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 machine makes 4000 light bulbs a day and runs 5 days a week. Two inspectors test bulbs from this machine. Inspector A tests 40 bulbs and finds 4 faulty. Inspector B tests 500 bulbs and finds 30 faulty. Using the better of the two estimates, work out how many faulty bulbs the machine is expected to make in one week.
- 2.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?
- 3.A frequency tree records 320 calls to a helpline. The calls first split into calls about billing and calls about technical support. Of the technical support calls, the probability that a call was resolved on the first contact is 0.75. If 180 technical support calls were resolved on the first contact, work out how many calls were about billing.
- 4.At a fair, a game costs £2 to play. The probability of winning the game is 0.15, and each win pays out £10. Amir plays the game 200 times. Work out how much money Amir should expect to lose in total.
- 5.A spinner is spun 50 times. It lands on green 18 times. Work out the relative frequency of landing on green.
- 6.A survey asked 80 people to name their favourite drink. The table shows the results: Tea 20 people, Coffee 32 people, Juice 16 people, Water 12 people. Work out the percentage of people whose favourite drink was Coffee.
- 7.In a class experiment a fair six-sided dice is rolled 60 times and lands on a six 14 times. The whole school then rolls the same fair dice 3000 times. Work out the best estimate of the number of sixes the school should expect.
- 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 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.
- 10.A fair six-sided dice is rolled 150 times. The table shows how many times each number came up: 1 came up 22 times, 2 came up 27 times, 3 came up 24 times, 4 came up 34 times, 5 came up 21 times and 6 came up 22 times. The theoretical probability of each number is 1/6. Which number is most over-represented compared with its theoretical probability?
- 11.In a game, Zara says the probability of scoring is 3/8, the probability of missing is 5/12 and the probability of a rebound is 1/6, and that these are the only three outcomes. Is Zara correct that her three probabilities are valid?
- 12.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.
- 13.On the probability scale from 0 (impossible) to 1 (certain), evens is exactly halfway. An event is 3/10 more likely than evens. Work out the probability of that event, giving your answer as a decimal.
- 14.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.
- 15.An ordinary six-sided dice, numbered 1 to 6, is rolled 30 times and lands on a 6 seven times. Ravi says the theoretical probability of rolling a 6 and the relative frequency of rolling a 6 in this trial are the same number. Is Ravi right?
Answer key
- (a) 1200 — Method: take the estimate from the larger sample, because an unbiased relative frequency tends towards the true probability as the sample grows, then multiply by the number of bulbs made in a week. Working: Inspector B tested 500 bulbs, far more than Inspector A's 40, so use B's relative frequency: 30 ÷ 500 = 0.06. A week's production is 4000 × 5 = 20000 bulbs. The expected number of faulty bulbs is 20000 × 0.06 = 1200. Answer: about 1200 faulty bulbs a week. The distractors: 2000 uses Inspector A's estimate, 4 ÷ 40 = 0.1, giving 20000 × 0.1 = 2000, and so rests on a sample of only 40 bulbs; 1600 comes from averaging the two estimates of 0.1 and 0.06 to get 0.08, and 20000 × 0.08 = 1600, which gives the small sample equal weight with the large one; 240 uses the right estimate but stops at a single day, 4000 × 0.06 = 240.
- (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) 80 — Since 180 calls are 0.75 of all the technical support calls, the technical support total is 180 ÷ 0.75 = 240. The billing calls make up the rest of the 320 calls, so 320 − 240 = 80. Choosing 240 comes from stopping after finding the technical support total and forgetting the question asks for the billing calls, which are the rest. Choosing 185 comes from multiplying 180 × 0.75 = 135 instead of dividing, then working out 320 − 135 = 185. Choosing 140 comes from using 180 directly as the whole technical support total, ignoring the probability altogether, then working out 320 − 180 = 140.
- (b) £100 — Method: find the expected number of wins, turn that into the expected pay out, then compare it with what the games cost. Working: the expected number of wins is 200 × 0.15 = 30. Each win pays £10, so the expected pay out is 30 × 10 = 300 pounds. Playing 200 times at £2 a go costs 200 × 2 = 400 pounds. The expected loss is 400 − 300 = 100 pounds. Answer: Amir should expect to be about £100 down. The distractors: £300 is the expected winnings on their own, with the cost of playing never taken off; £400 is the total cost of playing, with the winnings never taken off; £700 comes from adding the two totals, 400 + 300 = 700, instead of subtracting one from the other.
- (a) 0.36 — Relative frequency is the number of times the event happened divided by the total number of trials: 18 ÷ 50 = 0.36. Dividing by 100 instead of the actual 50 spins gives 18 ÷ 100 = 0.18. Finding the relative frequency of NOT landing on green, using 50 − 18 = 32 spins, gives 32 ÷ 50 = 0.64. Misplacing the decimal point in the division, so that 18 ÷ 50 is carried out as 18 ÷ 500, gives 0.036 — a tenth of the correct value.
- (b) 40% — 32 out of the 80 people chose Coffee, so the percentage is (32 ÷ 80) × 100 = 40%. Choosing 32% comes from treating 32 as if it were already a percentage out of 100, instead of dividing by the actual total of 80 people surveyed. Choosing 25% comes from using the Tea row instead of the Coffee row, (20 ÷ 80) × 100 = 25%. Choosing 20% comes from using the Juice row instead of the Coffee row, (16 ÷ 80) × 100 = 20%.
- (b) 500 — Method: when a dice is known to be fair, the theoretical probability is the best thing to work from, and the more trials there are the closer the results tend to it. Working: for a fair dice the probability of a six is 1/6, so the expected number of sixes in 3000 rolls is 3000 × 1 ÷ 6 = 500. The class experiment gave a relative frequency of 14/60, but 60 trials is far too few to overturn a known theoretical value, and the school's 3000 rolls will tend towards 1/6 in any case. Answer: about 500 sixes. The distractors: 700 comes from using the class relative frequency instead of the theory, 3000 × 14 ÷ 60 = 700; 600 comes from splitting the difference between the two, since 1/6 is about 0.167 and 14/60 is about 0.233, whose mean is 0.2, and 3000 × 0.2 = 600; 2500 uses 5/6 instead of 1/6 and counts the rolls expected not to be a six.
- (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.
- (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.
- (b) 4 — With 150 rolls and probability 1/6 for each number, the expected count is 150 ÷ 6 = 25. Comparing each actual count with 25: 1 is 22 (3 below), 2 is 27 (2 above), 3 is 24 (1 below), 4 is 34 (9 above), 5 is 21 (4 below) and 6 is 22 (3 below). Number 4 is furthest above its expected count, so it is the most over-represented. Number 2 is also above its expected count, but by only 2, far less than 4's 9. Number 3's count of 24 is below the expected 25, so it is under-represented, not over. Number 6's count of 22 is also below the expected 25, so it too is under-represented.
- (d) No, because 3/8 + 5/12 + 1/6 = 23/24 — Using a common denominator of 24: 3/8 = 9/24, 5/12 = 10/24 and 1/6 = 4/24. Adding these numerators gives 9 + 10 + 4 = 23, so the three probabilities sum to 23/24, which is less than 1 — Zara is not correct. Adding the original numerators (3 + 5 + 1 = 9) over a denominator of 12 instead of converting each fraction properly gives 9/12 = 3/4, still less than 1 but the wrong fraction. Converting 1/6 to 5/24 instead of 4/24 (using the wrong scaling) makes the total 9/24 + 10/24 + 5/24 = 24/24 = 1, wrongly suggesting the probabilities are valid. Judging validity from the fact that each individual fraction lies between 0 and 1 ignores that an exhaustive set must sum to exactly 1, not merely contain valid individual values.
- (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.
- (d) 0.80 — Evens is exactly halfway along the scale, at 0.5, and 3/10 written as a decimal is 0.3. A probability 3/10 higher than evens is 0.5 + 0.3 = 0.80. Giving 0.30 as the answer converts the increase but never adds it to the value of evens. Adding the increase to 1, the 'certain' end of the scale, instead of to evens gives 1 + 0.3 = 1.30, which cannot be a probability. Writing 3/10 as 0.03 instead of 0.3, a place-value slip, gives 0.5 + 0.03 = 0.53.
- (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.
- (c) No — 7/30 is the relative frequency; theory stays 1/6. — The theoretical probability of rolling a 6 on an ordinary dice is fixed at 1/6, worked out from the number of equally likely outcomes, and does not change however the dice is actually rolled. The relative frequency from this trial is 7/30, found from what happened in these particular 30 rolls. Since 7/30 and 1/6 are different numbers, the correct statement is 'No — 7/30 is the relative frequency; theory stays 1/6.' Assuming the two values must always match because they describe the same event gives 'Yes — relative frequency always equals theory.' Believing that an observed result redefines the theoretical probability gives 'Yes — the theoretical probability has now become 7/30.' Refusing to work out either value at all gives 'Neither can be found — 30 rolls is too few to tell', which ignores that both numbers CAN be calculated from the information given.
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