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.
Probability worksheet — GCSE Foundation
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- 1.A fair six-sided dice is rolled once. Write down the probability that the score is an even number.
- 2.80 people are asked whether they drink tea and whether they drink coffee, and the results are recorded on a frequency tree. The first pair of branches splits the 80 people into 50 tea drinkers and 30 non-tea-drinkers. Of the 50 tea drinkers, 32 also drink coffee. Of the 30 non-tea-drinkers, 6 drink coffee. Work out n(T ∩ C), the number of people who drink both tea and coffee.
- 3.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.
- 4.A bag contains 5 letter tiles spelling the word MATHS, with one letter on each tile. Two tiles are drawn at random, one after the other, without the first tile being put back. Work out the probability that both tiles drawn are consonants.
- 5.A raffle has two independent draws. In the first, a winning ticket is picked at random from 5 red tickets and 3 blue tickets. In the second, a winning number is picked using a fair spinner with 4 equal sections, numbered 1 to 4. Work out the probability that the first winning ticket is blue and the second winning number is greater than 2.
- 6.A tree diagram shows two rounds of a game, played independently. On each round, the probability of losing is 0.7 and the probability of winning is 0.3. Work out the probability of losing both rounds.
- 7.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.
- 8.A factory finds that the probability a randomly chosen light bulb is defective is 0.035. In a batch of 4,000 bulbs, work out how many bulbs you would expect to work correctly.
- 9.Ellie drops a bottle top and records whether it lands open end up. In her first 20 drops it lands open end up 13 times. She carries on, and after 200 drops in total it has landed open end up 84 times. Work out the best estimate of the probability that the bottle top lands open end up.
- 10.A pupil must choose 2 different subjects at random from these five: maths, physics, biology, chemistry and history. Work out the probability that the two subjects chosen are maths and biology.
- 11.A phone network sends automatic text alerts to customers. On average, 1,500 alerts are sent each day, and the probability that a customer replies 'STOP' to an alert is 0.18. Work out how many replies of 'STOP' the network should expect over a 30-day month.
- 12.A bag contains 45 sweets. 18 of the sweets are lemon flavour and the rest are orange. A sweet is taken at random, its flavour is recorded, and it is put back in the bag. This is repeated 200 times. Work out how many times you would expect an orange sweet to be taken.
- 13.A bag contains one red counter and one blue counter. Josh takes a counter, notes its colour, puts it back, then takes a counter again. He draws a tree diagram to show every possible pair of colours. Work out how many outcomes are on his tree diagram.
- 14.Grace rolls two ordinary fair dice at the same time and works out the total of the two scores. Work out the probability that the total is greater than 10.
- 15.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.
Answer key
- (b) 1/2 — Method: for equally likely outcomes the probability is the number of favourable outcomes divided by the total number of outcomes, written in its simplest form. Working: the dice can show 1, 2, 3, 4, 5 or 6, so there are 6 equally likely scores; the even scores are 2, 4 and 6, which is 3 of them, giving 3/6. Answer: 3/6 cancels to 1/2, which sits halfway along the 0 to 1 probability scale. The distractors: 1/3 comes from listing only 2 and 4 as the even scores and writing 2/6; 1/6 comes from giving the probability of one particular even score, such as the 2, rather than any even score; 2/3 comes from listing the even numbers as 0, 2, 4 and 6 and writing 4/6, forgetting that a dice has no 0 on it.
- (d) 32 — The frequency tree already shows the tea-and-coffee branch directly: of the 50 tea drinkers, 32 also drink coffee, so n(T ∩ C) = 32. Adding both coffee branches together, 32 + 6 = 38, gives n(C), the total number of coffee drinkers, not just those who also drink tea. Using the non-tea branch's figure, 6, describes people who drink coffee but NOT tea. Subtracting to get 50 − 32 = 18 finds the tea drinkers who do NOT drink coffee, the opposite region to the one asked for.
- (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.
- (d) 3/5 — The first tile drawn is a consonant with probability 4/5, since 4 of the 5 letters in MATHS are consonants. Since it is not replaced, 3 consonants remain out of the 4 tiles left, so the second tile is a consonant with probability 3/4. Multiplying gives 4/5 × 3/4 = 3/5. Choosing 4/5 comes from only using the probability for the first draw and forgetting the second draw changes once a tile is removed. Choosing 3/4 comes from only using the probability for the second draw on its own. Choosing 16/25 comes from treating the two draws as independent, using 4/5 × 4/5 = 16/25, which would only be correct if the first tile were replaced before the second draw.
- (d) 3/16 — Method: work out each draw's own probability first, then multiply them together since the two draws are independent. Working: there are 8 tickets in all, 3 of them blue, so P(blue) = 3/8. P(spinner number greater than 2) = 2/4 = 1/2, since 3 and 4 qualify. Multiplying gives 3/8 × 1/2, which comes to 3/16. Answer: 3/16. Watch out: writing down 3/8 stops after the first draw and never brings in the spinner at all. Writing down 1/2 does the opposite, using only the spinner and ignoring the ticket draw. And writing down 5/16 uses 5/8, the probability of a RED ticket, instead of 3/8 for blue — reading the wrong colour off the raffle.
- (c) 0.49 — The two rounds are independent, so multiply the probability of losing each round: 0.7 × 0.7 = 0.49. Choosing 0.7 comes from giving the probability of losing just one round, forgetting there are two rounds to combine. Choosing 1.4 comes from adding the two probabilities instead of multiplying, 0.7 + 0.7 = 1.4. Choosing 0.09 comes from using the probability of WINNING instead of losing, 0.3 × 0.3 = 0.09.
- (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.
- (a) 3,860 — The probability a bulb works correctly is the complement of being defective: 1 − 0.035 = 0.965. Expected number working correctly = 0.965 × 4,000 = 3,860. Using the probability of being defective instead of its complement gives 4,000 × 0.035 = 140, the expected number of DEFECTIVE bulbs, not working ones. Shifting the decimal point in the complement, using 0.0965 instead of 0.965, gives 4,000 × 0.0965 = 386. Assuming every bulb works, ignoring the 0.035 probability altogether, gives the full batch of 4,000.
- (d) 0.420 — Method: use the relative frequency worked out from the larger number of trials as the best estimate of the probability, since a bigger sample tends to sit closer to the true probability. Working: 84 out of 200 drops land open end up, so the relative frequency from the total is 84 ÷ 200 = 0.420. Answer: 0.420. Watch out: writing down 0.650 comes from 13 ÷ 20, using only the first, much smaller sample instead of the total. Writing down 0.535 comes from averaging 0.650 and 0.420, treating the 20-drop run and the 200-drop run as equally reliable instead of using the larger sample on its own. And writing down 0.580 comes from 1 − 0.420, working out the probability that the bottle top lands the other way up instead of open end up.
- (b) 1/10 — Method: list every possible pair of subjects systematically, so that no pair is missed and no pair is counted twice, then compare the number of successful pairs with the size of the list. Working: pairing maths with each of the other four gives 4 pairs, physics with each subject after it gives 3, biology gives 2 and chemistry gives 1, so there are 4 + 3 + 2 + 1 = 10 pairs. Exactly one of them is maths with biology. Answer: the probability is 1/10. The distractors: 1/5 comes from counting maths-then-biology and biology-then-maths as two separate successes while still dividing by the 10 unordered pairs; 1/15 comes from a list that also pairs each subject with itself, giving 15 entries instead of 10; 1/4 comes from working out the second step alone, that 1 of the 4 subjects left after maths is biology, without allowing for the chance that maths is picked at all.
- (b) 8,100 — First find the total number of alerts sent in the month: 1,500 × 30 = 45,000. Then apply the probability of a 'STOP' reply: 45,000 × 0.18 = 8,100. Stopping after finding only one day's expected replies, 1,500 × 0.18 = 270, forgets to scale up to the whole month. Multiplying the number of days by the probability instead of by the daily total of alerts gives 30 × 0.18 = 5.4, which rounds to 5. Shifting the decimal point in the probability, using 0.018 instead of 0.18, gives 45,000 × 0.018 = 810.
- (c) 120 — 27 of the 45 sweets are orange (45 − 18 = 27), so the probability of taking an orange sweet is 27/45 = 3/5, and 200 × 3/5 = 120. Writing 80 is wrong because 200 × 18/45 = 80 uses the LEMON sweets' fraction instead of orange. Writing 182 is wrong because it takes the 18 lemon sweets away from the 200 repeats (200 − 18 = 182), applying the ‘the rest are orange’ subtraction to the number of goes instead of to the 45 sweets in the bag. Writing 27 is wrong because it is simply the number of orange sweets in the bag — it has not been scaled up to account for the 200 repeats. The expected number of times an orange sweet is taken is 120.
- (d) 4 — Method: each of the first draw's 2 outcomes can be paired with each of the second draw's 2 outcomes, since the counter is put back before the second draw, so the tree has one branch for every combination. Working: 2 × 2 = 4 outcomes: red-red, red-blue, blue-red, blue-blue. Answer: 4. Watch out: writing down 2 lists only the colours of a single draw and never branches out to a second draw at all. Writing down 3 treats red-then-blue and blue-then-red as the same branch, when the tree diagram shows them as two separate paths, since the counter is put back and either colour could come first or second. And writing down 16 comes from working out 2 × 2 × 2 × 2, as though the counter were drawn four times instead of twice.
- (d) 3/36 — Method: list the results as ordered pairs, decide which totals satisfy the condition, count the pairs that give those totals and divide by the number of pairs there are. Working: there are 6 × 6 = 36 equally likely ordered pairs. Greater than 10 means a total of 11 or a total of 12. A total of 11 comes from (5, 6) and (6, 5); a total of 12 comes from (6, 6) alone, because both dice must show a 6. That is 2 + 1 = 3 pairs out of the 36. Answer: the probability is 3/36. The distractors: 2/36 comes from counting the two ways of making 11 and forgetting that 12 is greater than 10 as well; 4/36 comes from writing (6, 6) down twice, applying the rule that a pair can be turned round to a double that can only happen one way; 33/36 comes from reading the condition the wrong way round and giving the probability that the total is 10 or less.
- (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.
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