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.
Probability worksheet — GCSE Foundation
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- 1.A café offers 3 types of sandwich, cheese, ham and egg, and 4 types of drink, tea, coffee, juice and water. A customer chooses one sandwich and one drink at random. Work out the probability that the customer chooses egg and water.
- 2.A bag contains one red counter and one blue counter. Josh takes a counter, notes its colour, puts it back, then takes a counter again. He draws a tree diagram to show every possible pair of colours. Work out how many outcomes are on his tree diagram.
- 3.A biased six-sided dice is rolled once. The probability that it lands on 6 is 0.25. The other five scores are all equally likely. Work out the probability that it lands on 3.
- 4.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.
- 5.A fair coin is flipped and a fair three-sided spinner, labelled X, Y and Z, is spun. Every outcome is listed as a pair, such as (H, X). Work out the probability that the outcome is a head and Z.
- 6.Four pupils each flip the same coin a number of times and record the results. Sam flips it 40 times and gets 24 heads. Priti flips it 25 times and gets 10 heads. Leo flips it 20 times and gets 9 heads. Fatima flips it 15 times and gets 6 heads. Combine all four pupils' results to estimate the probability that the coin lands on heads.
- 7.The probability that a plant grown from a seed packet germinates is 7/10. Work out the probability that it does not germinate.
- 8.A basketball player takes two free throws, and the throws are independent. The probability of scoring on each throw is 0.6, and the probability of missing is 0.4. Work out the probability that she scores exactly one of the two throws.
- 9.150 pupils are asked about swimming and cycling, and the results are recorded on a frequency tree. The first pair of branches splits the 150 pupils into 90 who swim and 60 who do not swim. Of the 90 who swim, 54 also cycle. Of the 60 who do not swim, 18 cycle. Work out P(S ∩ C), the probability that a pupil, chosen at random, swims and cycles. Give your answer as a fraction in its simplest form.
- 10.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.
- 11.A biased spinner is spun 40 times and lands on red 16 times. It is then spun a further 60 times and lands on red 21 times. Work out the best estimate of the probability that the spinner lands on red, using the results of all 100 spins together.
- 12.A box contains 1 white bead, 4 purple beads and 4 red beads. One bead is taken at random from the box. Work out the probability that the bead is white or purple.
- 13.A train company runs 25 trains a day, every day. The probability that any one train is late is 0.08. Work out how many late trains the company should expect over a period of 4 weeks.
- 14.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.
- 15.The probability that a seed fails to germinate is 1/5. Three seeds are planted independently. Work out the probability that at least one of the three seeds germinates.
Answer key
- (d) 1/12 — Method: list the full possibility space of sandwich-and-drink pairs, then divide the one matching pair by the size of the whole space. Working: there are 3 × 4 = 12 equally likely sandwich-and-drink pairs, and exactly one of them is egg and water. Answer: 1/12. Watch out: writing down 1/7 comes from adding the two counts, 3 + 4 = 7, instead of multiplying them to build the possibility space. Writing down 1/3 uses only the chance of choosing egg out of 3 sandwiches and ignores the drink altogether. And writing down 1/4 uses only the chance of choosing water out of 4 drinks and ignores the sandwich altogether.
- (d) 4 — Method: each of the first draw's 2 outcomes can be paired with each of the second draw's 2 outcomes, since the counter is put back before the second draw, so the tree has one branch for every combination. Working: 2 × 2 = 4 outcomes: red-red, red-blue, blue-red, blue-blue. Answer: 4. Watch out: writing down 2 lists only the colours of a single draw and never branches out to a second draw at all. Writing down 3 treats red-then-blue and blue-then-red as the same branch, when the tree diagram shows them as two separate paths, since the counter is put back and either colour could come first or second. And writing down 16 comes from working out 2 × 2 × 2 × 2, as though the counter were drawn four times instead of twice.
- (b) 0.150 — The six scores are exhaustive, so their probabilities sum to 1. The probability of not landing on 6 is 1 − 0.25 = 0.75, and this is shared equally between the other five scores, so each has probability 0.75 ÷ 5 = 0.150. Giving 0.750 as the answer stops after finding the probability of not landing on 6, without sharing it out between the five remaining scores. Dividing 0.75 by 6 instead of by 5 gives 0.125, wrongly including the score of 6 among the equally likely scores. Ignoring the bias completely and dividing 1 by all six scores gives 1 ÷ 6 = 0.167.
- (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) 1/6 — There are 2 outcomes for the coin and 3 outcomes for the spinner, giving 2 × 3 = 6 equally likely pairs: (H,X), (H,Y), (H,Z), (T,X), (T,Y), (T,Z). Only one of these, (H,Z), matches both conditions, so the probability is 1/6. A candidate who answers 1/3 has listed only the spinner's 3 outcomes and ignored the coin. A candidate who answers 1/2 has considered only the coin and ignored the spinner. A candidate who answers 1/5 has missed one pair when listing the possibility space, treating it as 5 outcomes instead of 6.
- (b) 0.49 — Adding all four pupils' flips gives 40 + 25 + 20 + 15 = 100 flips in total, and adding their heads gives 24 + 10 + 9 + 6 = 49 heads in total, so the combined estimate is 49/100 = 0.49. Writing 0.46 is wrong because it averages the four pupils' individual rates (0.60, 0.40, 0.45 and 0.40) as if they all came from the same number of flips, which they do not — this ignores that Sam and Priti flipped far more times than Leo and Fatima. Writing 0.60 is wrong because it only uses Sam's own result (24/40 = 0.60), ignoring the other three pupils. Writing 0.40 is wrong because it only combines Priti and Fatima's results (16/40 = 0.40), leaving out Sam and Leo entirely. The combined estimate from all four pupils is 0.49.
- (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.
- (a) 0.48 — There are two ways to score exactly one throw: scoring on the first and missing the second, 0.6 × 0.4 = 0.24, or missing the first and scoring the second, 0.4 × 0.6 = 0.24. Adding these gives 0.24 + 0.24 = 0.48. Choosing 0.24 comes from working out only one of the two paths and forgetting the other one also gives exactly one score. Choosing 0.36 comes from working out the probability of scoring BOTH throws, 0.6 × 0.6 = 0.36, instead of exactly one. Choosing 0.84 comes from working out the probability of scoring AT LEAST one throw, 1 − 0.4 × 0.4 = 0.84, instead of exactly one.
- (c) 9/25 — The frequency tree already shows the swim-and-cycle branch directly: 54 out of 150, which simplifies to 9/25. Adding both cycling branches together, 54 + 18 = 72, gives the total number of cyclists, so 72/150 = 12/25, not just those who also swim. Using the non-swimmers' cycling figure, 18, gives 18/150 = 3/25, the probability of cycling WITHOUT swimming. Using the swimmers' total of 90, before splitting by cycling, gives 90/150 = 3/5, the probability of swimming on its own.
- (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.
- (c) 0.37 — Method: pool the two runs into one combined set of results, then find the relative frequency of red across all of the spins together. Working: total reds = 16 + 21 = 37. Total spins = 40 + 60 = 100. Relative frequency = 37 ÷ 100 = 0.37. Answer: 0.37. Watch out: writing down 0.40 uses only the first run, 16 ÷ 40, and throws away the extra evidence from the second 60 spins. Writing down 0.35 uses only the second run, 21 ÷ 60, and throws away the first run instead. And writing down 0.375 averages the two runs' separate rates, (0.40 + 0.35) ÷ 2, which treats a run of 40 spins and a run of 60 spins as equally weighted, when pooling the actual counts gives the larger run its fair share of influence.
- (a) 5/9 — Method: a bead cannot be two colours at once, so white and purple are mutually exclusive and their probabilities are added over the common total. Working: there are 1 + 4 + 4 = 9 beads, so P(white) = 1/9 and P(purple) = 4/9; adding gives 1/9 + 4/9, and 1 + 4 = 5 ninths. Answer: 5/9. The distractors: 4/9 comes from giving the probability of a purple bead alone and forgetting to include the white one; 5/8 comes from counting 5 favourable beads but using 8 as the total, leaving the single white bead out of the count of the box; 5/18 comes from adding 1/9 and 4/9 by adding the denominators as well as the numerators.
- (d) 56 — Method: count the trials over the whole period first, then multiply the number of trials by the probability. Working: 4 weeks is 4 × 7 = 28 days, and at 25 trains a day that is 25 × 28 = 700 trains. The expected number of late trains is 700 × 0.08 = 56. Answer: about 56 late trains over the 4 weeks. The distractors: 2 is the expected number for a single day, 25 × 0.08 = 2, with the 28 days never brought in; 14 uses one week instead of four, 25 × 7 × 0.08 = 14; 644 is 700 − 56 and counts the trains expected to be on time.
- (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.
- (d) 124/125 — The probability that a seed germinates is 1 − 1/5 = 4/5, so the probability that all three seeds fail to germinate is 1/5 × 1/5 × 1/5 = 1/125. The probability that at least one germinates is 1 − 1/125 = 124/125. Choosing 4/5 comes from giving the probability that a single seed germinates, forgetting to combine all three seeds. Choosing 64/125 comes from working out the probability that ALL three seeds germinate, 4/5 × 4/5 × 4/5 = 64/125, instead of at least one. Choosing 12/125 comes from working out the probability that EXACTLY one seed germinates, 3 × 4/5 × 1/5 × 1/5 = 12/125, instead of at least one.
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