Printable · GCSE Foundation · ages 14-16
Probabilities of exhaustive events sum to one worksheet — GCSE Foundation
Fifteen questions on "probabilities of exhaustive events sum to one" — DfE statement P4. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.
Probabilities of exhaustive events sum to one worksheet — GCSE Foundation
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- 1.At a school fete, each raffle ticket wins a prize, wins a voucher or wins nothing, and no ticket can win more than one of these. The probability that a ticket wins a prize is 0.4 and the probability that it wins a voucher is 0.35. Work out the probability that a ticket wins a prize or a voucher.
- 2.A bag contains 2 white counters and 8 black counters. One counter is taken at random from the bag. Work out the probability that the counter is not black.
- 3.A spinner can land on red, blue, green or yellow. The probability that it lands on red is 0.24 and the probability that it lands on yellow is 0.16. The probabilities of landing on blue and on green are equal, and each is called x. Work out the value of x.
- 4.A bag contains counters that are red, blue or green only. The probability that a counter taken at random is not red is 0.8. The bag contains 25 counters in total. Work out how many of the counters are red.
- 5.A spinner lands on one of four colours: amber, black, cyan or damson. The probability it lands on amber is 0.18, on black is 0.22 and on cyan is 0.24. Work out the probability that it lands on amber or damson.
- 6.A bag contains beads that are exactly one of red, white or black. The probability of taking a red bead is 3/10 and the probability of taking a white bead is 1/4. Work out the probability of taking a bead that is red or white.
- 7.The probability that a plant grown from a seed packet germinates is 7/10. Work out the probability that it does not germinate.
- 8.At a coffee shop, each customer buys tea, coffee, water or juice, and never more than one of these. The probability that a customer buys tea is 0.36 and the probability that they buy juice is 0.04. The probability that a customer buys coffee is three times the probability that they buy water. Work out the probability that a customer buys water.
- 9.At a school car park, each car is a saloon, an estate or a hatchback, and no car is more than one of these. Among 200 cars, the probability that a car chosen at random is a saloon is 0.28 and the probability that it is an estate is 0.37. Work out how many of the 200 cars are hatchbacks.
- 10.At a fun run, each runner finishes, retires or is disqualified, and cannot do more than one of these. The probability that a runner finishes is 68.5% and the probability that a runner retires is 24.75%. Work out the probability, as a percentage, that a runner is disqualified.
- 11.A box contains 1 white bead, 4 purple beads and 4 red beads. One bead is taken at random from the box. Work out the probability that the bead is white or purple.
- 12.A spinner can land on red, blue, green or yellow, and it cannot land on more than one colour. The probability that it lands on red is 0.05 and the probability that it lands on yellow is 0.35. The probability that it lands on blue is twice the probability that it lands on green. Work out the probability that it lands on green.
- 13.A jar contains counters that are pink, gold or grey only. The probability that a counter taken at random is pink is 0.2 and the probability that it is gold is 0.45. The jar contains 60 counters in total. Work out the number of grey counters.
- 14.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.
- 15.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.
Answer key
- (a) 0.75 — Method: a ticket cannot win both a prize and a voucher, so the two events are mutually exclusive and their probabilities are added. Working: 0.4 + 0.35, lining the decimal points up, gives 0.40 + 0.35. Answer: 0.75. The distractors: 0.25 comes from going one step too far and giving the probability that a ticket wins nothing, which is what is left when the total is taken from 1; 0.14 comes from multiplying 0.4 by 0.35 instead of adding them; 0.05 comes from subtracting 0.35 from 0.4 instead of adding them.
- (a) 1/5 — Method: find the probability that the counter is black, then use the fact that an event and its complement add to 1. Working: there are 2 + 8 = 10 counters, so P(black) = 8/10 = 4/5; the complement is 1 − 4/5, and writing 1 as 5/5 gives 5/5 − 4/5. Answer: 1/5. The distractors: 4/5 comes from giving the probability that the counter is black instead of its complement; 1/2 comes from assuming the two colours are equally likely because there are only two of them; 1/4 comes from writing the 2 white counters over the 8 black counters rather than over all 10 counters.
- (a) 0.30 — Red, blue, green and yellow are exhaustive, so all four probabilities sum to 1: 0.24 + 0.16 + x + x = 1, so 2x + 0.40 = 1, giving 2x = 0.60 and x = 0.30. Stopping at 2x = 0.60 without dividing by 2 leaves 0.60, the combined probability of both blue and green together, not the value of x on its own. Sharing the 0.60 across all four colours instead of just the two unknown ones gives 0.60 ÷ 4 = 0.15. Leaving out the 0.16 for yellow gives 2x + 0.24 = 1, so 2x = 0.76 and x = 0.38.
- (d) 5 — Being red and not being red are exhaustive, so their probabilities sum to 1: the probability of red is 1 − 0.8 = 0.2. The number of red counters is 0.2 × 25 = 5. Using 0.8 directly as the probability of red, without taking the complement, gives 0.8 × 25 = 20 — the number of counters that are NOT red. Sharing the 25 counters equally between the three colours, ignoring the given probability altogether, gives 25 ÷ 3 ≈ 8. Misreading the total as 20 counters instead of 25 gives 0.2 × 20 = 4.
- (a) 0.54 — The four colours are exhaustive, so all four probabilities sum to 1: the probability of damson is 1 − 0.18 − 0.22 − 0.24 = 0.36. Amber and damson cannot both happen on one spin, so the probability of amber or damson is 0.18 + 0.36 = 0.54. Stopping after finding the probability of damson alone, without adding the probability of amber, gives 0.36. Adding the three given probabilities together, 0.18 + 0.22 + 0.24 = 0.64, and treating that total as the answer never finds the probability of damson at all. Subtracting only the probability of amber from 1, 1 − 0.18 = 0.82, ignores black, cyan and damson completely.
- (c) 11/20 — 'Red or white' combines two mutually exclusive events, so add their probabilities: 3/10 = 6/20 and 1/4 = 5/20, giving 6/20 + 5/20 = 11/20. Multiplying the two probabilities instead of adding them, 3/10 × 1/4, gives 3/40, which would be the probability of red and white together, not red or white — and a bead can't be both colours. Subtracting the sum from 1, 1 − 11/20 = 9/20, gives the probability of the bead being black instead of red or white. Converting 1/4 as 4/20 instead of 5/20 (dividing 20 by 4 but forgetting to scale the numerator) gives 6/20 + 4/20 = 1/2.
- (a) 3/10 — Germinating and not germinating are exhaustive, so their probabilities sum to 1: 1 − 7/10 = 3/10. Writing 7/10 again gives the probability that the plant DOES germinate, not its complement. Putting the difference 10 − 7 = 3 over the original numerator instead of the original denominator gives 3/7. Inverting the correct answer, swapping its numerator and denominator, gives 10/3, which is impossible as a probability since it is greater than 1.
- (a) 0.15 — Let P(water) = x, so P(coffee) = 3x. The four outcomes are exhaustive: 0.36 + 0.04 + x + 3x = 1, so 0.4 + 4x = 1, giving 4x = 0.6 and x = 0.15. So P(water) = 0.15. Reporting 3x, the coffee probability, instead of water gives 0.45. Splitting the remaining 0.6 evenly between coffee and water, ignoring the 3:1 ratio, gives 0.30. Stopping once the remaining probability 0.6 is found, without dividing by the four equal shares, gives 0.60.
- (b) 70 — Saloon, estate and hatchback are exhaustive, so their probabilities sum to 1: the probability of a hatchback is 1 − 0.28 − 0.37 = 0.35. The number of hatchbacks is 0.35 × 200 = 70. Treating the SUM of the other two probabilities, 0.28 + 0.37 = 0.65, as the probability of a hatchback instead of its complement gives 0.65 × 200 = 130. Multiplying the correct probability, 0.35, by 100 instead of the 200 cars actually surveyed gives 35. Averaging the two given probabilities, (0.28 + 0.37) ÷ 2 = 0.325, instead of subtracting them from 1, and then multiplying by 200 gives 65.
- (a) 6.75% — Finishing, retiring and being disqualified are exhaustive, so the three percentages sum to 100%: 100% − 68.5% − 24.75% = 6.75%. Adding the two given percentages instead of subtracting them from 100% gives 68.5% + 24.75% = 93.25%, the combined probability of finishing or retiring, not of being disqualified. Subtracting only the retiring percentage from 100% and forgetting the finishing percentage gives 100% − 24.75% = 75.25%. Subtracting only the finishing percentage and forgetting the retiring percentage gives 100% − 68.5% = 31.50%.
- (a) 5/9 — Method: a bead cannot be two colours at once, so white and purple are mutually exclusive and their probabilities are added over the common total. Working: there are 1 + 4 + 4 = 9 beads, so P(white) = 1/9 and P(purple) = 4/9; adding gives 1/9 + 4/9, and 1 + 4 = 5 ninths. Answer: 5/9. The distractors: 4/9 comes from giving the probability of a purple bead alone and forgetting to include the white one; 5/8 comes from counting 5 favourable beads but using 8 as the total, leaving the single white bead out of the count of the box; 5/18 comes from adding 1/9 and 4/9 by adding the denominators as well as the numerators.
- (c) 0.2 — Let P(green) = x, so P(blue) = 2x. Red, blue, green and yellow are exhaustive: 0.05 + 0.35 + x + 2x = 1, so 0.4 + 3x = 1, giving 3x = 0.6 and x = 0.2. So P(green) = 0.2. Splitting the remaining 0.6 evenly between blue and green, ignoring the 2:1 ratio, gives 0.3. Working out x correctly but then reporting 2x, the probability of blue, gives 0.4. Stopping after finding that blue and green together account for 0.6, without dividing by the three equal shares of x, gives 0.6.
- (d) 21 — Pink, gold and grey cover every counter, so their probabilities sum to 1: P(grey) = 1 − 0.2 − 0.45 = 0.35. Number of grey counters = 0.35 × 60 = 21. Multiplying the pink probability by 60 gives 12, the number of pink counters, not grey. Multiplying the gold probability by 60 gives 27, the number of gold counters. Adding 0.2 and 0.45 without subtracting from 1 gives 0.65, and 0.65 × 60 = 39 counts the pink and gold counters together, not the grey ones.
- (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.
- (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.
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