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
MathsUKwww.geekhero.co.uk
- 1.Two games are offered at a school fair, and every one of the 40 pupils in a class plays each game once. In Game A, the probability of winning a prize is 0.2 and the prize is worth £5. In Game B, the probability of winning a prize is 0.1 and the prize is worth £12. Work out which game gives a higher expected total value of prizes for the class, and by how much.
- 2.Oliver flips three fair coins at the same time. Work out the probability that exactly two of the three coins land on heads.
- 3.In a class of 32 pupils, 19 study geography and 15 study history. 8 study both subjects. Work out n(G ∪ H), the number of pupils who study geography or history or both.
- 4.Five balls numbered 1, 2, 3, 4 and 5 are in a bag. Two of them are taken out together at random. Work out the probability that the two numbers on them add up to more than 7.
- 5.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.
- 6.A fair spinner has 10 equal sections, numbered 1 to 10. The spinner is spun 150 times. Work out how many times you would expect it to land on a number greater than 7.
- 7.A two-way table records how 80 students travel to school. 45 of the students are boys and the rest are girls. 27 of the boys walk to school. Work out the probability, as a fraction, that a student chosen at random from the boys walks to school.
- 8.Two fair coins are thrown together 400 times. Work out how many of the 400 throws you would expect to give one head and one tail.
- 9.A fair spinner has 8 equal sections. 3 of the sections are red. The spinner is spun 240 times. Work out how many times you would expect it to land on red.
- 10.At a gym, 70 members are asked whether they use the pool or the sauna. 38 use the pool, 30 use the sauna, and 12 use neither. Work out n(P ∩ S), the number of members who use both the pool and the sauna.
- 11.A fair four-sided dice, numbered 1 to 4, is rolled twice. Every outcome is listed as an ordered pair, first roll then second roll, such as (2, 3). Work out how many different outcomes are in the list.
- 12.A drawer contains 4 red socks and 2 blue socks. Two socks are taken out at random, one after the other, without the first being put back. Work out the probability that both socks are the same colour.
- 13.Two ordinary fair dice are rolled. Work out the probability that the difference between the two scores is 1.
- 14.The probability that Priya scores with a netball shot is 3/4. She takes two shots, and the shots are independent. Work out the probability that she scores with her first shot but misses her second shot.
- 15.A box contains counters that are silver, bronze or copper only. The probability that a counter taken at random is silver is 1/2 and the probability that it is bronze is 1/8. The box contains 64 counters in total. Work out the number of copper counters.
Answer key
- (a) Game B, by £8 — Game A's expected total is 40 × 0.2 × £5 = £40. Game B's expected total is 40 × 0.1 × £12 = £48. Game B is higher, by £48 − £40 = £8. Writing 'Game A, by £8' is wrong because it has the right difference but the wrong game — Game A's total (£40) is actually LOWER than Game B's, not higher. Writing 'Game B, by £48' is wrong because £48 is Game B's whole expected total, not the DIFFERENCE between the two games. Writing 'Game A, by £40' is wrong in the same way, using Game A's whole total as if it were the margin, and naming the wrong game as the winner. Game B gives the higher expected total, by £8.
- (c) 3/8 — Method: write out every result of the three coins as a string of three letters, H for heads and T for tails, count the results that match the description and divide by how many results the list holds. Working: each coin lands two ways and no coin affects another, so the list holds 2 × 2 × 2 = 8 equally likely results. Exactly two heads means one coin lands on tails and the other two on heads, so the results are HHT, HTH and THH — 3 of the 8. Answer: the probability is 3/8. The distractors: 4/8 comes from reading 'exactly two heads' as 'at least two heads' and counting HHH as well; 2/8 comes from a list made without a system, in which HHT and THH are written down and HTH, the result with the tail between the two heads, is missed; 6/8 comes from counting 3 × 2 = 6 ways of picking which two of the three coins show heads, which counts every pair of coins twice, once in each order.
- (b) 26 — Method: n(G ∪ H) = n(G) + n(H) − n(G ∩ H), taking off the overlap once so the pupils who study both subjects are not counted twice. Working: 19 + 15 − 8 = 26. Answer: 26. Watch out: adding 19 and 15 without taking off the overlap gives 34, which counts the 8 pupils who study both subjects twice. Taking the 8 off both totals before adding, 19 − 8 + 15 − 8 = 18, counts only the pupils who study exactly one of the two subjects and leaves out the 8 who study both. And writing down 11, which is 19 − 8, gives the number who study geography only, not the number who study geography or history or both.
- (d) 1/5 — Method: list the pairs systematically, work out each total, then count the pairs that meet the condition and compare that count with the length of the list. Working: the pairs and their totals are 1 and 2 giving 3, 1 and 3 giving 4, 1 and 4 giving 5, 1 and 5 giving 6, 2 and 3 giving 5, 2 and 4 giving 6, 2 and 5 giving 7, 3 and 4 giving 7, 3 and 5 giving 8, and 4 and 5 giving 9. That is 10 pairs, of which 2 have a total of more than 7. Answer: the probability is 1/5. The distractors: 2/5 comes from counting the totals of exactly 7 as well, reading 'more than 7' as '7 or more'; 1/10 comes from finding only the pair 4 and 5 and missing that 3 and 5 also beat 7; 4/5 comes from counting the pairs on the wrong side of the condition, the 8 pairs whose total is 7 or less.
- (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.
- (c) 45 — Three of the ten numbers (8, 9 and 10) are greater than 7, so the probability is 3/10, and 150 × 3/10 = 45. Writing 105 is wrong because 150 × 7/10 = 105 uses the seven numbers that are NOT greater than 7 (1 to 7), the opposite of what is asked. Writing 60 is wrong because it counts 7, 8, 9 and 10 as four numbers greater than 7, wrongly including 7 itself: 150 × 4/10 = 60. Writing 50 is wrong because 150 ÷ 3 = 50 divides by the count of favourable numbers instead of multiplying by the correct fraction of the spinner. The expected number of spins landing on a number greater than 7 is 45.
- (d) 3/5 — 'Chosen from the boys' restricts the sample space to the 45 boys, of whom 27 walk. So the probability is 27/45 = 3/5. Using all 80 students as the denominator instead of just the boys gives 27/80, which is the probability that a student chosen from everyone is a boy who walks — not what was asked. Using the 18 boys who do NOT walk (45 − 27) as the numerator instead of the 27 who do gives 18/45 = 2/5. Dividing the 27 walking boys by the number of girls (80 − 45 = 35) instead of by the number of boys gives 27/35.
- (b) 200 — Method: list the equally likely outcomes for the two coins before writing any probability, then multiply by the number of throws. Working: the equally likely outcomes are head then head, head then tail, tail then head, and tail then tail, so there are 4 of them. Two of those 4 give one head and one tail, so the probability is 2/4, which is 1/2. Over 400 throws the expected number is 400 × 1 ÷ 2 = 200. Answer: about 200 of the throws would be expected to give one head and one tail. The distractors: 133 comes from treating two heads, two tails and one of each as three equally likely results and working out 400 ÷ 3 = 133.3, then rounding; 100 comes from counting only head then tail as a success, giving 400 × 1 ÷ 4 = 100; 300 is the expected number of throws that do not give two heads, 400 × 3 ÷ 4 = 300.
- (c) 90 — Method: over many future trials the expected number of successes is the number of trials multiplied by the probability of a success. Working: the sections are equal, so each is equally likely and P(red) is 3 out of 8. Over 240 spins the expected number of reds is 240 × 3 ÷ 8 = 90. Answer: about 90 of the spins would be expected to land on red. The distractors: 150 uses the 5 sections that are not red, 240 × 5 ÷ 8 = 150, which is the expected number of spins that do not land on red; 30 is 240 ÷ 8 and is the expected count for one single section; 80 comes from dividing by the number of red sections instead of by the number of sections, 240 ÷ 3 = 80.
- (b) 10 — The number who use the pool or the sauna (or both) is the total minus those who use neither: 70 − 12 = 58. Since pool + sauna double-counts the overlap, n(P ∩ S) = 38 + 30 − 58 = 10. Adding the pool and sauna counts without subtracting the overlap at all gives 38 + 30 = 68, more members than are in the whole gym. Subtracting the sauna count from the union, 58 − 30 = 28, actually finds the number who use ONLY the pool, not both. Reporting the 'neither' count, 12, confuses it with the 'both' region — they describe opposite corners of the diagram.
- (a) 16 — Method: keep the two rolls in order, first roll then second roll, and count every ordered pair the dice can land on, so a (2, 3) result is counted separately from a (3, 2) result. Working: the first roll can land on any of 4 numbers, and for each of those the second roll can also land on any of 4 numbers, giving 4 × 4 = 16 ordered pairs. Answer: 16. Watch out: writing down 8 comes from adding the two rolls' counts, 4 + 4, instead of multiplying them. Writing down 10 comes from listing unordered pairs, so (2, 3) and (3, 2) are counted as the same entry — that gives the 4 doubles plus 6 mixed pairs, 10 in total, instead of 16 ordered pairs. And writing down 4 comes from counting only one roll's outcomes and never pairing the two rolls together at all.
- (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.
- (d) 5/18 — Method: list every ordered pair of dice scores whose difference is 1, then divide by the 36 equally likely pairs. Working: the pairs with a difference of 1 are (1, 2), (2, 1), (2, 3), (3, 2), (3, 4), (4, 3), (4, 5), (5, 4), (5, 6) and (6, 5), which is 10 pairs out of 36, cancelling down to 5/18. Answer: 5/18. Watch out: writing down 5/36 lists only the 5 pairs going up, (1, 2), (2, 3), (3, 4), (4, 5) and (5, 6), and misses that each one has a matching pair the other way round, such as (2, 1) — the two dice are different objects, and order matters, so each of those 5 gaps counts twice. Writing down 1/6 treats the six possible differences, 0 to 5, as equally likely and picks 1 out of 6 of them, but a difference of 1 is reached by far more pairs of scores than a difference of 5 is, so the six differences are not equally likely. And writing down 1/4 comes from adding the two dice's outcome counts instead of multiplying them, 6 + 6 = 12, and grouping the six scores into just three non-overlapping pairs one apart, {1, 2}, {3, 4} and {5, 6}, giving 3 out of that wrong pool of 12.
- (c) 3/16 — The probability that Priya misses a shot is 1 − 3/4 = 1/4. The two shots are independent, so multiply the probability of scoring the first shot by the probability of missing the second shot: 3/4 × 1/4 = 3/16. Choosing 9/16 comes from using the probability of scoring, 3/4, for both shots instead of switching to the miss probability for the second shot, 3/4 × 3/4 = 9/16, which is the probability that she scores with both shots. Choosing 1/2 comes from subtracting the two probabilities instead of multiplying them, 3/4 − 1/4 = 1/2. Choosing 3/4 comes from giving only the probability that she scores with the first shot, forgetting to combine it with what happens on the second shot.
- (c) 24 — Silver, bronze and copper cover every counter, so their probabilities sum to 1: P(copper) = 1 − 1/2 − 1/8 = 3/8. Number of copper counters = 3/8 × 64 = 24. Multiplying the silver probability by 64 gives 32, the number of silver counters, not copper. Multiplying the bronze probability by 64 gives 8, the number of bronze counters. Adding the silver and bronze probabilities (1/2 + 1/8 = 5/8) and multiplying by 64 gives 40, the combined number of silver and bronze counters, not the copper count.
Build your own mix at the worksheet builder.