Printable · GCSE Foundation · ages 14-16
Possibility spaces for single and combined events worksheet — GCSE Foundation
Fifteen questions on "possibility spaces for single and combined events" — DfE statement P7. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.
Possibility spaces for single and combined events worksheet — GCSE Foundation
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- 1.A fair coin is flipped and a fair three-sided spinner, labelled X, Y and Z, is spun. Every outcome is listed as a pair, such as (H, X). Work out the probability that the outcome is a head and Z.
- 2.At a school fête, a stall invites visitors to spin a fair spinner with 8 equal sections numbered 1 to 8, and separately toss a fair coin. A visitor wins a small prize only if the spinner lands on a multiple of 3 and the coin lands on heads. Work out the probability that a visitor wins a prize, giving your answer as a fraction in its simplest form.
- 3.A spinner has three equal sections, coloured red, blue and green. It is spun twice, and each outcome is listed as an ordered pair of colours, such as (red, blue). Work out how many outcomes are in the full list.
- 4.Two fair five-sided dice, numbered 1 to 5, are rolled. Work out the probability that the two scores differ by at least 2.
- 5.In a board game Mia moves forward by the total of two ordinary fair dice. Work out the probability that her total is 7. Give your answer as a fraction in its simplest form.
- 6.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.
- 7.A small pack contains 5 cards, numbered 1 to 5. One card is drawn at random from the pack, and a fair coin is flipped. Every outcome is listed as a pair, such as (3, H). Work out how many equally likely outcomes there are in total.
- 8.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.
- 9.Oliver flips three fair coins at the same time. Work out the probability that exactly two of the three coins land on heads.
- 10.Spinner C has 3 equal sections, numbered 2, 3 and 5. Spinner D has 6 equal sections, numbered 1 to 6. Both spinners are spun once, and the two numbers are multiplied together. Work out the probability that the product is a multiple of 5.
- 11.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.
- 12.Jack rolls two ordinary fair dice and adds the two scores together. Work out the probability that the total is 6.
- 13.Two ordinary fair dice are rolled. Work out the probability that the difference between the two scores is 1.
- 14.Spinner A has 4 equal sections, numbered 1, 2, 3 and 4. Spinner B has 5 equal sections, numbered 1, 2, 3, 4 and 5. Both spinners are spun once, and the two numbers are multiplied together. Work out the probability that the product is 12.
- 15.Spinner E has 4 equal sections, numbered 1 to 4. Spinner F has 6 equal sections, numbered 1 to 6. Both spinners are spun once. Work out the probability of getting at least one 3.
Answer key
- (b) 1/6 — There are 2 outcomes for the coin and 3 outcomes for the spinner, giving 2 × 3 = 6 equally likely pairs: (H,X), (H,Y), (H,Z), (T,X), (T,Y), (T,Z). Only one of these, (H,Z), matches both conditions, so the probability is 1/6. A candidate who answers 1/3 has listed only the spinner's 3 outcomes and ignored the coin. A candidate who answers 1/2 has considered only the coin and ignored the spinner. A candidate who answers 1/5 has missed one pair when listing the possibility space, treating it as 5 outcomes instead of 6.
- (d) 1/8 — The multiples of 3 from 1 to 8 are 3 and 6, so the probability of that event is 2/8, which simplifies to 1/4. The probability of the coin landing on heads is 1/2. Since the spin and the toss are independent, multiply the two probabilities: 1/4 × 1/2 = 1/8. A candidate who answers 1/4 has considered only the spinner and forgotten to combine it with the coin toss. A candidate who answers 1/2 has considered only the coin and forgotten the spinner condition entirely. A candidate who answers 1/16 has counted only one number, 6, as a multiple of 3 instead of two, giving 1/8 × 1/2.
- (d) 9 — Method: each of the first spin's 3 outcomes can be paired with each of the second spin's 3 outcomes, so the two counts are multiplied together. Working: 3 × 3 = 9 ordered pairs. Answer: 9. Watch out: writing down 6 comes either from adding the two spins' counts, 3 + 3, instead of multiplying them, or from listing colour pairs without order, treating (red, blue) as the same entry as (blue, red) — both shortcuts miss outcomes that the ordered list contains. Writing down 3 lists only a single spin's outcomes and never pairs the two spins up at all. And writing down 27 treats the spinner as though it had been spun three times, 3 × 3 × 3, one spin too many.
- (b) 12/25 — List every outcome as an ordered pair (first dice, second dice) out of the 25 equally likely outcomes. The pairs with a difference of exactly 2 are (1, 3), (3, 1), (2, 4), (4, 2), (3, 5) and (5, 3), the pairs with a difference of 3 are (1, 4), (4, 1), (2, 5) and (5, 2), and the pairs with a difference of 4 are (1, 5) and (5, 1), giving 6 + 4 + 2 = 12 outcomes with a difference of at least 2, so the probability is 12/25. Choosing 13/25 comes from finding the probability that the two scores differ by LESS than 2 instead, using the remaining 13 outcomes, the opposite of what was asked. Choosing 10/25 comes from forgetting the two outcomes where the difference is 4, (1, 5) and (5, 1), and adding only the difference-2 and difference-3 outcomes, 6 + 4 = 10 out of 25. Choosing 6/25 comes from counting only the pairs with a difference of exactly 2, forgetting that differences of 3 and 4 also count as at least 2.
- (a) 1/6 — Method: write the results of the two dice as ordered pairs, count the pairs whose scores add to the total asked for, divide by the number of ordered pairs there are, then cancel the fraction down. Working: there are 6 × 6 = 36 equally likely ordered pairs. The pairs whose scores add to 7 are (1, 6), (2, 5), (3, 4), (4, 3), (5, 2) and (6, 1), which is 6 pairs, so the probability is 6/36. Dividing the top and the bottom by 6 gives 1/6. Answer: the probability is 1/6. The distractors: 7/36 comes from taking the number of favourable pairs to be 7 because 7 is the total asked for, confusing the size of a total with the number of ways of making it; 1/7 comes from using the 21 different combinations of two scores as the equally likely results, finding the 3 combinations 1 and 6, 2 and 5, 3 and 4, and cancelling 3/21; 6/11 comes from counting the 6 favourable pairs correctly but dividing by the 11 possible totals from 2 to 12 rather than by the 36 pairs.
- (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.
- (c) 10 — There are 5 possible cards and 2 possible coin results, so listing every pair gives 5 × 2 = 10 equally likely outcomes. Choosing 5 comes from listing only the card outcomes and forgetting the coin flip adds a second stage to each one. Choosing 7 comes from adding the two stages instead of combining them, 5 + 2 = 7, rather than pairing every card with every coin result. Choosing 20 comes from counting each coin result twice for every card, 5 × 2 × 2 = 20, effectively pairing every card with the coin twice over.
- (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.
- (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.
- (a) 4/9 — There are 3 × 6 = 18 equally likely outcomes. The product is a multiple of 5 whenever Spinner C shows 5, whatever Spinner D shows, which is 6 outcomes, or whenever Spinner D shows 5 and Spinner C shows 2 or 3, which is 2 more outcomes. This gives 6 + 2 = 8 outcomes, so the probability is 8/18 = 4/9. Choosing 1/3 comes from only counting the 6 outcomes where Spinner C shows 5 and forgetting the 2 outcomes where Spinner D shows 5 instead. Choosing 1/9 comes from only counting the 2 outcomes where Spinner D shows 5 and forgetting the 6 outcomes where Spinner C shows 5. Choosing 1/2 comes from counting 9 outcomes instead of 8, by listing the case where Spinner C shows 5 and Spinner D shows 5 twice over.
- (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.
- (a) 5/36 — Method: list every result of the two dice as an ordered pair, first score then second score, count the pairs that give the total asked for and divide by how many pairs the list holds. Working: each dice can show 6 scores, so there are 6 × 6 = 36 equally likely ordered pairs. The pairs whose scores add to 6 are (1, 5), (2, 4), (3, 3), (4, 2) and (5, 1), which is 5 pairs out of the 36. Answer: the probability is 5/36. The distractors: 4/36 comes from listing (1, 5), (5, 1), (2, 4) and (4, 2) and leaving (3, 3) out, because a double does not look like a pair that can be turned round; 5/12 comes from finding the 5 pairs but taking the number of possible results to be 6 + 6 = 12, adding the two dice instead of multiplying them; 1/11 comes from treating the eleven possible totals 2, 3, 4 and so on up to 12 as equally likely, so that a total of 6 is one result out of eleven.
- (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.
- (a) 1/10 — There are 4 × 5 = 20 equally likely outcomes in total. The pairs whose product is 12 are Spinner A showing 3 with Spinner B showing 4, and Spinner A showing 4 with Spinner B showing 3, which is 2 outcomes, giving a probability of 2/20 = 1/10. Choosing 1/20 comes from finding only one of the two pairs, (3, 4), and missing (4, 3) as a separate outcome. Choosing 1/8 comes from using 16 as the total number of outcomes, 4 × 4, forgetting that Spinner B has 5 sections rather than 4. Choosing 1/5 comes from listing the factor pairs of 12 as 2 × 6 and 3 × 4 and counting each one in both orders, (2, 6), (6, 2), (3, 4) and (4, 3), giving 4 outcomes out of 20 without checking that neither spinner has a 6 on it.
- (a) 3/8 — There are 4 × 6 = 24 equally likely outcomes. The outcomes with no 3 at all have Spinner E showing 1, 2 or 4 and Spinner F showing 1, 2, 4, 5 or 6, giving 3 × 5 = 15 outcomes. So the outcomes with at least one 3 are 24 − 15 = 9, and the probability is 9/24 = 3/8. Choosing 5/12 comes from adding the two individual probabilities of a 3, 1/4 + 1/6, which counts the outcome where both spinners show 3 twice over. Choosing 1/4 comes from only counting the case where Spinner E shows 3, and forgetting the outcomes where Spinner F shows 3 instead. Choosing 5/8 comes from working out the probability of getting no 3 at all, 15/24 = 5/8, and giving that as the final answer instead of subtracting it from 1.
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