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This capstone lesson brings together every probability skill in the Edexcel GCSE Mathematics (1MA1) specification. It contains eight full exam-style questions — a mix of Foundation (F) and Higher (H) tier — with complete mark-scheme-style solutions, mark allocations, and an examiner's note after each question identifying where students most often lose marks. The lesson finishes with a top-ten tips checklist and a final strategy summary.
Work through each question first on your own, then check the solution carefully. Where your working diverges from the mark scheme, identify whether it was a method error, an arithmetic slip or a presentation issue. Paper-3 and Paper-2 (calculator) questions still reward clean working — method marks often carry the question when the final answer is wrong.
| Paper | Calculator | Typical probability content | Typical marks |
|---|---|---|---|
| Paper 1 | No | Probability scale, simple P(A) = fav/total, complement, listing, small tree diagrams, Venn diagrams | 4–6 |
| Paper 2 | Yes | Two-way tables, expected outcomes, relative frequency, conditional probability, Venn diagrams | 4–7 |
| Paper 3 | Yes | Tree diagrams (with/without replacement), conditional probability, algebraic "show that" proofs | 4–8 |
Across all three papers, expect 8–15 marks on probability overall. Nearly every paper contains at least one tree-diagram or Venn-diagram question and one "expected number" or relative-frequency item.
A bag contains 4 red, 3 blue and 5 green sweets. A sweet is taken at random.
(a) Write down the probability the sweet is blue. (1) (b) Work out the probability the sweet is not green. (2)
Mark-scheme solution:
Total sweets = 4 + 3 + 5 = 12.
(a) P(blue) = 3/12 = 1/4 (1 mark — B1 for any correct equivalent: 3/12, 0.25, 25%)
(b) P(green) = 5/12, so P(not green) = 1 − 5/12 = 7/12. 7/12 (M1 for subtracting from 1; A1 for answer)
Examiner's note: Students lose marks here by writing 3/12 without simplifying (still accepted, but weak) or by trying to count "not green" directly and miscounting the 7. Always use 1 − P(green) when a complement is asked for.
A biased four-sided spinner has sections labelled 1, 2, 3 and 4. The probability distribution is:
| Score | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| Probability | 0.15 | 0.25 | 0.35 | ? |
(a) Work out the probability of a score of 4. (2) (b) The spinner is spun 200 times. Work out the expected number of 3s. (2)
Mark-scheme solution:
(a) 0.15 + 0.25 + 0.35 = 0.75. P(4) = 1 − 0.75 = 0.25 (M1 for subtracting from 1; A1 for answer)
(b) Expected 3s = 0.35 × 200 = 70 (M1 for multiplying; A1 for answer)
Examiner's note: Part (b) uses the probability of 3, not 4. Students frequently pick up the latest number they calculated. Read the subscript carefully.
The two-way table shows information about 100 adults and their transport to work.
| Car | Bus | Walk | Total | |
|---|---|---|---|---|
| Men | 22 | ? | 8 | 50 |
| Women | ? | 14 | 10 | 50 |
| Total | ? | ? | ? | 100 |
(a) Complete the two-way table. (3) (b) An adult is chosen at random. Find the probability the adult is a woman who takes the bus. (1) (c) Find the probability the adult walks to work. (1)
Mark-scheme solution:
(a) Men Bus = 50 − 22 − 8 = 20 Women Car = 50 − 14 − 10 = 26 Total Car = 22 + 26 = 48 Total Bus = 20 + 14 = 34 Total Walk = 8 + 10 = 18 Check: 48 + 34 + 18 = 100 ✓ (B1 for each of: one correct missing row, one correct missing column-total pair, whole table consistent; 3 max)
(b) P(woman ∩ bus) = 14/100 = 7/50 (B1)
(c) P(walk) = 18/100 = 9/50 (B1)
Examiner's note: Students often leave the table partially filled or forget to cross-check totals. The grand total check (100) catches most arithmetic slips before they cost marks.
A coin is tossed 400 times. It lands heads 256 times.
(a) Work out the relative frequency of heads. (1) (b) Do you think the coin is fair? You must justify your answer. (2) (c) The coin is tossed another 1000 times. Estimate how many heads will appear. (1)
Mark-scheme solution:
(a) 256/400 = 0.64 (B1 — accept 256/400, 0.64, 64%, 16/25)
(b) If the coin were fair, P(heads) = 0.5. Getting 0.64 is significantly higher. This gives evidence that the coin is biased towards heads. (M1 for comparison with 0.5 or expected 200; A1 for correct conclusion with justification)
(c) Expected heads = 0.64 × 1000 = 640 (B1)
Examiner's note: For (b) you must do two things: compare to 0.5 (or expected frequency 200) AND draw a clear conclusion. One without the other scores 1 of 2. Cautious phrases like "there is evidence to suggest" are accepted.
60 students are surveyed about whether they study Geography (G) and History (H). The results are:
(a) Draw a Venn diagram to show this information. (3) (b) Find P(G∪H). (1) (c) A student is chosen at random from those who study Geography. Find the probability that they also study History. (2)
Mark-scheme solution:
(a) G only = 34 − 12 = 22. H only = 28 − 12 = 16. Neither = 60 − 22 − 12 − 16 = 10.
| Region | G only | G ∩ H | H only | Neither |
|---|---|---|---|---|
| Number | 22 | 12 | 16 | 10 |
Total: 22 + 12 + 16 + 10 = 60 ✓ (B1 for each region correctly placed, max 3)
(b) P(G∪H) = (22 + 12 + 16)/60 = 50/60 = 5/6 (B1)
(c) Given Geography — restrict to 34 Geography students; 12 of them also study History. P(H∣G)=3412 = 6/17 (M1 for correct denominator of 34; A1 for 6/17 or equivalent)
Examiner's note: The commonest mistake in (c) is dividing by 60, not 34. The phrase "from those who study Geography" is the signal that the denominator is n(G).
A box contains 4 white counters and 6 black counters. Two counters are taken out at random, one after the other, without replacement.
(a) Complete a tree diagram for this information. (2) (b) Work out the probability that both counters are the same colour. (2) (c) Given that the first counter is black, find the probability that the second is also black. (2)
Mark-scheme solution:
(a) First pick: P(W)=104 = 2/5, P(B)=106 = 3/5. Second pick after W: P(W)=93 = 1/3, P(B)=96 = 2/3. Second pick after B: P(W)=94, P(B)=95. (B1 for first-stage probabilities; B1 for both second-stage pairs correct)
(b) P(same colour) = P(WW) + P(BB) = (2/5)(1/3) + (3/5)(5/9) = 2/15 + 15/45 = 6/45 + 15/45 = 21/45 = 7/15 (M1 for sum of both "same" probabilities; A1 for 7/15 or equivalent)
(c) This is read directly from the second set of tree branches after B: P(B2∣B1)=95 (M1 for identifying the "after B" branch; A1 for 5/9)
Examiner's note: Students confuse (c) with P(both black) = (3/5)(5/9) = 1/3. Given the first is black, we do not multiply by P(first black). The 5/9 is the conditional probability all on its own.
A bag contains n counters. 7 of them are yellow and the rest are purple. Two counters are drawn at random without replacement. The probability that both are yellow is 1/5.
(a) Show that n² − n − 210 = 0. (3) (b) Find the value of n. (2)
Mark-scheme solution:
(a) P(both yellow) = (7/n) × (6/(n − 1)) = 42/(n(n − 1))
Set equal to 1/5: 42/(n(n − 1)) = 1/5
Cross-multiply: 5 × 42 = n(n − 1) 210 = n² − n n² − n − 210 = 0 ✓ (M1 for correct product of probabilities; M1 for cross-multiplication; A1 for the given equation derived)
(b) Factorise: (n − 15)(n + 14) = 0 n = 15 or n = −14. Since n is a positive count: n = 15 (M1 for factorising correctly; A1 for n = 15 with the negative rejected)
Check: (7/15)(6/14) = 42/210 = 1/5 ✓
Examiner's note: "Show that" questions require every algebraic step to be visible. Leaps will lose marks even if the final equation is correct. Always reject the negative root explicitly ("n must be positive, so n = 15").
A company tests 500 batteries. The results are:
| Pass | Fail | Total | |
|---|---|---|---|
| Brand A | 276 | 24 | 300 |
| Brand B | 170 | 30 | 200 |
| Total | 446 | 54 | 500 |
(a) A battery is chosen at random. Find P(Fail). (1) (b) A battery is chosen at random. Find P(Brand A | Fail). (2) (c) Are the events "Brand A" and "Pass" independent? Justify using probability. (3) (d) Two batteries are picked at random without replacement. Find the probability that both are Brand B. (1)
Mark-scheme solution:
(a) P(Fail) = 54/500 = 27/250 (B1 — accept 0.108 or 10.8%)
(b) Given Fail — restrict to the 54 failed batteries, 24 of which are Brand A. P(Brand A | Fail) = 24/54 = 4/9 (M1 for denominator 54; A1 for 4/9)
(c) Test: P(Pass | Brand A) versus P(Pass). P(Pass | Brand A) = 276/300 = 0.92 P(Pass) = 446/500 = 0.892. Since 0.92 ≠ 0.892, the events are not independent. (M1 for calculating P(Pass | Brand A); M1 for calculating P(Pass); A1 for comparing and concluding)
(d) P(both Brand B) = (200/500) × (199/499) = 39800/249500 ≈ 0.1597 (to 4 d.p.) (B1 — accept the unsimplified product or any equivalent decimal)
Examiner's note: For (c), an equivalent valid check is P(Brand A) × P(Pass) vs P(Brand A ∩ Pass): (300/500)(446/500) = 0.5352 but P(Brand A ∩ Pass) = 276/500 = 0.552. They differ, so not independent. Either test scores full marks — but you must present both quantities being compared, not just one.
A warehouse receives a component from Supplier X with probability 0.6 and from Supplier Y with probability 0.4. The probability that a component from Supplier X is defective is 0.02; the probability that a component from Supplier Y is defective is 0.05.
(a) Draw a tree diagram. (2) (b) Find the probability that a randomly chosen component is defective. (2) (c) Of 2000 components received, estimate how many will be defective. (1)
Mark-scheme solution:
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