Printable · GCSE Foundation · ages 14-16
Probability worksheet — GCSE Foundation
Fifteen questions across the probability statements at Foundation tier. Choose the non-calculator filter to rehearse Paper 1, which counts for a third of the marks.
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Probability worksheet — GCSE Foundation
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- 1.The probability that a machine completes a task successfully is 0.85. Work out the probability that the machine does not complete the task successfully.
- 2.Spinner A is numbered 1 to 3 and Spinner B is numbered 1 to 5. Both spinners are spun once, and every outcome is listed as a pair, one number from each spinner. Work out how many equally likely outcomes there are in total.
- 3.A two-way table records how 130 pupils travel to school. 70 of the pupils are girls and the rest are boys. 42 of the girls walk to school and the rest of the girls cycle. 33 of the boys walk to school. Work out what fraction of the girls walk to school.
- 4.Four students each drop a drawing pin repeatedly to estimate the probability that it lands point-up. Freya makes 20 drops, Malik makes 50 drops, Priya makes 200 drops and Tom makes 80 drops. Which student's relative frequency is likely to give the best estimate of the true probability?
- 5.A fair coin is flipped again and again. After the first 10 flips there have been 7 heads. After 1000 flips there have been 528 heads. Which statement best describes what these results show?
- 6.Ellie drops a bottle top and records whether it lands open end up. In her first 20 drops it lands open end up 13 times. She carries on, and after 200 drops in total it has landed open end up 84 times. Work out the best estimate of the probability that the bottle top lands open end up.
- 7.A pack contains 6 cards, numbered 1 to 6. A card is drawn at random from the pack, and a fair coin is flipped. Work out the probability of drawing an even-numbered card and the coin landing on tails.
- 8.Oliver flips three fair coins at the same time. Work out the probability that exactly two of the three coins land on heads.
- 9.120 people took a driving theory test. A frequency tree records the results. The first pair of branches splits the 120 people into 70 men and 50 women. On the men's branch, 45 passed and the rest failed. On the women's branch, 38 passed and the rest failed. Work out how many of the 120 people failed the test.
- 10.A school surveys 60 pupils. 32 say they walk to school, 24 say they cycle to school, and 12 say they do both, on different days. The rest travel only by car. One of the 60 pupils is chosen at random. Work out the probability that this pupil travels only by car.
- 11.In Manchester the probability of rain on any day in November is taken to be 0.3, and whether it rains on one day is independent of whether it rains on the next. Work out the probability that it rains on both the 10th and the 11th of November.
- 12.The probability that a spinner lands on blue is 2/5. Write this probability as a percentage.
- 13.A fair spinner has 8 equal sections: 5 are red and 3 are blue. Freya spins it twice, and each spin is independent of the other. Work out the probability that she gets one red and one blue in the two spins, in either order.
- 14.A spinner has three equal sections, coloured red, blue and green. It is spun twice, and each outcome is listed as an ordered pair of colours, such as (red, blue). Work out how many outcomes are in the full list.
- 15.A bag contains counters that are red, blue or green only. The probability that a counter taken at random is not red is 0.8. The bag contains 25 counters in total. Work out how many of the counters are red.
Answer key
- (b) 0.15 — Method: success and failure are the only two outcomes of the task, so they form an exhaustive set and their probabilities add to 1; subtract the given probability from 1. Working: writing 1 as 1.00 so that both numbers have two decimal places gives 1.00 − 0.85; exchanging once, the hundredths give 10 − 5 = 5 and the tenths, now 9, give 9 − 8 = 1. Answer: 0.15, close to the left-hand end of the 0 to 1 scale because the machine nearly always succeeds. The distractors: 0.85 comes from giving back the probability of success instead of the probability of failure; 0.25 comes from taking each decimal column from 10 on its own in 1.00 − 0.85, writing 10 − 5 = 5 in the hundredths and 10 − 8 = 2 in the tenths instead of reducing the tenths to 9 after the exchange; 0.5 comes from assuming that success and failure must be equally likely because there are only two outcomes.
- (d) 15 — Method: each of Spinner A's outcomes can be paired with each of Spinner B's outcomes, so the two counts are combined by multiplying, not adding. Working: Spinner A has 3 outcomes and Spinner B has 5, so there are 3 × 5 = 15 equally likely outcomes in total. Answer: 15. Watch out: adding the two counts, 3 + 5 = 8, undercounts drastically — every one of Spinner A's 3 outcomes pairs with all 5 of Spinner B's outcomes, not just one each. Writing down 3 counts Spinner A alone and leaves Spinner B out completely. And writing down 25 comes from squaring Spinner B's own outcome count, 5 × 5, and forgetting Spinner A altogether.
- (b) 3/5 — The question asks about the girls only, so use the girls' total of 70 as the denominator: 42 out of 70 girls walk, giving 42/70 = 3/5. Choosing 21/65 comes from using the whole survey of 130 pupils as the denominator instead of just the 70 girls, 42/130 = 21/65. Choosing 33/70 comes from using the boys' walking count, 33, over the girls' total of 70, mixing up the two rows of the table. Choosing 2/5 comes from using the number of girls who CYCLE, 70 − 42 = 28, instead of the number who walk, giving 28/70 = 2/5.
- (b) Priya — The larger the number of trials, the closer a relative frequency tends to be to the true probability. Priya made 200 drops, more than Freya's 20, Malik's 50 or Tom's 80, so her relative frequency gives the best estimate. Freya's estimate is based on only 20 drops, the smallest sample, so it is the least reliable of the four. Malik's 50 drops and Tom's 80 drops are both larger than Freya's but still well short of Priya's 200.
- (b) The relative frequency is settling near 0.5 — Method: turn each result into a relative frequency before comparing them, because it is the relative frequency, and not the difference between the two counts, that tends towards the theoretical probability. Working: after 10 flips the relative frequency of a head is 7 ÷ 10 = 0.7, which is a long way from 0.5. After 1000 flips it is 528 ÷ 1000 = 0.528, which is much closer to 0.5. Meanwhile the gap between the two counts has grown rather than shrunk: it was 7 − 3 = 4 after 10 flips and is 528 − 472 = 56 after 1000 flips. Answer: the relative frequency is settling near 0.5, which is what an unbiased experiment does as the sample grows. The distractors: saying the counts are levelling out is the usual form of this idea and the figures contradict it, since the gap went from 4 to 56; saying the coin is biased treats 28 extra heads in 1000 flips as proof, when 0.528 sits close to 0.5 and a fair coin gives results like this often; saying the next flip is more likely to be a tail is the gambler's fallacy, since each flip stays at 1/2 whatever came before.
- (d) 0.420 — Method: use the relative frequency worked out from the larger number of trials as the best estimate of the probability, since a bigger sample tends to sit closer to the true probability. Working: 84 out of 200 drops land open end up, so the relative frequency from the total is 84 ÷ 200 = 0.420. Answer: 0.420. Watch out: writing down 0.650 comes from 13 ÷ 20, using only the first, much smaller sample instead of the total. Writing down 0.535 comes from averaging 0.650 and 0.420, treating the 20-drop run and the 200-drop run as equally reliable instead of using the larger sample on its own. And writing down 0.580 comes from 1 − 0.420, working out the probability that the bottle top lands the other way up instead of open end up.
- (b) 1/4 — There are 6 × 2 = 12 equally likely outcomes. The even-numbered cards are 2, 4 and 6, so there are 3 × 1 = 3 outcomes with an even card and tails, giving a probability of 3/12 = 1/4. Choosing 1/2 comes from working out only the probability of drawing an even card, 3/6, and forgetting to combine it with the coin landing on tails. Choosing 1/12 comes from treating only one specific outcome, such as card 6 with tails, as the only one that counts, instead of all three even cards paired with tails. Choosing 1/8 comes from doubling the coin stage when counting the total, using 6 × 2 × 2 = 24 outcomes instead of 6 × 2 = 12, and giving 3/24 = 1/8.
- (c) 3/8 — Method: write out every result of the three coins as a string of three letters, H for heads and T for tails, count the results that match the description and divide by how many results the list holds. Working: each coin lands two ways and no coin affects another, so the list holds 2 × 2 × 2 = 8 equally likely results. Exactly two heads means one coin lands on tails and the other two on heads, so the results are HHT, HTH and THH — 3 of the 8. Answer: the probability is 3/8. The distractors: 4/8 comes from reading 'exactly two heads' as 'at least two heads' and counting HHH as well; 2/8 comes from a list made without a system, in which HHT and THH are written down and HTH, the result with the tail between the two heads, is missed; 6/8 comes from counting 3 × 2 = 6 ways of picking which two of the three coins show heads, which counts every pair of coins twice, once in each order.
- (b) 37 — Method: on a frequency tree each pair of branches adds back up to the number it came from, so work out the failures on each branch and then add the two end branches. Working: on the men's branch 70 − 45 = 25 men failed. On the women's branch 50 − 38 = 12 women failed. Adding the two failing end branches gives 25 + 12 = 37. Answer: 37 of the 120 people failed the test. The distractors: 83 comes from adding the two passing branches, 45 + 38 = 83, and so reads the tree for the wrong outcome; 25 is the men's failing branch on its own, with the women never added; 12 is the women's failing branch on its own, with the men never added.
- (a) 4/15 — Method: first find how many pupils travel only by car, then write that as a fraction of the 60 pupils surveyed. Working: pupils who walk or cycle or both = 32 + 24 − 12 = 44. Only by car = 60 − 44 = 16. P(only by car) = 16/60 = 4/15. Answer: 4/15. Watch out: writing down 1/5 takes the overlap of 12 pupils on its own, 12/60, mistaking the group who do both for the group who travel only by car. Writing down 4/5 comes from 60 − 12 = 48, subtracting only the overlap from the total instead of the whole walk-or-cycle count, so cyclists and walkers who are not in the overlap are wrongly swept into the only-car group. And writing down 7/15 comes from 60 − 32 = 28, subtracting the walkers alone and forgetting the cyclists altogether.
- (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.
- (b) 40% — Method: a fraction becomes a percentage by scaling it so that its denominator is 100, or equivalently by dividing the numerator by the denominator and multiplying the decimal by 100. Working: 5 × 20 = 100, so the numerator is scaled by 20 as well, 2 × 20 = 40, giving 40/100; the same figure comes from 2 ÷ 5 = 0.4 and 0.4 × 100 = 40. Answer: 40%. The distractors: 20% comes from dividing 100 by the denominator alone, 100 ÷ 5 = 20, which is the percentage for 1/5 and never uses the numerator 2; 4% comes from finding the decimal 0.4 correctly and then writing the 4 down as the percentage instead of multiplying by 100; 2% comes from scaling only the denominator up to 100 and leaving the numerator as it was, giving 2/100.
- (b) 15/32 — Method: there are two ways to get one of each colour, red then blue and blue then red. Work out the probability of each path by multiplying, then add the two paths. Working: red then blue is 5/8 × 3/8 = 15/64, and blue then red is 3/8 × 5/8 = 15/64. Adding the two paths gives 30/64. Answer: the probability is 15/32. The distractors: 15/64 comes from working out red then blue only and forgetting that blue then red also gives one of each; 15/28 comes from doubling correctly but reducing the total to 7 for the second spin, which is what happens to a bag when an item is kept out, not to a spinner; 39/64 comes from working from the opposite event and subtracting only the two-red case, 1 − 25/64, leaving the two-blue case inside the answer.
- (d) 9 — Method: each of the first spin's 3 outcomes can be paired with each of the second spin's 3 outcomes, so the two counts are multiplied together. Working: 3 × 3 = 9 ordered pairs. Answer: 9. Watch out: writing down 6 comes either from adding the two spins' counts, 3 + 3, instead of multiplying them, or from listing colour pairs without order, treating (red, blue) as the same entry as (blue, red) — both shortcuts miss outcomes that the ordered list contains. Writing down 3 lists only a single spin's outcomes and never pairs the two spins up at all. And writing down 27 treats the spinner as though it had been spun three times, 3 × 3 × 3, one spin too many.
- (d) 5 — Being red and not being red are exhaustive, so their probabilities sum to 1: the probability of red is 1 − 0.8 = 0.2. The number of red counters is 0.2 × 25 = 5. Using 0.8 directly as the probability of red, without taking the complement, gives 0.8 × 25 = 20 — the number of counters that are NOT red. Sharing the 25 counters equally between the three colours, ignoring the given probability altogether, gives 25 ÷ 3 ≈ 8. Misreading the total as 20 counters instead of 25 gives 0.2 × 20 = 4.
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