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.
Non-calculator
Answer key: Probability worksheet — GCSE Foundation
MathsUKwww.geekhero.co.uk
- (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.
- (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.
- (b) 60 — The probability of landing on purple in a single spin is 2/6. The expected number of times it lands on purple in 180 spins is 180 × 2/6 = 60. A candidate who answers 90 has used 3/6 instead of 2/6, miscounting the purple sections as 3. A candidate who answers 30 has used 1/6 instead of 2/6, forgetting one of the two purple sections. A candidate who answers 120 has used the probability of NOT landing on purple, 4/6, by mistake.
- (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.
- (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) 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.
- (b) 41/160 — In total, 27 + 14 = 41 of the 160 employees cycle to work, so the probability is 41/160 (41 and 160 share no common factor, so this is already in its simplest form). Writing 27/160 is wrong because it only counts the full-time cyclists and leaves out the 14 part-time cyclists. Writing 41/90 is wrong because it uses the full-time total (90) as the denominator instead of the whole survey (160). Writing 1/5 is wrong because it only uses the part-time branch, simplifying 14/70 to 1/5 and ignoring the full-time cyclists completely. The probability is 41/160.
- (d) 7/15 — Both socks are the same colour either if both are red or if both are blue. The probability both are red is 4/6 × 3/5 = 12/30. The probability both are blue is 2/6 × 1/5 = 2/30. Adding these gives 12/30 + 2/30 = 14/30 = 7/15. Choosing 2/5 comes from only working out the 'both red' path, 12/30, and forgetting the 'both blue' path also counts. Choosing 5/9 comes from treating the draws as if the first sock were replaced, using 4/6 × 4/6 + 2/6 × 2/6 = 20/36 = 5/9, instead of reducing the totals for the second draw. Choosing 7/18 comes from reducing the number of socks removed but not the number left to choose from, using 4/6 × 3/6 + 2/6 × 1/6 = 14/36 = 7/18, instead of 5 remaining socks for the second draw.
- (c) 0.15 — Method: for two independent events, multiply along the branches of the tree to find the probability of both outcomes happening together. Working: P(red and heads) = P(red) × P(heads) = 0.3 × 0.5 = 0.15. Answer: 0.15. Watch out: adding the two probabilities, 0.3 + 0.5 = 0.8, does not give the probability of both — probabilities along one path of a tree are multiplied, not added. Writing down 0.5 ignores the spinner altogether and gives only the coin's probability. And writing down 0.65 is the probability of red OR heads, which is 0.3 + 0.5 − 0.15 = 0.65, a different question from the one asked here.
- (a) 1/10 — There are 4 × 5 = 20 equally likely outcomes in total. The pairs whose product is 12 are Spinner A showing 3 with Spinner B showing 4, and Spinner A showing 4 with Spinner B showing 3, which is 2 outcomes, giving a probability of 2/20 = 1/10. Choosing 1/20 comes from finding only one of the two pairs, (3, 4), and missing (4, 3) as a separate outcome. Choosing 1/8 comes from using 16 as the total number of outcomes, 4 × 4, forgetting that Spinner B has 5 sections rather than 4. Choosing 1/5 comes from listing the factor pairs of 12 as 2 × 6 and 3 × 4 and counting each one in both orders, (2, 6), (6, 2), (3, 4) and (4, 3), giving 4 outcomes out of 20 without checking that neither spinner has a 6 on it.
- (c) 7/8 — Method: 'at least one head' is the opposite of 'no heads at all', so work out the probability of three tails and take it away from 1. Working: a flip that is not a head has probability 1/2, and the flips are independent, so three tails in a row has probability 1/2 × 1/2 × 1/2 = 1/8. Taking this from 8/8 leaves 7/8. Answer: the probability is 7/8. The distractors: 1/8 is the probability of three tails, written down without the final subtraction; 3/8 is the probability of exactly one head, which comes from reading 'at least one' as 'exactly one'; 1/2 comes from giving the probability of a head on a single flip and ignoring that three flips are made.
- (d) 315 — The relative frequency from the survey is 42 ÷ 120 = 0.35, so the expected number who prefer paper bags among 900 shoppers is 0.35 × 900 = 315. Giving 42 as the answer reuses the original survey count without scaling it up to 900 shoppers at all. Finding the expected number who prefer PLASTIC bags instead of paper, using the relative frequency 78 ÷ 120 = 0.65, gives 0.65 × 900 = 585. Using 1000 shoppers instead of the 900 actually stated gives 0.35 × 1000 = 350.
- (d) 1/15 — The probability that the first bead is red is 3/10. Since the first bead is not put back, there are now only 2 red beads left out of 9 beads in total, so the probability that the second bead is also red is 2/9. Multiplying these, 3/10 × 2/9 = 6/90 = 1/15. A candidate who answers 9/100 has treated the beads as replaced, using 3/10 twice. A candidate who answers 3/50 has correctly reduced the red count to 2 for the second pick but forgotten that the total also falls to 9, using 2/10 instead. A candidate who answers 5/19 has added the numerators and added the denominators, (3+2)/(10+9), instead of multiplying.
- (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.
- (a) 1/6 — Method: for equally likely outcomes, write the number of favourable outcomes over the total number of outcomes. Working: the dice has one face showing 6, and there are 6 equally likely faces in total, so the probability is 1 over 6. Answer: 1/6, which is close to the left-hand end of the 0 to 1 scale because the outcome is unlikely. The distractors: 5/6 comes from giving the probability of not rolling a 6; 1/2 comes from treating 'a 6' and 'not a 6' as two equally likely outcomes; 1/5 comes from writing the one favourable face over the 5 faces that are not a 6 instead of over all 6 faces.
Build your own mix at the worksheet builder.