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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.

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Probability worksheet — GCSE Foundation

Maths · GCSE Foundation · Probability · Non-calculator · 15 questions

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1.The probability that Kofi passes his driving test on any attempt is 0.6, and each attempt is independent of the others. Work out the probability that he fails both his first two attempts.
(a)0.16
(b)0.36
(c)0.24
(d)0.4
2.A fair spinner has 10 equal sections, numbered 1 to 10. The spinner is spun 150 times. Work out how many times you would expect it to land on a number greater than 7.
(a)60
(b)50
(c)45
(d)105
3.Priya spins a fair spinner with P(red) = 0.3, and separately flips a fair coin with P(heads) = 0.5. Using a tree diagram, work out the probability that she gets red AND heads.
(a)0.5
(b)0.8
(c)0.15
(d)0.65
4.Two ordinary fair dice are rolled. Work out the probability that both dice show the same score.
(a)1/3
(b)1/6
(c)1/36
(d)1/2
5.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.
(a)1/5
(b)1/2
(c)4/5
(d)1/4
6.A survey of 160 employees at a company recorded whether they cycle to work, using a frequency tree. The first branch splits them into 90 who work full-time and 70 who work part-time. Of the full-time employees, 27 cycle to work. Of the part-time employees, 14 cycle to work. Work out the probability that an employee, chosen at random from the 160, cycles to work. Give your answer as a fraction in its simplest form.
(a)27/160
(b)41/160
(c)1/5
(d)41/90
7.A charity tombola has 30 tickets, 6 of which win a prize. Priya buys a ticket at random and does not return it. Her friend Tom then buys a second ticket at random from the remaining tickets. Work out the probability that both Priya and Tom win a prize.
(a)1/29
(b)11/59
(c)1/25
(d)1/30
8.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.
(a)1/10
(b)1/8
(c)1/5
(d)1/20
9.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
10.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.
(a)16
(b)2
(c)3
(d)4
11.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
12.An ordinary fair dice is rolled once. Set A is the set of even scores and set B is the set of scores greater than 4. Work out the probability that the score rolled is in set A or in set B or in both.
(a)2/3
(b)1/2
(c)1/6
(d)5/6
13.An ordinary six-sided dice, numbered 1 to 6, is rolled 30 times and lands on a 6 seven times. Ravi says the theoretical probability of rolling a 6 and the relative frequency of rolling a 6 in this trial are the same number. Is Ravi right?
(a)Yes — relative frequency always equals theory.
(b)Yes — the theoretical probability has now become 7/30.
(c)No — 7/30 is the relative frequency; theory stays 1/6.
(d)Neither can be found — 30 rolls is too few to tell.
14.A bag contains 3 red beads and 7 blue beads. A bead is taken out at random and not put back. A second bead is then taken out at random. Work out the probability that the first bead is red and the second bead is also red.
(a)3/50
(b)5/19
(c)9/100
(d)1/15
15.The probability that a seed fails to germinate is 1/5. Three seeds are planted independently. Work out the probability that at least one of the three seeds germinates.
(a)12/125
(b)4/5
(c)64/125
(d)124/125

Answer key

(a) 0.16 — The probability that Kofi fails a single attempt is 1 − 0.6 = 0.4. Since the attempts are independent, the probability he fails both is 0.4 × 0.4 = 0.16. Choosing 0.36 comes from squaring the probability of PASSING instead, 0.6 × 0.6 = 0.36, which is the probability of passing both attempts, not failing both. Choosing 0.4 comes from giving the probability of failing just one attempt, forgetting to combine two attempts. Choosing 0.24 comes from multiplying the fail probability by the pass probability, 0.4 × 0.6 = 0.24, mixing up passing and failing between the two attempts.
(c) 45 — Three of the ten numbers (8, 9 and 10) are greater than 7, so the probability is 3/10, and 150 × 3/10 = 45. Writing 105 is wrong because 150 × 7/10 = 105 uses the seven numbers that are NOT greater than 7 (1 to 7), the opposite of what is asked. Writing 60 is wrong because it counts 7, 8, 9 and 10 as four numbers greater than 7, wrongly including 7 itself: 150 × 4/10 = 60. Writing 50 is wrong because 150 ÷ 3 = 50 divides by the count of favourable numbers instead of multiplying by the correct fraction of the spinner. The expected number of spins landing on a number greater than 7 is 45.
(c) 0.15 — Method: for two independent events, multiply along the branches of the tree to find the probability of both outcomes happening together. Working: P(red and heads) = P(red) × P(heads) = 0.3 × 0.5 = 0.15. Answer: 0.15. Watch out: adding the two probabilities, 0.3 + 0.5 = 0.8, does not give the probability of both — probabilities along one path of a tree are multiplied, not added. Writing down 0.5 ignores the spinner altogether and gives only the coin's probability. And writing down 0.65 is the probability of red OR heads, which is 0.3 + 0.5 − 0.15 = 0.65, a different question from the one asked here.
(b) 1/6 — Method: list the ordered pairs where the two scores match, and divide by the 36 equally likely pairs. Working: the matching pairs are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6), which is 6 pairs out of 36, cancelling down to 1/6. Answer: 1/6. Watch out: writing down 1/36 finds the probability of one particular double, such as (6, 6), rather than any double at all. Guessing 1/2 treats 'same' and 'different' as equally likely events, when there are only 6 matching pairs against 30 non-matching ones. And writing down 1/3 comes from listing each double twice, once for each order of the two dice, giving 12 pairs out of 36 — but (1, 1) is a single outcome, and swapping the two dice over does not produce a second one.
(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) 41/160 — In total, 27 + 14 = 41 of the 160 employees cycle to work, so the probability is 41/160 (41 and 160 share no common factor, so this is already in its simplest form). Writing 27/160 is wrong because it only counts the full-time cyclists and leaves out the 14 part-time cyclists. Writing 41/90 is wrong because it uses the full-time total (90) as the denominator instead of the whole survey (160). Writing 1/5 is wrong because it only uses the part-time branch, simplifying 14/70 to 1/5 and ignoring the full-time cyclists completely. The probability is 41/160.
(a) 1/29 — The probability that Priya's ticket wins is 6/30. Since her ticket is not returned, there are now only 5 winning tickets left out of 29 tickets in total, so the probability that Tom's ticket also wins is 5/29. Multiplying these, 6/30 × 5/29 = 30/870 = 1/29. A candidate who answers 1/25 has treated Priya's ticket as returned, using 6/30 twice. A candidate who answers 1/30 has correctly reduced the winning tickets to 5 for Tom but forgotten to reduce the total number of tickets, using 5/30 instead of 5/29. A candidate who answers 11/59 has added the numerators and added the denominators, (6+5)/(30+29), instead of multiplying.
(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.
(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.
(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) 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) 2/3 — Method: 'in A or in B' means every score that belongs to at least one of the two sets; a score that belongs to both is still only one outcome, so it is listed once. Working: the even scores are 2, 4 and 6; the scores greater than 4 are 5 and 6. Listing the scores that appear in either set gives 2, 4, 5 and 6, with 6 written once. That is 4 of the 6 faces, or 4/6. Answer: the probability is 2/3. The distractors: 5/6 comes from adding the sizes of the two sets, 3 + 2, so that the score 6 is counted in both and appears twice; 1/2 comes from using the even scores alone; 1/6 comes from giving the probability that the score is in both sets, which is the single score 6, rather than in either of them.
(c) No — 7/30 is the relative frequency; theory stays 1/6. — The theoretical probability of rolling a 6 on an ordinary dice is fixed at 1/6, worked out from the number of equally likely outcomes, and does not change however the dice is actually rolled. The relative frequency from this trial is 7/30, found from what happened in these particular 30 rolls. Since 7/30 and 1/6 are different numbers, the correct statement is 'No — 7/30 is the relative frequency; theory stays 1/6.' Assuming the two values must always match because they describe the same event gives 'Yes — relative frequency always equals theory.' Believing that an observed result redefines the theoretical probability gives 'Yes — the theoretical probability has now become 7/30.' Refusing to work out either value at all gives 'Neither can be found — 30 rolls is too few to tell', which ignores that both numbers CAN be calculated from the information given.
(d) 1/15 — The probability that the first bead is red is 3/10. Since the first bead is not put back, there are now only 2 red beads left out of 9 beads in total, so the probability that the second bead is also red is 2/9. Multiplying these, 3/10 × 2/9 = 6/90 = 1/15. A candidate who answers 9/100 has treated the beads as replaced, using 3/10 twice. A candidate who answers 3/50 has correctly reduced the red count to 2 for the second pick but forgotten that the total also falls to 9, using 2/10 instead. A candidate who answers 5/19 has added the numerators and added the denominators, (3+2)/(10+9), instead of multiplying.
(d) 124/125 — The probability that a seed germinates is 1 − 1/5 = 4/5, so the probability that all three seeds fail to germinate is 1/5 × 1/5 × 1/5 = 1/125. The probability that at least one germinates is 1 − 1/125 = 124/125. Choosing 4/5 comes from giving the probability that a single seed germinates, forgetting to combine all three seeds. Choosing 64/125 comes from working out the probability that ALL three seeds germinate, 4/5 × 4/5 × 4/5 = 64/125, instead of at least one. Choosing 12/125 comes from working out the probability that EXACTLY one seed germinates, 3 × 4/5 × 1/5 × 1/5 = 12/125, instead of at least one.

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