Printable · GCSE Higher · ages 14-16
Probability worksheet — GCSE Higher
Fifteen questions across the probability statements at Higher tier. Choose the non-calculator filter to rehearse Paper 1, which counts for a third of the marks.
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Answer key: Probability worksheet — GCSE Higher
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- (b) 57/100 — Pooling both trials: total heads = 24 + 33 = 57, total flips = 40 + 60 = 100, so the combined relative frequency is 57/100, which is already in its simplest form since 57 and 100 share no common factor. Averaging the two separate relative frequencies instead, (24/40 + 33/60) ÷ 2 = (0.6 + 0.55) ÷ 2 = 0.575 = 23/40, treats the two trials as equally weighted even though Ben made more flips, which is not correct. Using only Leah's data gives 24/40 = 3/5. Using only Ben's data gives 33/60 = 11/20.
- (b) 1/10 — Method: list every possible pair of subjects systematically, so that no pair is missed and no pair is counted twice, then compare the number of successful pairs with the size of the list. Working: pairing maths with each of the other four gives 4 pairs, physics with each subject after it gives 3, biology gives 2 and chemistry gives 1, so there are 4 + 3 + 2 + 1 = 10 pairs. Exactly one of them is maths with biology. Answer: the probability is 1/10. The distractors: 1/5 comes from counting maths-then-biology and biology-then-maths as two separate successes while still dividing by the 10 unordered pairs; 1/15 comes from a list that also pairs each subject with itself, giving 15 entries instead of 10; 1/4 comes from working out the second step alone, that 1 of the 4 subjects left after maths is biology, without allowing for the chance that maths is picked at all.
- (d) 1/3 — List the outcomes for the two spinners systematically in a 3 × 3 grid: 1-1, 1-2, 1-3, 2-1, 2-2, 2-3, 3-1, 3-2, 3-3, where the first number is the score on spinner A and the second is the score on spinner B — 9 equally likely outcomes in total. The pairs where the two numbers are the same are 1-1, 2-2 and 3-3, so there are 3 favourable outcomes. P(same number) = 3/9 = 1/3. 2/3 comes from working out the probability that the two numbers are different and then forgetting to take the complement the right way round, so the probability of "different" is given instead of the probability of "same". 1/2 comes from listing only the 6 unordered pairs 1-1, 2-2, 3-3, 1-2, 1-3, 2-3 instead of all 9 ordered outcomes in the grid, then taking 3 out of that 6. 1/9 comes from spotting only one of the three matching pairs, such as 1-1, and missing 2-2 and 3-3.
- (d) 5/18 — Method: list every ordered pair of dice scores whose difference is 1, then divide by the 36 equally likely pairs. Working: the pairs with a difference of 1 are (1, 2), (2, 1), (2, 3), (3, 2), (3, 4), (4, 3), (4, 5), (5, 4), (5, 6) and (6, 5), which is 10 pairs out of 36, cancelling down to 5/18. Answer: 5/18. Watch out: writing down 5/36 lists only the 5 pairs going up, (1, 2), (2, 3), (3, 4), (4, 5) and (5, 6), and misses that each one has a matching pair the other way round, such as (2, 1) — the two dice are different objects, and order matters, so each of those 5 gaps counts twice. Writing down 1/6 treats the six possible differences, 0 to 5, as equally likely and picks 1 out of 6 of them, but a difference of 1 is reached by far more pairs of scores than a difference of 5 is, so the six differences are not equally likely. And writing down 1/4 comes from adding the two dice's outcome counts instead of multiplying them, 6 + 6 = 12, and grouping the six scores into just three non-overlapping pairs one apart, {1, 2}, {3, 4} and {5, 6}, giving 3 out of that wrong pool of 12.
- (b) (48/52) × (47/51) — Method: for two deals one after the other with nothing put back, multiply the probability of the first by the probability of the second worked out from the cards that are left. Working: 52 − 4 = 48 cards are not aces, so the first card is not an ace with probability 48/52. One card has now gone and it was not an ace, so 51 cards remain and 47 of them are not aces, giving 47/51. Answer: the probability is (48/52) × (47/51). The distractors: (48/52) × (48/52) comes from leaving the pack at 52 cards for the second deal, which is only true if the first card is replaced; (4/52) × (3/51) comes from working out the probability that both cards ARE aces instead of neither; (4/52) × (4/51) comes from the same misreading with the ace count left at 4 while the total is reduced, adjusting only half of the second fraction.
- (d) 7/18 — Method: the pupil picked is known to study French, so the sample space shrinks to the 18 French students; divide the number who study both languages by 18. Working: 7 of the pupils study both French and German, and all 7 of them are among the 18 French students, so the probability is 7/18, which will not cancel. Answer: the probability is 7/18. The distractors: 7/30 divides by the whole class, keeping the restricted numerator but the full denominator; 1/2 is 7/14, which conditions on the German students instead, answering the probability that a German student also studies French; 7/25 uses 18 + 14 minus 7 = 25, the number who study at least one language, which is a larger group than the one the question restricts you to.
- (c) 0.512 — Method: independent events that must all happen are combined by multiplying their probabilities. Working: the first two questions give 0.8 × 0.8 = 0.64. Bringing in the third gives 0.64 × 0.8, and since 64 × 8 = 512 with three decimal places in the product, this is 0.512. Answer: the probability is 0.512. The distractors: 0.64 comes from multiplying only two of the three probabilities and stopping; 0.8 comes from reading 'independent' as meaning the probability never changes and writing down the single-question figure; 0.0512 comes from a place-value slip in the last multiplication, counting four decimal places instead of three.
- (d) 4 — Method: list the elements of each set in full, then find which elements appear in both lists — that is A ∩ B. Working: factors of 12 = {1, 2, 3, 4, 6, 12}. Factors of 18 = {1, 2, 3, 6, 9, 18}. The elements in both lists are 1, 2, 3 and 6, so A ∩ B = {1, 2, 3, 6} and n(A ∩ B) = 4. Answer: 4. Watch out: writing down 6 gives n(A), the size of the factors-of-12 list on its own, not the size of the overlap. Writing down 8 comes from counting every element that appears in EITHER list, 1, 2, 3, 4, 6, 9, 12 and 18 — that is the union, a different set from the intersection. And writing down 3 misses that 1 is a factor of both 12 and 18, and so belongs in A ∩ B alongside 2, 3 and 6.
- (d) 1/8 — The multiples of 3 from 1 to 8 are 3 and 6, so the probability of that event is 2/8, which simplifies to 1/4. The probability of the coin landing on heads is 1/2. Since the spin and the toss are independent, multiply the two probabilities: 1/4 × 1/2 = 1/8. A candidate who answers 1/4 has considered only the spinner and forgotten to combine it with the coin toss. A candidate who answers 1/2 has considered only the coin and forgotten the spinner condition entirely. A candidate who answers 1/16 has counted only one number, 6, as a multiple of 3 instead of two, giving 1/8 × 1/2.
- (d) 0.80 — Evens is exactly halfway along the scale, at 0.5, and 3/10 written as a decimal is 0.3. A probability 3/10 higher than evens is 0.5 + 0.3 = 0.80. Giving 0.30 as the answer converts the increase but never adds it to the value of evens. Adding the increase to 1, the 'certain' end of the scale, instead of to evens gives 1 + 0.3 = 1.30, which cannot be a probability. Writing 3/10 as 0.03 instead of 0.3, a place-value slip, gives 0.5 + 0.03 = 0.53.
- (d) 5/11 — Method: the pen already taken was black, so update the contents of the box before working out the second probability. Working: the box held 12 pens and one black pen has gone, so 11 pens remain. None of the blue pens has been taken, so all 5 are still there, and the probability is 5/11, which will not cancel. Answer: the probability is 5/11. The distractors: 5/12 uses the box as it was at the start, which is only correct if the first pen is put back; 4/11 takes one off the blue count as well as the total, as though the pen removed had been blue; 6/11 gives the probability that the second pen is black, carrying on with the colour of the first pen instead of the colour asked for.
- (a) 195 — The relative frequency from the trial is 52 ÷ 80 = 0.65, and the expected number of point-up landings in 300 drops is 0.65 × 300 = 195. Giving 52 as the answer reuses the original count from the 80-drop trial without scaling it up to 300 drops at all. Misreading 52 out of 80 as 52% and finding 52% of 300 gives 156. Finding the expected number of point-DOWN landings instead of point-up, using the relative frequency 28 ÷ 80 = 0.35, gives 0.35 × 300 = 105.
- (a) 0.09 — Method: two independent events that must both happen are combined by multiplying their probabilities. Working: the same probability 0.3 applies to each day, so the calculation is 0.3 × 0.3. Written as fractions this is 3/10 × 3/10 = 9/100. Answer: the probability is 0.09. The distractors: 0.6 comes from adding 0.3 and 0.3 instead of multiplying them; 0.3 comes from quoting the single-day probability, as though the second day added no further condition; 0.9 comes from multiplying 3 by 3 correctly but keeping only one decimal place in the product instead of two.
- (a) 50 — To find the number of shots needed for an expected 12 hits, divide the number of hits wanted by the probability of a hit: 12 ÷ 0.24 = 50. Multiplying the number of hits by the probability instead of dividing gives 12 × 0.24 = 2.88, which rounds to 3 shots. Rounding 0.24 to 0.25 before dividing gives 12 ÷ 0.25 = 48. Using the probability of missing, 1 − 0.24 = 0.76, instead of the probability of hitting, gives 12 ÷ 0.76 = 15.79, which rounds to 16.
- (b) 5/12 — There are 36 equally likely ordered pairs. Listing the pairs where the red score is smaller than the blue score gives 15 outcomes, so the probability is 15/36 = 5/12. Choosing 1/2 comes from assuming the 36 outcomes split evenly between 'red smaller' and 'red bigger', which ignores the 6 outcomes where the two dice show the same score. Choosing 7/12 comes from including the outcomes where the two scores are equal, 15 + 6 = 21, giving 21/36 = 7/12, instead of using 'less than' strictly. Choosing 5/36 comes from counting only the outcomes where the two scores differ by exactly 1, such as (1, 2) and (2, 3), which is 5 outcomes, and missing the pairs that differ by more than 1.
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