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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.

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Probabilities of exhaustive events sum to one worksheet — GCSE Foundation

Maths · GCSE Foundation · P4 · Probabilities of exhaustive events sum to one · Non-calculator · 15 questions

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1.At a quiz night, each team's result is a win, a draw or a loss, and no team can have more than one of these results. The probability that a team wins is 42% and the probability that a team draws is 18.5%. Work out the probability, as a percentage, that a team does not win and does not draw.
(a)58.0%
(b)60.5%
(c)39.5%
(d)81.5%
2.A biased six-sided dice is rolled once. The probability that it lands on 6 is 0.25. The other five scores are all equally likely. Work out the probability that it lands on 3.
(a)0.167
(b)0.150
(c)0.125
(d)0.750
3.The probability that it rains in Manchester tomorrow is 1/12. Work out the probability that it does not rain in Manchester tomorrow.
(a)11/11
(b)11/12
(c)1/11
(d)1/12
4.At a tombola stall, each ticket wins a toy, wins a sweet or wins nothing, and no ticket can win more than one of these. The probability that a ticket wins a toy is 0.15 and the probability that it wins a sweet is 0.4. Work out the probability that a ticket wins neither a toy nor a sweet.
(a)0.55
(b)0.45
(c)0.6
(d)0.85
5.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.
(a)0.75
(b)0.14
(c)0.05
(d)0.25
6.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.
(a)0.15
(b)0.30
(c)0.60
(d)0.45
7.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.
(a)0.4
(b)0.6
(c)0.2
(d)0.3
8.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.
(a)8
(b)40
(c)24
(d)32
9.The probability that the Hughes family go away during the summer holidays is 3/7. Work out the probability that they do not go away during the summer holidays.
(a)4/7
(b)1/7
(c)3/7
(d)1/2
10.The probability that Oliver's train is delayed on any morning is 0.3. Work out the probability that his train is not delayed.
(a)0.5
(b)0.97
(c)0.3
(d)0.7
11.A charity raffle has three types of ticket: winning, near-miss and losing, and every ticket is exactly one of these. The probability that a ticket is winning is 1/8 and the probability that it is a near-miss is 1/4. Work out the probability that a ticket is losing.
(a)5/8
(b)3/8
(c)3/4
(d)7/8
12.The probability that a machine completes a task successfully is 0.85. Work out the probability that the machine does not complete the task successfully.
(a)0.85
(b)0.15
(c)0.25
(d)0.5
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.
(a)12
(b)39
(c)27
(d)21
14.A spinner lands on red, on blue or on green, and it cannot land on more than one colour at a time. The probability that it lands on red is 0.3 and the probability that it lands on blue is 0.5. Work out the probability that it lands on red or on blue.
(a)0.8
(b)0.2
(c)0.15
(d)0.4
15.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.
(a)4
(b)8
(c)20
(d)5

Answer key

(c) 39.5% — Win, draw and lose are mutually exclusive and exhaustive, so their probabilities sum to 100%: 42% + 18.5% = 60.5% is the percentage that wins or draws. 100% − 60.5% = 39.5% is the percentage that neither wins nor draws. Adding 42% and 18.5% and stopping there, 60.5%, is the probability of winning or drawing, not of neither. Subtracting only the 42% from 100% gives 58.0%, ignoring the draw percentage. Subtracting only the 18.5% from 100% gives 81.5%, ignoring the win percentage.
(b) 0.150 — The six scores are exhaustive, so their probabilities sum to 1. The probability of not landing on 6 is 1 − 0.25 = 0.75, and this is shared equally between the other five scores, so each has probability 0.75 ÷ 5 = 0.150. Giving 0.750 as the answer stops after finding the probability of not landing on 6, without sharing it out between the five remaining scores. Dividing 0.75 by 6 instead of by 5 gives 0.125, wrongly including the score of 6 among the equally likely scores. Ignoring the bias completely and dividing 1 by all six scores gives 1 ÷ 6 = 0.167.
(b) 11/12 — Method: raining and not raining are the only two outcomes, so their probabilities add to 1; subtract the given probability from 1. Working: writing 1 as 12/12 gives 12/12 − 1/12, and only the numerators are subtracted, 12 − 1 = 11. Answer: 11/12, close to the right-hand end of the 0 to 1 scale because rain is unlikely. The distractors: 1/12 comes from giving back the probability that it does rain; 1/11 comes from subtracting the 1 from the denominator instead of subtracting the fraction from 1; 11/11 comes from subtracting 1 from the numerator and from the denominator of 12/12 rather than from the numerator alone.
(b) 0.45 — Winning a toy, winning a sweet and winning neither are mutually exclusive and exhaustive, so their probabilities sum to 1. P(toy) + P(sweet) = 0.15 + 0.4 = 0.55. P(neither) = 1 − 0.55 = 0.45. Adding 0.15 and 0.4 and stopping there gives 0.55, which is the probability of winning a toy or a sweet, not of winning neither. Subtracting only 0.15 from 1 gives 0.85, and ignores the sweet probability entirely. Subtracting only 0.4 from 1 gives 0.6, and ignores the toy probability entirely.
(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) 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.
(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.
(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) 4/7 — Method: going away and not going away are the only two possibilities, so the two probabilities form an exhaustive set and add to 1; subtract the given probability from 1. Working: writing 1 as 7/7 gives 7/7 − 3/7, and 7 − 3 = 4 sevenths. Answer: 4/7. The distractors: 3/7 comes from giving back the probability that the family do go away; 1/7 comes from taking the 1 in '1 − 3/7' as a numerator and writing it over the denominator 7; 1/2 comes from assuming that going away and not going away must be equally likely because there are only two possibilities.
(d) 0.7 — Method: delayed and not delayed are the only two outcomes, so the two probabilities form an exhaustive set and add to 1; subtract the given probability from 1. Working: 1 − 0.3, written as 1.0 − 0.3 with the decimal points lined up. Answer: 0.7. The distractors: 0.3 comes from giving back the probability that the train is delayed instead of its complement; 0.97 comes from lining the decimal points up wrongly and working out 1.00 − 0.03; 0.5 comes from assuming that delayed and not delayed must be equally likely because there are only two outcomes.
(a) 5/8 — Winning, near-miss and losing are exhaustive, so the three probabilities sum to 1. Writing 1/4 as 2/8 so every fraction has the same denominator, 1 − 1/8 − 2/8 = 8/8 − 1/8 − 2/8 = 5/8. Subtracting only the winning probability and forgetting the near-miss probability gives 1 − 1/8 = 7/8. Subtracting only the near-miss probability and forgetting the winning probability gives 1 − 1/4 = 3/4. Adding the two given probabilities and stopping there gives 1/8 + 2/8 = 3/8, the probability that a ticket is winning or a near-miss, not the probability that it is losing.
(b) 0.15 — Method: success and failure are the only two outcomes of the task, so they form an exhaustive set and their probabilities add to 1; subtract the given probability from 1. Working: writing 1 as 1.00 so that both numbers have two decimal places gives 1.00 − 0.85; exchanging once, the hundredths give 10 − 5 = 5 and the tenths, now 9, give 9 − 8 = 1. Answer: 0.15, close to the left-hand end of the 0 to 1 scale because the machine nearly always succeeds. The distractors: 0.85 comes from giving back the probability of success instead of the probability of failure; 0.25 comes from taking each decimal column from 10 on its own in 1.00 − 0.85, writing 10 − 5 = 5 in the hundredths and 10 − 8 = 2 in the tenths instead of reducing the tenths to 9 after the exchange; 0.5 comes from assuming that success and failure must be equally likely because there are only two outcomes.
(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.
(a) 0.8 — Method: the spinner cannot land on red and blue at the same time, so the two events are mutually exclusive and their probabilities are added. Working: 0.3 + 0.5, lining the decimal points up. Answer: 0.8, which also tells you that the remaining colour, green, has probability 0.2 because the three must add to 1. The distractors: 0.2 comes from subtracting 0.3 from 0.5 instead of adding the two probabilities; 0.15 comes from multiplying 0.3 by 0.5 instead of adding them; 0.4 comes from finding the mean of 0.3 and 0.5 rather than their total.
(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.

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