paper gives 75 minutes for 50 marks, a minute and a half per mark. A workable plan:

Block	Marks	Time
Question 1 (selections)	6	9 minutes
Question 2 (biased coins)	8	12 minutes
Question 3 (3-digit numbers)	4	6 minutes
Question 4 (binomial and normal)	8	12 minutes
Question 5 (pens)	7	10 minutes
Question 6 (stem-and-leaf)	8	11 minutes
Question 7 (normal distribution)	9	12 minutes
Check, especially cases and rounding	–	3 minutes

Question 6(a) is slow to draw neatly; sort the Zebras’ heights into order on scrap paper first, because the table gives them unsorted.

Other May/June 2026 9709 papers

• 9709 Paper 12 (Pure Mathematics 1)
• 9709 Paper 22 (Pure Mathematics 2)
• 9709 Paper 32 (Pure Mathematics 3)
• 9709 Paper 42 (Mechanics)
• 9709 Paper 62 (Probability & Statistics 2)

Practise and download

• Download 9709/52 and its mark scheme on the Learning Portal →
• All 9709 May/June 2026 papers: overview and marking patterns
• Try a free A Level Maths quiz

Need help with a 9709 topic? Book a free trial lesson with an A Level Maths tutor →

Frequently asked questions

What topics were on 9709 Paper 52 May/June 2026? +

Seven questions: selecting a team and dividing a group, the distribution of heads on three biased coins, counting 3-digit numbers, the binomial distribution and its normal approximation, conditional probability with pens taken from a box, a back-to-back stem-and-leaf diagram, and the normal distribution.

How many marks is 9709/52 and how long is it? +

It is out of 50 marks and lasts 1 hour 15 minutes. All seven questions are compulsory. Question 7 on the normal distribution is the biggest at 9 marks; Questions 2, 4 and 6 are worth 8 marks each.

What was the answer to the normal approximation question in 9709/52? +

0.113. The number of successes in 120 days is B(120, 0.6), approximated by N(72, 28.8). Fewer than 66 means 65 or fewer, so the continuity correction uses 65.5: z = (65.5 − 72) ÷ √28.8 = −1.211, giving 1 − 0.8871 = 0.113.

How do you find μ and σ in 9709/52 Question 7? +

20% below 3.60 gives (3.60 − μ)/σ = −0.842, and 70% above 5.40 means 30% below it, so (5.40 − μ)/σ = −0.524. Subtracting gives 1.8 = 0.318σ, so σ = 5.66 and μ = 8.37.

Do I need a key on the stem-and-leaf diagram in 9709/52? +

Yes. One of the 4 marks was for a single key that names both teams and states the units (cm), for example 2|17|9 meaning 172 cm for the Hippos and 179 cm for the Zebras.

Where can I download 9709/52 May/June 2026 and its mark scheme? +

Both are free to download and print on the Tutopiya Learning Portal, on the Cambridge AS & A Level Mathematics 9709 past papers page.

Ready to Excel in Your Studies?

Get personalised help from Tutopiya's expert tutors. Whether it's IGCSE, IB, A-Levels, or any other curriculum — we match you with the perfect tutor and your first session is free.

Book Your Free Trial