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 garage tested 200 cars in one week. 70% of the cars were more than 3 years old and the rest were 3 years old or less. Of the cars more than 3 years old, 1 in 4 failed the test. 12 of the cars that were 3 years old or less failed the test. Work out how many of the 200 cars failed the test altogether.
- 2.Each turn of a game ends in a win, a draw or a loss, and no turn can end in more than one of these. The probability of a win is 1/4 and the probability of a draw is 1/3. Work out the probability of a loss.
- 3.Two games at a school fete are played independently. The probability of winning game X is 3/8. The probability of winning game Y is 0.4. Which statement correctly compares the two games?
- 4.A bag contains 5 red counters and 5 green counters. Three counters are taken out one at a time and are not put back. Work out the probability that all three counters are red.
- 5.Priya and Ben each want to estimate the probability that a coin lands on heads. Priya flips the coin 50 times and works out her relative frequency. Ben flips the same coin 500 times and works out his relative frequency. Whose relative frequency is more likely to be close to the true probability? Give a reason for your answer.
- 6.A fair coin is flipped again and again. After the first 10 flips there have been 7 heads. After 1000 flips there have been 528 heads. Which statement best describes what these results show?
- 7.A garden centre recorded 240 plant sales using a frequency tree. The first stage splits the sales into three types: shrubs, bedding plants and trees. 96 sales were shrubs, 114 were bedding plants and the rest were trees. Work out the probability that a sale chosen at random from these 240 was a tree.
- 8.The universal set is {1, 2, 3, ..., 15}. A = {multiples of 3}. Work out n(A′), the number of elements not in A.
- 9.Spinner A has 4 equal sections, numbered 1, 2, 3 and 4. Spinner B has 3 equal sections, coloured red, red and blue. Both spinners are spun once. Work out the probability of getting an even number on Spinner A and red on Spinner B.
- 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.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.
- 12.At a summer fair, the probability of winning at the hoopla stall is 6/10 and the probability of winning at the coconut shy is 2/10. Work out how many times as likely a player is to win at the hoopla stall as at the coconut shy.
- 13.At a gym, 45 members are surveyed. 24 use the running machines, 20 use the weights machines, and 9 use both. Work out how many of the 45 members use neither machine.
- 14.Leo flips a fair coin twice. He makes a list of every possible result of the two flips, writing down what the first flip lands on and then what the second flip lands on. Work out how many of the results in his list contain at least one tail.
- 15.A fair six-sided dice is rolled once. Write down the probability that the score is an even number.
Answer key
- (c) 47 — Method: fill the first pair of branches of the frequency tree, then the failures on each branch, then add only the failing end branches. Working: 70% of 200 is 140, so 140 cars were more than 3 years old and 200 − 140 = 60 cars were 3 years old or less. One quarter of the older cars failed: 140 ÷ 4 = 35. The newer branch gives 12 failures. Adding the two failing branches gives 35 + 12 = 47. Answer: 47 of the cars failed the test. The distractors: 35 is the older branch on its own, with the 12 newer failures never added; 62 comes from taking 1 in 4 of all the cars, 200 ÷ 4 = 50, and then adding the 12; 153 is 200 − 47 and counts the cars that passed.
- (a) 5/12 — Method: a win, a draw and a loss are the only outcomes and no two can happen together, so the three probabilities form an exhaustive set of mutually exclusive events and add to 1; add the two given probabilities, then subtract from 1. Working: 1/4 + 1/3 over the common denominator 12 is 3/12 + 4/12 = 7/12, and 1 − 7/12 = 12/12 − 7/12. Answer: 5/12. The distractors: 7/12 comes from stopping at the probability of a win or a draw and never subtracting from 1; 1/12 comes from subtracting the two given probabilities from each other, 1/3 − 1/4, instead of adding them and taking the total from 1; 5/7 comes from adding 1/4 and 1/3 by adding the numerators and the denominators to get 2/7 and then subtracting that from 1.
- (c) Game Y — 0.4 is greater than 3/8. — Converting 3/8 to a decimal gives 3 ÷ 8 = 0.375, and 0.4 is greater than 0.375, so game Y gives the better chance of winning: 'Game Y — 0.4 is greater than 3/8.' Comparing the numerator 3 to the digit 0 in front of the decimal point in 0.4 compares two unrelated numbers and gives 'Game X — 3 is greater than 0, the units digit of 0.4.' Treating a larger denominator as meaning more chances, rather than smaller equal shares, gives 'Game X — 3/8 has the larger denominator.' Rounding 0.375 to 0.4 to one decimal place makes the two values look equal, but they are not exactly equal, so 'Equal — 3/8 rounds to 0.4 to one decimal place' mistakes a rounded match for a true one.
- (a) 1/12 — Method: for draws with nothing put back, multiply the probabilities of the three draws, reducing both the number of red counters and the total each time a red counter is removed. Working: the first counter is red with probability 5/10. One red counter has gone, so the second is red with probability 4/9, and then the third is red with probability 3/8. Multiplying gives 60/720. Answer: the probability is 1/12. The distractors: 1/8 comes from using 5/10 three times, which is what happens only if each counter is put back; 2/9 comes from stopping after two draws and giving 5/10 × 4/9; 3/50 comes from taking one off the red count each time but leaving the total at 10, giving 5/10 × 4/10 × 3/10.
- (a) Ben, because a larger sample is closer to the theory — Method: a relative frequency is an estimate of a probability, and for an unbiased experiment that estimate tends towards the theoretical value as the sample grows. Working: Priya's estimate rests on 50 results, so a few unexpected heads move it a long way; one extra head shifts her relative frequency by 1 ÷ 50 = 0.02. Ben's estimate rests on 500 results, where one extra head shifts his relative frequency by only 1 ÷ 500 = 0.002. The larger sample therefore swings far less around the true value. Answer: Ben's relative frequency is the one more likely to be close, because a larger unbiased sample tends closer to the theoretical probability. The distractors: saying a small sample is less affected by luck reverses the result, since it is the small sample that swings most; saying every flip is a separate random event is true of the flips themselves but says nothing about the estimates, and is often used to argue wrongly that the number of trials does not matter; saying 500 flips must give exactly 250 heads confuses an expected value with a guaranteed one, and 500 flips very rarely give exactly 250 heads.
- (b) The relative frequency is settling near 0.5 — Method: turn each result into a relative frequency before comparing them, because it is the relative frequency, and not the difference between the two counts, that tends towards the theoretical probability. Working: after 10 flips the relative frequency of a head is 7 ÷ 10 = 0.7, which is a long way from 0.5. After 1000 flips it is 528 ÷ 1000 = 0.528, which is much closer to 0.5. Meanwhile the gap between the two counts has grown rather than shrunk: it was 7 − 3 = 4 after 10 flips and is 528 − 472 = 56 after 1000 flips. Answer: the relative frequency is settling near 0.5, which is what an unbiased experiment does as the sample grows. The distractors: saying the counts are levelling out is the usual form of this idea and the figures contradict it, since the gap went from 4 to 56; saying the coin is biased treats 28 extra heads in 1000 flips as proof, when 0.528 sits close to 0.5 and a fair coin gives results like this often; saying the next flip is more likely to be a tail is the gambler's fallacy, since each flip stays at 1/2 whatever came before.
- (b) 1/8 — Shrubs, bedding plants and trees are the three branches at the first stage of the tree, so they must total 240: tree sales = 240 − 96 − 114 = 30. So P(tree) = 30/240 = 1/8. Using the shrub count instead, 96/240 = 2/5, is the probability of a shrub sale, not a tree sale. Using the bedding-plant count instead, 114/240 = 19/40, is the probability of a bedding-plant sale. Subtracting the bedding count from the shrub count (114 − 96 = 18) instead of subtracting both from 240 gives 18/240 = 3/40, which is not the number of tree sales at all.
- (b) 10 — Method: A′ means everything in the universal set that is NOT in A, so n(A′) = n(universal set) − n(A). Working: the universal set has 15 elements. A = {3, 6, 9, 12, 15}, so n(A) = 5. n(A′) = 15 − 5 = 10. Answer: 10. Watch out: writing down 5 gives n(A) itself, the size of the multiples-of-3 set, which is the opposite of its complement. Writing down 11 comes from missing 15 off the list of multiples of 3, treating A as only {3, 6, 9, 12}, so A is undercounted as 4 and A′ is overstated as 15 − 4. And writing down 12 comes from only listing the multiples of 3 up to 9 — 3, 6 and 9 — and missing that 12 and 15 also belong to A, undercounting A as 3 rather than 5.
- (a) 1/3 — Method: for two independent spinners, multiply the probability of each separate outcome, but first work out each spinner's own probability correctly, using how many of its equal sections actually carry that result. Working: Spinner A has 2 even numbers, 2 and 4, out of 4 sections, so P(even) = 2/4 = 1/2. Spinner B has 2 red sections out of 3, so P(red) = 2/3. Multiplying gives 1/2 × 2/3, which cancels down to 1/3. Answer: 1/3. Watch out: writing down 1/4 treats Spinner B's two colours as equally likely and uses 1/2 for red, when in fact 2 of its 3 sections are red — the sections are not split evenly between the two colours. Writing down 1/6 undercounts Spinner A's even numbers as just one out of four instead of two. And writing down 5/6 applies the 'at least one' formula, P(A) + P(B) − P(A)×P(B), which answers a different question about EITHER spinner landing the right way, not both together.
- (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.
- (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.
- (b) 3 times as likely — Method: to say how many times as likely one event is as another, divide the larger probability by the smaller one; subtracting them gives the gap between the two probabilities, not the multiple. Working: both probabilities are counted in tenths, so 6/10 ÷ 2/10 compares 6 tenths with 2 tenths, and 6 ÷ 2 = 3. Answer: winning at the hoopla stall is 3 times as likely, which is why 6/10 sits three times as far along the 0 to 1 scale as 2/10. The distractors: 4 times as likely comes from subtracting the two counts, 6 − 2, instead of dividing them, which measures the gap rather than the multiple; 6 times as likely comes from reading the larger probability's 6 tenths straight off as the multiple without ever comparing it with the 2 tenths at the other stall; 12 times as likely comes from multiplying the two counts, 6 × 2, instead of dividing one by the other.
- (b) 10 — Method: the number using at least one machine is the running total plus the weights total minus the overlap, since the overlap would otherwise be added in twice; the number using neither is the survey total minus that. Working: at least one machine = 24 + 20 − 9 = 35. Neither = 45 − 35 = 10. Answer: 10. Watch out: subtracting 24 + 20 from 45 without adding the 9 back, 45 − 24 − 20 = 1, removes the overlap twice over instead of once. Writing down 15, which is 24 − 9, gives the number who use ONLY the running machines, not the number who use neither. And writing down 9 mistakes the overlap region for the region outside both circles altogether.
- (b) 3 — Method: build the list of possible results systematically, taking the first flip as a head and then as a tail, and keeping the two flips in order so that a head then a tail is a different result from a tail then a head; then count the results that match the description. Working: with a head at the first flip the results are HH and HT, and with a tail at the first flip they are TH and TT, so the list is HH, HT, TH, TT — four results in all. The results containing at least one tail are HT, TH and TT. Answer: 3 of the results in the list contain at least one tail. The distractors: 2 comes from treating a head and a tail as one result however they are ordered, which shortens the list to HH, HT, TT so that only two entries hold a tail; 1 comes from reading 'at least one tail' as a tail at both flips, which is the single result TT; 4 comes from counting the tails written across the list — one in HT, one in TH and two in TT — instead of counting the results that contain a tail.
- (b) 1/2 — Method: for equally likely outcomes the probability is the number of favourable outcomes divided by the total number of outcomes, written in its simplest form. Working: the dice can show 1, 2, 3, 4, 5 or 6, so there are 6 equally likely scores; the even scores are 2, 4 and 6, which is 3 of them, giving 3/6. Answer: 3/6 cancels to 1/2, which sits halfway along the 0 to 1 probability scale. The distractors: 1/3 comes from listing only 2 and 4 as the even scores and writing 2/6; 1/6 comes from giving the probability of one particular even score, such as the 2, rather than any even score; 2/3 comes from listing the even numbers as 0, 2, 4 and 6 and writing 4/6, forgetting that a dice has no 0 on it.
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