GCSE Higher — Probability
This is the probability practice for GCSE Higher, the content area that carries 7.5% of the paper. 9 statements make up probability here. Each has its own page with the specification wording, the common mistakes, and questions. Expect frequency of outcomes, tables and frequency trees, expected outcomes and fairness, relative frequency and the probability scale, probabilities of exhaustive events sum to one, empirical samples and sample size and sets, Venn diagrams and tree diagrams. The Higher-only part of this area is conditional probability. The assessment objectives split 40% AO1 against 60% AO2 and AO3 combined, which is why so many questions ask you to explain or to justify rather than simply to calculate. Most of this area is natural Paper 1 material, so practise it without a calculator before you practise it with one.
- P1 — Frequency of outcomes, tables and frequency trees
- P2 — Expected outcomes and fairness
- P3 — Relative frequency and the probability scale
- P4 — Probabilities of exhaustive events sum to one
- P5 — Empirical samples and sample size
- P6 — Sets, Venn diagrams and tree diagrams
- P7 — Possibility spaces for single and combined events
- P8 — Independent and dependent combined events
- P9 — Conditional probability (Higher only)
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Probability at GCSE Higher: the 9 DfE statements
Every question on this site is filed against one of these statements. Pick one to practise it on its own.
Sample probability questions for GCSE Higher
- Four bags each contain counters, some of which are winning counters. The probability of taking a winning counter is 5/12 from the red bag, 3/8 from the blue bag, 7/24 from the green bag and 1/3 from the yellow bag. Work out which bag has the smallest probability of giving a winning counter.(a)The blue bag (3/8)(b)The red bag (5/12)(c)The green bag (7/24)(d)The yellow bag (1/3)
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The green bag (7/24) — Method: fractions can only be ordered once they share a denominator, so rewrite all four over the lowest common denominator and compare the numerators. Working: the lowest common denominator of 12, 8, 24 and 3 is 24, and scaling gives 5/12 = 10/24, 3/8 = 9/24, 7/24 stays as it is, and 1/3 = 8/24; the numerators are then 10, 9, 7 and 8. Answer: the smallest numerator is 7, so the green bag, with 7/24, is the least likely and sits furthest to the left on the 0 to 1 scale. The distractors: the red bag (5/12) comes from finding the largest of the four probabilities instead of the smallest; the blue bag (3/8) comes from scaling 3/8 by changing only the denominator to 24, which turns it into 3/24 and makes it look the smallest; the yellow bag (1/3) comes from comparing numerators alone and assuming the fraction with the numerator 1 must be the smallest. - The probability that a pupil chosen at random has a nut allergy is 1/10. There are 30 pupils in a class. Work out how many of the 30 pupils would be expected to have a nut allergy.(a)1 pupil(b)3 pupils(c)27 pupils(d)10 pupils
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3 pupils — Method: an expected frequency is the probability multiplied by the number of trials, so multiply the probability by the number of pupils. Working: 30 × 1/10 means finding one tenth of 30, and 30 ÷ 10 = 3. Answer: 3 pupils would be expected to have a nut allergy. The distractors: 27 pupils comes from working out how many are expected NOT to have the allergy, 30 − 3, instead of how many are; 10 pupils comes from reading the 10 in the fraction 1/10 as the number of pupils; 1 pupil comes from reading the numerator of the fraction as the expected number. - Each turn of a game ends in a win, a draw or a loss, and no turn can end in more than one of these. The probability of a win is 1/4 and the probability of a draw is 1/3. Work out the probability of a loss.(a)5/12(b)5/7(c)1/12(d)7/12
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5/12 — Method: a win, a draw and a loss are the only outcomes and no two can happen together, so the three probabilities form an exhaustive set of mutually exclusive events and add to 1; add the two given probabilities, then subtract from 1. Working: 1/4 + 1/3 over the common denominator 12 is 3/12 + 4/12 = 7/12, and 1 − 7/12 = 12/12 − 7/12. Answer: 5/12. The distractors: 7/12 comes from stopping at the probability of a win or a draw and never subtracting from 1; 1/12 comes from subtracting the two given probabilities from each other, 1/3 − 1/4, instead of adding them and taking the total from 1; 5/7 comes from adding 1/4 and 1/3 by adding the numerators and the denominators to get 2/7 and then subtracting that from 1. - An ordinary pack of 52 playing cards contains 13 hearts. One card is taken at random from the pack. Work out the probability that the card is not a heart. Give your answer as a percentage.(a)39%(b)50%(c)25%(d)75%
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75% — Method: count how many cards are not hearts, write that count over the total number of cards, cancel the fraction down and then turn it into a percentage. Working: 52 − 13 = 39 cards are not hearts, so the probability is 39/52; dividing the numerator and the denominator by 13 gives 3/4, and 3/4 = 0.75, so 0.75 × 100 = 75. Answer: 75%, three quarters of the way along the 0 to 1 scale. The distractors: 25% comes from giving the probability that the card is a heart, 13 out of 52, which cancels to 1/4; 50% comes from reading 'not a heart' as 'not a red card' and halving the pack; 39% comes from writing the count of 39 cards straight down as the percentage without comparing it with the 52 cards in the pack. - At a summer fair, the probability of winning at the hoopla stall is 6/10 and the probability of winning at the coconut shy is 2/10. Work out how many times as likely a player is to win at the hoopla stall as at the coconut shy.(a)4 times as likely(b)3 times as likely(c)12 times as likely(d)6 times as likely
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3 times as likely — Method: to say how many times as likely one event is as another, divide the larger probability by the smaller one; subtracting them gives the gap between the two probabilities, not the multiple. Working: both probabilities are counted in tenths, so 6/10 ÷ 2/10 compares 6 tenths with 2 tenths, and 6 ÷ 2 = 3. Answer: winning at the hoopla stall is 3 times as likely, which is why 6/10 sits three times as far along the 0 to 1 scale as 2/10. The distractors: 4 times as likely comes from subtracting the two counts, 6 − 2, instead of dividing them, which measures the gap rather than the multiple; 6 times as likely comes from reading the larger probability's 6 tenths straight off as the multiple without ever comparing it with the 2 tenths at the other stall; 12 times as likely comes from multiplying the two counts, 6 × 2, instead of dividing one by the other.
Frequently asked questions
How much of the GCSE Higher paper is probability?
7.5% of the total marks, and the three papers each carry an equal share of the qualification, so it is spread across all of them rather than concentrated in one.
How many topics are there in probability at GCSE Higher?
9 DfE content statements, coded P1 to P9 within this area. Every question in the bank is tagged to one of them.
Which probability topics are Higher only?
Conditional probability (P9).
Can I use a calculator for probability questions?
Paper 1 is non-calculator; Papers 2 and 3 allow one. Of the 9 statements in this area, 5 are naturally non-calculator and 0 naturally calculator, with the rest workable either way — so practise both.
What kind of marks does probability carry?
The GCSE Higher split is 40% AO1 (use and apply standard techniques), 30% AO2 (reason, interpret and communicate) and 30% AO3 (solve problems in and out of context). Questions in this area are set across all three.
Are these past paper questions?
No. Every question at /gcse-higher is original, written to the DfE content statements and checked before it is published. We do not host past papers or mark schemes.
Do I need an account?
No — you can start practising straight away. Progress is saved automatically in your browser.
Is it free?
Yes. The questions, worked answers and printable worksheets are all free, with no adverts.
Tips for probability at GCSE Higher
- Conditional probability P(A | B) is the probability of A given that B has already happened: P(A | B) = P(A and B) ÷ P(B). Restrict the sample space to the cases where B is true.
- In a two-way table or a Venn diagram, 'given that' tells you which row, column or region to divide by — the denominator is the total of that group, not the whole table.
- For dependent events use a tree diagram: the second-branch probabilities change according to the first outcome. Multiply along a path, add the paths that satisfy the event.
- The complement is your shortcut: P(at least one) = 1 − P(none). It is usually far quicker than adding every route that contains a success.
- P(A | B) and P(B | A) are usually different. Read the question twice to see which one is being asked.
Common mistakes (and how to avoid them)
- Dividing by the whole population instead of by the 'given' group when a question says 'given that'.
- Treating events as independent when there is dependence, such as drawing without replacement.
- Confusing P(A | B) with P(B | A).
- Working out 'at least one' by listing every route instead of using 1 − P(none), and dropping a route.
Worked examples
- P(first red) = 3/5. After a red is removed, 2 red remain out of 4, so P(second red | first red) = 2/4 = ½.
- P(both red) = 3/5 × ½ = 3/10.
- 'Given that she is a girl' restricts the sample space to the 12 girls.
- 3 of those 12 are taller than 170 cm.
- P(taller than 170 | girl) = 3/12 = ¼.
- P(no six on one roll) = 5/6, so P(no six on either) = 5/6 × 5/6 = 25/36.
- P(at least one six) = 1 − 25/36 = 11/36.
Probability lets you put a number on uncertainty, and Higher rewards students who know when to multiply and when to add. The distinction between dependent and independent events, the correct use of conditional probability, and a smart choice of complement are the tools that carry you through multi-stage problems. A tidy tree diagram or two-way table organises every route and its probability; practise dice, cards, counters, with and without replacement, until you recognise the structure of a question at a glance.
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