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.Priya spins a fair spinner with P(red) = 0.3, and separately flips a fair coin with P(heads) = 0.5. Using a tree diagram, work out the probability that she gets red AND heads.
- 2.A survey of 160 employees at a company recorded whether they cycle to work, using a frequency tree. The first branch splits them into 90 who work full-time and 70 who work part-time. Of the full-time employees, 27 cycle to work. Of the part-time employees, 14 cycle to work. Work out the probability that an employee, chosen at random from the 160, cycles to work. Give your answer as a fraction in its simplest form.
- 3.A charity bake sale tracked 240 items sold using a frequency tree. The first branch splits sales into 150 cake sales and 90 biscuit sales. Of the cake sales, 100 were bought by adults and the rest by children. Of the biscuit sales, 54 were bought by children. Each item sold for £1.20. Work out the total amount of money raised from sales to children.
- 4.On the probability scale from 0 to 1, which word best describes an event with probability 0.9?
- 5.A survey found that, of 250 shoppers questioned, 68% said they had used a self-checkout in the past month. A different store expects 1400 shoppers this week. Using this relative frequency, work out how many of the 1400 shoppers would be expected to have used a self-checkout in the past month.
- 6.A bag is known to contain red, blue and green counters in equal numbers, so the theoretical probability of taking each colour is 1/3. In 90 trials with replacement, red was taken 38 times, blue was taken 26 times and green was taken 26 times. Which colour is over-represented compared with its theoretical probability?
- 7.A fair six-sided dice is rolled once. Work out the probability that the score is an odd number. Give your answer as a decimal.
- 8.At a school fête, a tombola stall costs £1.50 to play. The probability of winning is 0.2, and the prize is worth £6. Work out the stall's expected profit, on average, from each game played.
- 9.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.
- 10.A two-way table records 150 customers at a café. 90 of the customers bought a hot drink and the rest did not. 54 of the hot-drink customers also bought a cake. 21 of the customers who did not buy a hot drink bought a cake. Work out the percentage of all 150 customers who bought a cake.
- 11.A fair spinner has 6 equal sections, and 2 of the sections are coloured purple. The spinner is spun 180 times. Work out how many times you would expect it to land on purple.
- 12.A weather station records whether it rains each day for 40 days: it rains on 9 of the days and does not rain on the rest. A local forecaster claims that the probability of rain on any day is 0.3. Work out the relative frequency of rain from the recorded data.
- 13.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.
- 14.A factory finds that the probability a randomly chosen light bulb is defective is 0.035. In a batch of 4,000 bulbs, work out how many bulbs you would expect to work correctly.
- 15.A padlock code is formed by arranging three different digits chosen from 2, 3, 4 and 5, with no digit repeated. Work out how many different three-digit codes can be made.
Answer key
- (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.
- (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.
- (a) £124.80 — On the cake branch, 150 − 100 = 50 cakes were bought by children. Adding the 54 biscuits bought by children gives 50 + 54 = 104 items sold to children in total, and at £1.20 each that raises 104 × £1.20 = £124.80. Writing £60.00 is wrong because 50 × £1.20 = £60.00 only counts the cake sales to children and leaves out the 54 biscuits. Writing £163.20 is wrong because it uses the ADULT sales instead of children's: 100 cake adults plus 90 − 54 = 36 biscuit adults gives 136 × £1.20 = £163.20. Writing £136.80 is wrong because it finds the cake children's number by subtracting the wrong branch (150 − 90 = 60 instead of 150 − 100 = 50), giving 60 + 54 = 114 items and 114 × £1.20 = £136.80. The total raised from sales to children is £124.80.
- (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.
- (a) 952 — 68% = 0.68. The relative frequency from the survey applies to the new group of 1400 shoppers, so the expected number is 0.68 × 1400 = 952. Working out 1 − 0.68 = 0.32 and applying that instead, 0.32 × 1400 = 448, finds the number who have NOT used a self-checkout, not the number who have. Applying 68% to the original sample size of 250 instead of the new total of 1400 gives 0.68 × 250 = 170. Applying the complement percentage to the original sample size, 0.32 × 250 = 80, compounds both mistakes.
- (c) Red — Theoretical probability is 1/3 ≈ 0.333 for each colour. Red's relative frequency is 38/90 ≈ 0.422, above 1/3, so red is over-represented. Blue's relative frequency is 26/90 ≈ 0.289, below 1/3, so blue is under-represented, not over. Green's relative frequency is also 26/90 ≈ 0.289, below 1/3 for the same reason. Since red's relative frequency clearly exceeds 1/3, it is not true that none of the colours are over-represented.
- (c) 0.5 — Method: count the favourable outcomes, write them over the total number of equally likely outcomes and then divide to turn the fraction into a decimal. Working: the odd scores are 1, 3 and 5, which is 3 of the 6 equally likely scores, so the probability is 3/6, and 3 ÷ 6 = 0.5. Answer: 0.5, the middle of the 0 to 1 probability scale. The distractors: 0.33 comes from listing only 3 and 5 as odd and working out 2 ÷ 6; 0.17 comes from giving the probability of one particular odd score, 1 ÷ 6; 0.3 comes from writing '3 out of 6' as 0.3, reading the 3 straight off as tenths instead of dividing.
- (b) £0.30 profit for the stall — The stall keeps the £1.50 entry fee whatever happens, and expects to pay out prize × probability of winning = £6 × 0.2 = £1.20 on average. So its expected profit per game is £1.50 − £1.20 = £0.30. Reporting the expected pay-out of £1.20 itself as the profit forgets that the stall also keeps the entry fee. Assuming the player always wins gives an expected cost of £6 − £1.50 = £4.50, treated as a loss for the stall. Using the probability of NOT winning, 0.8, to find the expected pay-out gives £6 × 0.8 = £4.80, and £1.50 − £4.80 = −£3.30, a £3.30 loss.
- (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.
- (b) 50% — The total number of customers who bought a cake is 54 + 21 = 75, combining both hot-drink and non-hot-drink customers. As a percentage of all 150 customers, this is (75 ÷ 150) × 100 = 50%. Choosing 36% comes from only counting the hot-drink customers who bought a cake, (54 ÷ 150) × 100 = 36%, and forgetting the 21 non-hot-drink customers who also bought a cake. Choosing 14% comes from only counting the non-hot-drink customers who bought a cake, (21 ÷ 150) × 100 = 14%, and forgetting the 54 hot-drink customers who also bought a cake. Choosing 60% comes from dividing by the hot-drink total of 90 instead of the grand total of 150, (54 ÷ 90) × 100 = 60%.
- (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) 0.225 — The relative frequency of rain is the number of rainy days out of all days recorded: 9 ÷ 40 = 0.225, which is noticeably less than the forecaster's claimed 0.3. Using the number of dry days, 40 − 9 = 31, as the denominator instead of the total of 40 gives 9 ÷ 31 = 0.29 (2 d.p.). Simply reporting the forecaster's claimed value, 0.3, without calculating anything from the data at all, ignores the recorded results completely. Misplacing the decimal point, treating 9 out of 40 as 9%, gives 0.09 instead of 0.225.
- (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.
- (a) 3,860 — The probability a bulb works correctly is the complement of being defective: 1 − 0.035 = 0.965. Expected number working correctly = 0.965 × 4,000 = 3,860. Using the probability of being defective instead of its complement gives 4,000 × 0.035 = 140, the expected number of DEFECTIVE bulbs, not working ones. Shifting the decimal point in the complement, using 0.0965 instead of 0.965, gives 4,000 × 0.0965 = 386. Assuming every bulb works, ignoring the 0.035 probability altogether, gives the full batch of 4,000.
- (d) 24 — There are 4 choices for the first digit. Once that digit is used, 3 digits remain for the second position, and then 2 digits remain for the third position: 4 × 3 × 2 = 24 codes. Choosing 64 comes from allowing a digit to be reused at every position, 4 × 4 × 4 = 64, which is not allowed here since no digit repeats. Choosing 12 comes from multiplying only the first two positions, 4 × 3 = 12, and forgetting that a third digit is also chosen from the digits that remain. Choosing 6 comes from counting only the arrangements of one single set of three digits, 3 × 2 × 1 = 6, and forgetting that there are 4 different sets of three digits that can be chosen from 2, 3, 4 and 5.
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