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.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.
- 2.The probability that Priya's dentist appointment overruns is 0.35. Work out the probability that it does not overrun.
- 3.The probability that it rains in Manchester tomorrow is 1/12. Work out the probability that it does not rain in Manchester tomorrow.
- 4.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.
- 5.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.
- 6.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.
- 7.A card is taken at random from an ordinary pack of 52 playing cards. The probability that the card is a diamond is 1/4. Work out the probability that the card is not a diamond.
- 8.A card is taken at random from an ordinary pack of 52 playing cards. The pack contains 4 aces and 4 kings. Work out the probability that the card is an ace or a king.
- 9.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.
- 10.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.
- 11.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.
- 12.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.
- 13.In a game, Freddie says the probability of winning is 0.45, the probability of drawing is 0.3 and the probability of losing is 0.35, and that all three of his probabilities are correct. Which statement about Freddie's probabilities is correct?
- 14.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.
- 15.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.
Answer key
- (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.
- (c) 0.65 — Overrunning and not overrunning are exhaustive: between them they cover every outcome, so their probabilities sum to 1. Work out 1 − 0.35 = 0.65. Writing 0.35 again is the probability that the appointment overruns, not its complement — the subtraction was never done. Adding instead of subtracting gives 1 + 0.35 = 1.35, which cannot be a probability at all. Subtracting each digit from 10 instead of borrowing from the 1, so 10 − 3 = 7 tenths and 10 − 5 = 5 hundredths, gives 0.75.
- (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) 0.15 — 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.
- (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.
- (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.
- (b) 3/4 — Method: a card either is a diamond or is not a diamond, so those two outcomes form an exhaustive set and their probabilities add to 1; subtract the given probability from 1. Working: P(diamond) = 1/4, so P(not a diamond) = 1 − 1/4; writing 1 as 4/4 gives 4/4 − 1/4. Answer: 3/4. The distractors: 1/4 comes from giving back the probability that the card is a diamond instead of its complement; 1/2 comes from reading 'not a diamond' as 'not a red card' and halving the pack; 3/52 comes from doing the subtraction 4 − 1 = 3 on the suits but then writing that 3 over the 52 cards in the pack instead of over the 4 suits.
- (b) 8/52 — Method: a card cannot be an ace and a king at the same time, so the two events are mutually exclusive and their probabilities are added, keeping the denominator the same. Working: P(ace) = 4/52 and P(king) = 4/52, so P(ace or king) = 4/52 + 4/52, and 4 + 4 = 8 fifty-seconds. Answer: 8/52. The distractors: 4/52 comes from giving the probability of just one of the two events and forgetting to add the other; 16/52 comes from multiplying the two counts, 4 × 4, instead of adding them; 1/52 comes from giving the probability of one particular named card rather than any of the eight.
- (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.
- (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) 0.75 — 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.
- (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.
- (a) No — the three probabilities sum to 1.10, over 1. — Winning, drawing and losing are exhaustive and mutually exclusive, so their probabilities must sum to exactly 1. Adding Freddie's three values gives 0.45 + 0.3 + 0.35 = 1.10, which is more than 1, so his probabilities cannot all be correct: 'No — the three probabilities sum to 1.10, over 1.' Checking only that each value lies between 0 and 1 accepts them as 'Yes — each probability lies between 0 and 1' without ever adding the three together. Judging by which outcome sounds most likely leads to 'Yes — winning has the highest single probability', which never checks the total either. Noting that a runner cannot win, draw and lose at once, and treating that alone as enough, gives 'Yes — the three outcomes are mutually exclusive' — but mutually exclusive outcomes that are also exhaustive must still sum to 1, and 1.10 does not.
- (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.
- (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.
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