Probability: Exam Practice

This lesson pulls the whole OCR GCSE Mathematics (J560) probability strand together into exam-style practice. Below are twelve numbered questions spanning the strand, from Foundation grades $1$1–$5$5 to Higher grades $4$4–$9$9. Each question states its marks and tier, and each is followed by a full worked solution with notes on where the method marks (M) and accuracy marks (A) fall. Use these as you would a real paper: cover the solutions, attempt each question first under timed conditions, and only then check your method line by line against the commentary that follows.

The questions exercise all three Assessment Objectives — AO1 (accurate technique), AO2 (reasoning and justification) and AO3 (multi-step problem solving) — and use OCR command words throughout: Work out, Calculate, Show that, Give a reason for your answer, Write down, Find and Complete.

Before you start, a word on how marks are awarded, because it shapes how you should write your answers. Most longer probability questions carry a mix of method marks (M) and accuracy marks (A). A method mark rewards setting up the correct calculation — writing $P \times n$P×n, multiplying along the right tree branches, or forming the equation "(sum of probabilities) $= 1$=1" — even if you then make an arithmetic slip. An accuracy mark rewards the correct final value. The practical lesson is to show every step: a wrong final answer with correct method can still earn most of the marks, whereas a bare answer with no working earns nothing if it is wrong. Throughout the solutions below, the points where method and accuracy marks fall are flagged so you can see exactly where the credit lies.

Pay attention, too, to the command words, because they tell you what the examiner expects. Write down means no working is needed — just the answer. Work out and Calculate expect a method. Show that means you must display every step that leads to the stated result, never skipping to it. Give a reason for your answer needs a precise written justification, not just a number. And Complete (a tree, table or diagram) is marked on the diagram itself, so fill in every region or branch. Matching your response to the command word is one of the easiest ways to avoid dropping marks you have actually earned.

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Exam-Style Questions

Question 1 — Foundation (Grade 1–2), 2 marks

A bag contains $3$3 red, $4$4 blue and $5$5 green counters. A counter is taken at random. Work out the probability that it is blue.

Question 2 — Foundation (Grade 2–3), 3 marks

A spinner has $5$5 equal sections coloured red, red, blue, green and green. (a) Write down the probability of spinning green. (b) Work out the probability of not spinning red.

Question 3 — Foundation (Grade 3), 3 marks

A fair six-sided dice is rolled $180$180 times. Work out the expected number of times it lands on a number greater than $4$4.

Question 4 — Foundation (Grade 3–4), 4 marks

A spinner can land on red, blue or green. $P(\text{red}) = 0.45$P(red)=0.45 and $P(\text{blue}) = 0.2$P(blue)=0.2. (a) Work out $P(\text{green})$P(green). (b) The spinner is spun $300$300 times. Work out the expected number of greens.

Question 5 — Foundation (Grade 4), 4 marks

A drawing pin is dropped $200$200 times and lands point-up $130$130 times. (a) Work out the relative frequency of landing point-up. (b) The pin is dropped $50$50 more times. Estimate the number of point-up landings. (c) Give a reason why your answer to (b) is only an estimate.

Question 6 — Foundation (Grade 4–5), 4 marks

$60$60 people were asked if they own a car, then if they have a driving licence.

• $42$42 own a car.
• All car owners have a licence.
• Of those who do not own a car, $9$9 have a licence.

Complete a frequency tree and work out how many people have a licence in total.

Question 7 — Foundation/Higher (Grade 5), 5 marks

A bag contains $4$4 red and $6$6 yellow counters. A counter is taken at random, its colour noted, and replaced. A second counter is taken. (a) Work out the probability that both counters are red. (b) Work out the probability that the counters are different colours.

Question 8 — Higher (Grade 5–6), 4 marks

$\xi = 50$ξ=50 students. $30$30 study History ($H$H), $22$22 study Geography ($G$G) and $12$12 study both. (a) Work out the number who study neither. (b) Work out $P(H \cup G)$P(H∪G).

Question 9 — Higher (Grade 6), 5 marks [H]

A bag contains $5$5 green and $3$3 blue counters. Two counters are taken at random without replacement. (a) Work out the probability that both are green. (b) Work out the probability that exactly one is green.

Question 10 — Higher (Grade 6–7), 4 marks

A biased four-sided spinner has $P(1) = 0.3$P(1)=0.3, $P(2) = k$P(2)=k, $P(3) = 0.2$P(3)=0.2 and $P(4) = 2k$P(4)=2k. Work out the exact value of $k$k, giving your answer as a fraction.

Question 11 — Higher (Grade 7), 5 marks [H]

$200$200 people were tested for allergy to a food. $50$50 are actually allergic; of these, $45$45 test positive. Of those not allergic, $30$30 test positive. (a) Complete a frequency tree. (b) A person tests positive. Work out the probability they are actually allergic.

Question 12 — Higher (Grade 8–9), 5 marks [H]

A bag contains $n$n counters, of which $4$4 are white. Two counters are taken without replacement. The probability that both are white is $\dfrac{2}{15}$152​. Show that $n^2 - n - 90 = 0$n2−n−90=0 and find $n$n.

Question 13 — Higher (Grade 7), 5 marks [H]

$\xi = 80$ξ=80 people. $46$46 read fiction ($F$F), $38$38 read non-fiction ($N$N) and $9$9 read neither. (a) Work out the number who read both. (b) A person is chosen at random. Work out $P(F \cap N)$P(F∩N). (c) A person who reads fiction is chosen. Work out the probability they also read non-fiction.

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Worked Solutions

Solution 1

Total $= 3 + 4 + 5 = 12$=3+4+5=12 (M1 for the total). $P(\text{blue}) = \dfrac{4}{12} = \dfrac{1}{3} \quad \textbf{(A1)}$P(blue)=124​=31​(A1)

Solution 2

(a) Two of the five sections are green: $P(\text{green}) = \dfrac{2}{5}$P(green)=52​ (B1). (b) Red sections $= 2$=2, so $P(\text{red}) = \dfrac{2}{5}$P(red)=52​; then $P(\text{not red}) = 1 - \dfrac{2}{5} = \dfrac{3}{5}$P(not red)=1−52​=53​ (M1 for $1 - P(\text{red})$1−P(red), A1).

Solution 3

Numbers greater than $4$4 are $\{5, 6\}${5,6}, so $P = \dfrac{2}{6} = \dfrac{1}{3}$P=62​=31​ (M1). $\text{expected} = \dfrac{1}{3} \times 180 = 60 \quad \textbf{(M1 for } P \times n\textbf{, A1)}$expected=31​×180=60(M1 for P×n, A1)

Solution 4

(a) $P(\text{green}) = 1 - (0.45 + 0.2) = 0.35$P(green)=1−(0.45+0.2)=0.35 (M1 for sum-to-$1$1, A1). (b) Expected greens $= 0.35 \times 300 = 105$=0.35×300=105 (M1 for $P \times n$P×n, A1).

Solution 5

(a) Relative frequency $= \dfrac{130}{200} = 0.65$=200130​=0.65 (M1, A1). (b) Expected point-up $= 0.65 \times 50 = 32.5$=0.65×50=32.5, so about $32$32 or $33$33 (M1, A1). (c) Relative frequency only estimates the true probability, and a different set of trials would give a slightly different value (B1).

Solution 6

Do not own a car $= 60 - 42 = 18$=60−42=18 (M1). Car owners with a licence $= 42$=42 (all of them); non-owners with a licence $= 9$=9, without $= 18 - 9 = 9$=18−9=9.

graph LR
    S["60 people"] -->|"owns a car"| C["42 own"]
    S -->|"no car"| NC["18 do not"]
    C -->|"licence"| CL["42 licence"]
    C -->|"no licence"| CX["0 no licence"]
    NC -->|"licence"| NL["9 licence"]
    NC -->|"no licence"| NX["9 no licence"]

Total with a licence $= 42 + 9 = 51$=42+9=51 (A1 for tree, A1 for $51$51).

Solution 7

With replacement, $P(\text{red}) = \dfrac{4}{10} = \dfrac{2}{5}$P(red)=104​=52​ each draw. (a) $P(\text{R,R}) = \dfrac{2}{5} \times \dfrac{2}{5} = \dfrac{4}{25}$P(R,R)=52​×52​=254​ (M1 for multiplying same probabilities, A1). (b) Different colours $=$= red–yellow $+$+ yellow–red: $\left(\dfrac{2}{5} \times \dfrac{3}{5}\right) + \left(\dfrac{3}{5} \times \dfrac{2}{5}\right) = \dfrac{6}{25} + \dfrac{6}{25} = \dfrac{12}{25} \quad \textbf{(M1, M1, A1)}$(52​×53​)+(53​×52​)=256​+256​=2512​(M1, M1, A1)

Solution 8

(a) History only $= 30 - 12 = 18$=30−12=18; Geography only $= 22 - 12 = 10$=22−12=10; both $= 12$=12. Neither $= 50 - (18 + 12 + 10) = 10$=50−(18+12+10)=10 (M1 for intersection working, A1). (b) $H \cup G = 18 + 12 + 10 = 40$H∪G=18+12+10=40, so $P(H \cup G) = \dfrac{40}{50} = \dfrac{4}{5}$P(H∪G)=5040​=54​ (M1, A1).