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 pack contains 6 cards, numbered 1 to 6. A card is drawn at random from the pack, and a fair coin is flipped. Work out the probability of drawing an even-numbered card and the coin landing on tails.
- 2.A basketball player takes two free throws, and the throws are independent. The probability of scoring on each throw is 0.6, and the probability of missing is 0.4. Work out the probability that she scores exactly one of the two throws.
- 3.An archer hits the bullseye with probability 0.24 on any one shot. She wants to know how many shots she must take to expect to hit the bullseye 12 times. Work out how many shots this is.
- 4.A fair six-sided dice is rolled once. Write down the probability of rolling a 6.
- 5.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?
- 6.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.
- 7.A school recorded what 60 students chose for lunch. 35 of the students are in Year 10 and the rest are in Year 11. 20 of the Year 10 students chose a sandwich and 9 of the Year 11 students chose a sandwich. Everyone else chose a hot meal. Work out how many Year 11 students chose a hot meal.
- 8.The probability that a seed fails to germinate is 1/5. Three seeds are planted independently. Work out the probability that at least one of the three seeds germinates.
- 9.Two ordinary fair dice are rolled, one after the other. Work out the probability that the first dice shows a 6 and the second dice shows an even number.
- 10.Two fair four-sided dice, numbered 1 to 4, are rolled and the two scores are added together. Work out the probability that the total is 5.
- 11.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.
- 12.Two fair spinners are each numbered 1, 2 and 3. Priya spins both spinners together and records the two numbers as a pair, listing every possible outcome systematically in a grid. Work out the probability that the two numbers are the same.
- 13.A two-way table records how 130 pupils travel to school. 70 of the pupils are girls and the rest are boys. 42 of the girls walk to school and the rest of the girls cycle. 33 of the boys walk to school. Work out what fraction of the girls walk to school.
- 14.The probability that a spinner lands on blue is 2/5. Write this probability as a percentage.
- 15.Amir has 3 different coloured pens, red, blue and green, and 2 different types of paper, lined and plain. He wants to choose one pen and one type of paper, and draws a grid to list every combination systematically. Work out how many combinations are in his grid.
Answer key
- (b) 1/4 — There are 6 × 2 = 12 equally likely outcomes. The even-numbered cards are 2, 4 and 6, so there are 3 × 1 = 3 outcomes with an even card and tails, giving a probability of 3/12 = 1/4. Choosing 1/2 comes from working out only the probability of drawing an even card, 3/6, and forgetting to combine it with the coin landing on tails. Choosing 1/12 comes from treating only one specific outcome, such as card 6 with tails, as the only one that counts, instead of all three even cards paired with tails. Choosing 1/8 comes from doubling the coin stage when counting the total, using 6 × 2 × 2 = 24 outcomes instead of 6 × 2 = 12, and giving 3/24 = 1/8.
- (a) 0.48 — There are two ways to score exactly one throw: scoring on the first and missing the second, 0.6 × 0.4 = 0.24, or missing the first and scoring the second, 0.4 × 0.6 = 0.24. Adding these gives 0.24 + 0.24 = 0.48. Choosing 0.24 comes from working out only one of the two paths and forgetting the other one also gives exactly one score. Choosing 0.36 comes from working out the probability of scoring BOTH throws, 0.6 × 0.6 = 0.36, instead of exactly one. Choosing 0.84 comes from working out the probability of scoring AT LEAST one throw, 1 − 0.4 × 0.4 = 0.84, instead of exactly one.
- (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.
- (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.
- (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.
- (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.
- (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.
- (d) 124/125 — The probability that a seed germinates is 1 − 1/5 = 4/5, so the probability that all three seeds fail to germinate is 1/5 × 1/5 × 1/5 = 1/125. The probability that at least one germinates is 1 − 1/125 = 124/125. Choosing 4/5 comes from giving the probability that a single seed germinates, forgetting to combine all three seeds. Choosing 64/125 comes from working out the probability that ALL three seeds germinate, 4/5 × 4/5 × 4/5 = 64/125, instead of at least one. Choosing 12/125 comes from working out the probability that EXACTLY one seed germinates, 3 × 4/5 × 1/5 × 1/5 = 12/125, instead of at least one.
- (a) 1/12 — The probability that the first dice shows a 6 is 1/6. The probability that the second dice shows an even number, 2, 4 or 6, is 3/6 = 1/2. Since the two dice are independent, multiply the probabilities: 1/6 × 1/2 = 1/12. A candidate who answers 1/6 has considered only the first dice and forgotten the condition on the second dice. A candidate who answers 1/2 has considered only the second dice and forgotten the condition on the first dice. A candidate who answers 1/36 has treated 'an even number' as a single specific value rather than three possible values, using 1/6 × 1/6.
- (b) 1/4 — There are 4 × 4 = 16 equally likely ordered outcomes for the two dice. The pairs that total 5 are: first dice 1 with second dice 4; first dice 2 with second dice 3; first dice 3 with second dice 2; and first dice 4 with second dice 1 — which is 4 outcomes, so the probability is 4/16 = 1/4. A candidate who answers 1/8 has listed only 2 of the four pairs, forgetting that first dice 1 with second dice 4 and first dice 4 with second dice 1 are separate outcomes because the dice are different. A candidate who answers 3/16 has found 3 pairs instead of 4, missing one from the list. A candidate who answers 1/16 has counted only a single pair, such as first dice 2 with second dice 3, and treated the order of the dice as not mattering.
- (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.
- (d) 1/3 — List the outcomes for the two spinners systematically in a 3 × 3 grid: 1-1, 1-2, 1-3, 2-1, 2-2, 2-3, 3-1, 3-2, 3-3, where the first number is the score on spinner A and the second is the score on spinner B — 9 equally likely outcomes in total. The pairs where the two numbers are the same are 1-1, 2-2 and 3-3, so there are 3 favourable outcomes. P(same number) = 3/9 = 1/3. 2/3 comes from working out the probability that the two numbers are different and then forgetting to take the complement the right way round, so the probability of "different" is given instead of the probability of "same". 1/2 comes from listing only the 6 unordered pairs 1-1, 2-2, 3-3, 1-2, 1-3, 2-3 instead of all 9 ordered outcomes in the grid, then taking 3 out of that 6. 1/9 comes from spotting only one of the three matching pairs, such as 1-1, and missing 2-2 and 3-3.
- (b) 3/5 — The question asks about the girls only, so use the girls' total of 70 as the denominator: 42 out of 70 girls walk, giving 42/70 = 3/5. Choosing 21/65 comes from using the whole survey of 130 pupils as the denominator instead of just the 70 girls, 42/130 = 21/65. Choosing 33/70 comes from using the boys' walking count, 33, over the girls' total of 70, mixing up the two rows of the table. Choosing 2/5 comes from using the number of girls who CYCLE, 70 − 42 = 28, instead of the number who walk, giving 28/70 = 2/5.
- (b) 40% — Method: a fraction becomes a percentage by scaling it so that its denominator is 100, or equivalently by dividing the numerator by the denominator and multiplying the decimal by 100. Working: 5 × 20 = 100, so the numerator is scaled by 20 as well, 2 × 20 = 40, giving 40/100; the same figure comes from 2 ÷ 5 = 0.4 and 0.4 × 100 = 40. Answer: 40%. The distractors: 20% comes from dividing 100 by the denominator alone, 100 ÷ 5 = 20, which is the percentage for 1/5 and never uses the numerator 2; 4% comes from finding the decimal 0.4 correctly and then writing the 4 down as the percentage instead of multiplying by 100; 2% comes from scaling only the denominator up to 100 and leaving the numerator as it was, giving 2/100.
- (c) 6 — Method: to list every combination of one item from a group of 3 and one item from a group of 2 systematically, multiply the two numbers of choices together. Working: 3 × 2 = 6. Answer: 6. Watch out: adding the two totals instead of multiplying, 3 + 2 = 5, misses combinations that a full grid would show — a grid with 3 rows and 2 columns has 6 cells, not 5. Writing down 3 counts only the pens, and writing down 2 counts only the types of paper — neither one pairs every pen with every type of paper.
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