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.A red dice and a blue dice are rolled. Both dice are ordinary and fair. Work out the probability of getting a 1 on the red dice and a 2 on the blue dice.
- 2.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.
- 3.A bag contains 5 letter tiles spelling the word MATHS, with one letter on each tile. Two tiles are drawn at random, one after the other, without the first tile being put back. Work out the probability that both tiles drawn are consonants.
- 4.A six-sided dice is rolled 12 times. It lands on a six 4 times. Erin says that the dice must be biased. Which of these is the best comment on Erin's statement?
- 5.An archer hits the bullseye with probability 0.24 on any one shot. She wants to know how many shots she must take to expect to hit the bullseye 12 times. Work out how many shots this is.
- 6.A box contains counters that are silver, bronze or copper only. The probability that a counter taken at random is silver is 1/2 and the probability that it is bronze is 1/8. The box contains 64 counters in total. Work out the number of copper counters.
- 7.A grower knows that the probability that one of their seeds germinates is 0.6. The grower wants to expect 300 of the seeds to germinate. Work out how many seeds the grower should plant.
- 8.The probability that Ben catches his bus on time is 0.8. The probability that Freya catches her bus on time is 0.5. The two events are independent. Work out the probability that both Ben and Freya catch their bus on time.
- 9.A spinner lands on one of four colours: amber, black, cyan or damson. The probability it lands on amber is 0.18, on black is 0.22 and on cyan is 0.24. Work out the probability that it lands on amber or damson.
- 10.A market stall sells umbrellas. Over the last 250 days, it rained on 70 of them. Using this as an estimate of the probability of rain, work out how many rainy days would be expected in the next 365 days.
- 11.An ordinary fair dice is rolled twice. Work out the probability of getting at least one 5.
- 12.On the probability scale from 0 to 1, which word best describes an event with probability 0.9?
- 13.A school recorded what 60 students chose for lunch. 35 of the students are in Year 10 and the rest are in Year 11. 20 of the Year 10 students chose a sandwich and 9 of the Year 11 students chose a sandwich. Everyone else chose a hot meal. Work out how many Year 11 students chose a hot meal.
- 14.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.
- 15.The probability that a plant grown from a seed packet germinates is 7/10. Work out the probability that it does not germinate.
Answer key
- (c) 1/36 — Method: the two dice do not affect each other, so the probability that both results happen is the product of the two separate probabilities. Working: a 1 on the red dice has probability 1/6 and a 2 on the blue dice has probability 1/6. Multiplying gives 1 × 1 = 1 on the top and 6 × 6 = 36 on the bottom. Answer: the probability is 1/36. The distractors: 1/18 comes from allowing either order, counting a 2 on the red dice with a 1 on the blue dice as well, although the stem names which dice shows which score; 1/3 comes from adding 1/6 and 1/6 instead of multiplying; 1/6 comes from giving the probability for one dice only and treating the other as making no difference.
- (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.
- (d) 3/5 — The first tile drawn is a consonant with probability 4/5, since 4 of the 5 letters in MATHS are consonants. Since it is not replaced, 3 consonants remain out of the 4 tiles left, so the second tile is a consonant with probability 3/4. Multiplying gives 4/5 × 3/4 = 3/5. Choosing 4/5 comes from only using the probability for the first draw and forgetting the second draw changes once a tile is removed. Choosing 3/4 comes from only using the probability for the second draw on its own. Choosing 16/25 comes from treating the two draws as independent, using 4/5 × 4/5 = 16/25, which would only be correct if the first tile were replaced before the second draw.
- (b) She is wrong; 12 rolls is too few to judge — Method: compare the result with what is expected, then ask whether the experiment is long enough for a difference to mean anything. Working: if the dice were fair the expected number of sixes in 12 rolls is 12 × 1 ÷ 6 = 2, so 4 sixes is 2 above what was expected. But over only 12 rolls a result like this turns up often by chance: the relative frequency here is 4/12, which is 1/3, and over so few trials a relative frequency can sit well away from 1/6 with no bias at all. Answer: Erin is wrong, because 12 rolls is far too few to decide; she should roll the dice many more times and see whether the relative frequency settles near 1/6. The distractors: saying she is right because 4 beats the expected 2 uses the correct expected value but treats any difference as proof, which so short an experiment cannot give; saying a fair dice gives each score twice in 12 rolls treats an expected value as a guaranteed one; saying that 4 sixes in 12 rolls cancels down to 1 in 6 mis-cancels the fraction, because 4/12 is 1/3, which is twice 1/6, so that comment reaches the right verdict from arithmetic that is wrong.
- (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.
- (c) 24 — Silver, bronze and copper cover every counter, so their probabilities sum to 1: P(copper) = 1 − 1/2 − 1/8 = 3/8. Number of copper counters = 3/8 × 64 = 24. Multiplying the silver probability by 64 gives 32, the number of silver counters, not copper. Multiplying the bronze probability by 64 gives 8, the number of bronze counters. Adding the silver and bronze probabilities (1/2 + 1/8 = 5/8) and multiplying by 64 gives 40, the combined number of silver and bronze counters, not the copper count.
- (c) 500 — Method: the expected number of successes is the number of trials multiplied by the probability, so to find the number of trials, divide the expected number by the probability. Working: let n be the number of seeds planted. Then n multiplied by 0.6 must come to 300, so n = 300 ÷ 0.6 = 500. Answer: the grower should plant 500 seeds. The distractors: 180 comes from multiplying instead of dividing, 300 × 0.6 = 180, which answers how many of 300 seeds would germinate; 750 comes from dividing by the probability of not germinating, 300 ÷ 0.4 = 750; 120 comes from multiplying by that same 0.4, 300 × 0.4 = 120.
- (b) 0.4 — The two events are independent, so multiply the two probabilities: 0.8 × 0.5 = 0.4. Choosing 1.3 comes from adding the two probabilities instead of multiplying, 0.8 + 0.5 = 1.3. Choosing 0.3 comes from subtracting the two probabilities, 0.8 − 0.5 = 0.3, instead of multiplying them. Choosing 0.1 comes from using the probability that Ben is NOT on time, 1 − 0.8 = 0.2, and multiplying that by Freya's probability instead, 0.2 × 0.5 = 0.1.
- (a) 0.54 — The four colours are exhaustive, so all four probabilities sum to 1: the probability of damson is 1 − 0.18 − 0.22 − 0.24 = 0.36. Amber and damson cannot both happen on one spin, so the probability of amber or damson is 0.18 + 0.36 = 0.54. Stopping after finding the probability of damson alone, without adding the probability of amber, gives 0.36. Adding the three given probabilities together, 0.18 + 0.22 + 0.24 = 0.64, and treating that total as the answer never finds the probability of damson at all. Subtracting only the probability of amber from 1, 1 − 0.18 = 0.82, ignores black, cyan and damson completely.
- (c) 102 — Method: turn the past record into a relative frequency, then use it as an estimate of the probability of rain and multiply by the number of days being predicted for. Working: relative frequency of rain = 70 ÷ 250 = 0.28. Expected rainy days in 365 days = 365 × 0.28 = 102.2, which rounds to about 102 days. Answer: about 102 days. Watch out: writing down 48 swaps which number is the sample and which is the target, working out 70 ÷ 365 × 250 instead of 70 ÷ 250 × 365. Writing down 70 just repeats the original count of rainy days without scaling it up to the new, longer period at all. And writing down 110 comes from rounding the relative frequency to 0.3 before multiplying, 365 × 0.3 = 109.5, when 70 ÷ 250 is exactly 0.28 and needs no rounding at all.
- (b) 11/36 — Method: 'at least one' is the opposite of 'none at all', so work out the probability of no 5 on either roll and take it away from 1. Working: a roll that is not a 5 has probability 5/6, and the rolls are independent, so no 5 at all has probability 5/6 × 5/6 = 25/36. Taking this from 36/36 leaves 11/36. Answer: the probability is 11/36. The distractors: 25/36 is the probability of no 5 at all, written down without the final subtraction; 12/36 comes from counting the 6 pairs with a 5 on the first roll and the 6 pairs with a 5 on the second and adding them, which counts the pair (5, 5) twice; 30/36 comes from working out 1 − 1/6 as though only one roll were made.
- (b) likely — On the probability scale, 'certain' is reserved for a probability of exactly 1, and 'evens' describes a probability of exactly 0.5. A probability of 0.9 is high but not equal to 1, so the correct word is 'likely'. Choosing 'certain' treats a probability close to 1 as if it were exactly 1, which it is not. Choosing 'evens' misjudges 0.9 as being close to the midpoint of the scale, when it is close to the 'certain' end instead. Choosing 'unlikely' reads the scale the wrong way round, as if a high probability meant a low chance of happening.
- (b) 16 — Method: put the counts into a two-way table, complete the row totals, then subtract along the Year 11 row. Working: there are 60 students altogether and 35 are in Year 10, so the number in Year 11 is 60 − 35 = 25. Of those 25 students, 9 chose a sandwich, so the number who chose a hot meal is 25 − 9 = 16. Answer: 16 Year 11 students chose a hot meal. The distractors: 15 comes from subtracting along the Year 10 row instead, 35 − 20 = 15, which is the number of Year 10 hot meals; 25 is the Year 11 row total, written down before the sandwiches are taken off; 31 comes from working with the sandwich figures for the whole school, 60 − 20 − 9 = 31, which counts the Year 10 hot meals as well.
- (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) 3/10 — Germinating and not germinating are exhaustive, so their probabilities sum to 1: 1 − 7/10 = 3/10. Writing 7/10 again gives the probability that the plant DOES germinate, not its complement. Putting the difference 10 − 7 = 3 over the original numerator instead of the original denominator gives 3/7. Inverting the correct answer, swapping its numerator and denominator, gives 10/3, which is impossible as a probability since it is greater than 1.
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