Printable · GCSE Higher · ages 14-16
Probability worksheet — GCSE Higher
Fifteen questions across the probability statements at Higher 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 Higher
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- 1.On a probability scale from 0 to 1, five descriptions are used: impossible (0), unlikely (between 0 and 0.5), evens (exactly 0.5), likely (between 0.5 and 1) and certain (1). The probability that a particular bus is late is 3/8. Which description best fits this probability?
- 2.Of the 70 students on the Year 10 geography trip, 42 are girls and the rest are boys. 24 of the girls and 10 of the boys brought a packed lunch. A student who did not bring a packed lunch is chosen at random. Work out the probability that this student is a girl.
- 3.A garage services 200 cars in one week. 120 of the cars are petrol cars and the rest are diesel cars. 30 of the petrol cars and 24 of the diesel cars fail the service. One of the cars that failed is picked at random. Work out the probability that it is a diesel car.
- 4.Bag A contains 3 red balls and 2 blue balls. Bag B contains 1 red ball and 3 blue balls. One ball is taken at random from each bag. Work out the probability that both balls are red.
- 5.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.
- 6.A survey of 60 readers records: 21 read only fiction, 17 read only non-fiction, 14 read both, and the rest read neither. Work out the probability that a randomly chosen reader reads exactly one of the two types, giving your answer as a fraction in its simplest form.
- 7.A red fair dice and a blue fair dice are rolled together. Every outcome is listed as an ordered pair, red score then blue score. Work out the probability that the red score is less than the blue score.
- 8.An ordinary fair dice is rolled twice. Work out the probability of getting at least one 5.
- 9.250 people took a theory test at one test centre. 150 of them had taken a preparation course and the rest had not. 120 of those who had taken the course passed and 50 of those who had not taken the course passed. One of the people who passed is picked at random. Work out the probability that this person had taken the preparation course.
- 10.At a youth club, 65 members are asked which sports they play. 27 play football, 21 play basketball, and 9 play both. Work out the number of members who play neither sport.
- 11.An archer hits the bullseye with probability 0.24 on any one shot. She wants to know how many shots she must take to expect to hit the bullseye 12 times. Work out how many shots this is.
- 12.Harry plays a game in which he flips a fair coin and rolls an ordinary fair dice. He wins a prize only if the coin shows heads and the dice shows a 6. Work out the probability that Harry wins a prize.
- 13.Spinner A has 4 equal sections, numbered 1, 2, 3 and 4. Spinner B has 3 equal sections, coloured red, red and blue. Both spinners are spun once. Work out the probability of getting an even number on Spinner A and red on Spinner B.
- 14.Spinner C has 3 equal sections, numbered 2, 3 and 5. Spinner D has 6 equal sections, numbered 1 to 6. Both spinners are spun once, and the two numbers are multiplied together. Work out the probability that the product is a multiple of 5.
- 15.At a summer fair, the probability of winning at the hoopla stall is 6/10 and the probability of winning at the coconut shy is 2/10. Work out how many times as likely a player is to win at the hoopla stall as at the coconut shy.
Answer key
- (a) unlikely — 3/8 = 3 ÷ 8 = 0.375, which is between 0 and 0.5, so the probability is best described as unlikely. Rounding 3/8 loosely to 'about a half' without converting it properly leads to evens, which is wrong since 0.375 is noticeably below 0.5. Reading 3/8 as the complement and using 1 − 3/8 = 5/8 = 0.625 instead gives likely, but the question asks about the probability of 3/8 itself, not its complement. Flipping the fraction to 8/3 ≈ 2.67, which is impossible for a probability, and rounding it towards the top of the scale gives certain.
- (c) 1/2 — Method: restrict to the students who did NOT bring a packed lunch, then find what fraction of that group are girls. Girls without lunch = 42 − 24 = 18. Boys = 70 − 42 = 28, so boys without lunch = 28 − 10 = 18. Total without lunch = 18 + 18 = 36. Working: P(girl | no lunch) = 18 ÷ 36 = 1/2. Answer: 1/2. Watch out: dividing 18 by 42 (the total number of girls) instead of by 36 finds P(no lunch | girl), the reverse conditional. Dividing by 70 (the whole trip) ignores that you already know the student did not bring a lunch. And using the 'brought a lunch' numbers (24 out of 34) answers the question for the wrong group entirely — you were asked about the students who did NOT bring one.
- (a) 4/9 — Method: two steps. Find how many cars failed altogether, because the car picked is known to be one of them, then divide the diesel failures by that total. Working: 30 petrol cars and 24 diesel cars failed, so 54 cars failed. The diesel failures give 24/54, and dividing the numerator and the denominator by 6 gives 4/9. Answer: the probability is 4/9. The distractors: 3/10 is 24/80, the probability that a car fails given that it is a diesel car, which is the condition and the event swapped; 3/25 is 24/200, dividing by every car serviced that week rather than by the 54 that failed; 2/5 is 80/200, the probability that a car chosen from the whole week is a diesel car, which ignores the fact that the car picked failed.
- (d) 3/20 — The probability of taking a red ball from Bag A is 3/5, and the probability of taking a red ball from Bag B is 1/4. Since the two picks are independent, the probabilities are multiplied: 3/5 × 1/4 = 3/20. A candidate who answers 4/9 has added the numerators and added the denominators, (3+1)/(5+4), instead of multiplying the two fractions. A candidate who answers 17/20 has added the two fractions, 3/5 + 1/4, instead of multiplying them. A candidate who answers 3/25 has misread Bag B as also containing 5 balls, using 1/5 instead of 1/4.
- (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.
- (d) 19/30 — Exactly one means only fiction or only non-fiction, not both: 21 + 17 = 38 out of the 60 readers, which simplifies to 19/30. Including the 14 who read both as well gives 21 + 17 + 14 = 52, so 52/60 = 13/15 — that is at least one, not exactly one. Using only the both-count, 14, as the numerator gives 14/60 = 7/30, the probability of reading both, not exactly one. Using 52, the number who read at least one type, as the denominator instead of the full 60 readers surveyed gives 38/52 = 19/26.
- (b) 5/12 — There are 36 equally likely ordered pairs. Listing the pairs where the red score is smaller than the blue score gives 15 outcomes, so the probability is 15/36 = 5/12. Choosing 1/2 comes from assuming the 36 outcomes split evenly between 'red smaller' and 'red bigger', which ignores the 6 outcomes where the two dice show the same score. Choosing 7/12 comes from including the outcomes where the two scores are equal, 15 + 6 = 21, giving 21/36 = 7/12, instead of using 'less than' strictly. Choosing 5/36 comes from counting only the outcomes where the two scores differ by exactly 1, such as (1, 2) and (2, 3), which is 5 outcomes, and missing the pairs that differ by more than 1.
- (b) 11/36 — Method: 'at least one' is the opposite of 'none at all', so work out the probability of no 5 on either roll and take it away from 1. Working: a roll that is not a 5 has probability 5/6, and the rolls are independent, so no 5 at all has probability 5/6 × 5/6 = 25/36. Taking this from 36/36 leaves 11/36. Answer: the probability is 11/36. The distractors: 25/36 is the probability of no 5 at all, written down without the final subtraction; 12/36 comes from counting the 6 pairs with a 5 on the first roll and the 6 pairs with a 5 on the second and adding them, which counts the pair (5, 5) twice; 30/36 comes from working out 1 − 1/6 as though only one roll were made.
- (d) 12/17 — Method: the person picked is known to have passed, so the sample space is everyone who passed; divide the course takers who passed by that total. Working: 120 course takers and 50 others passed, so 170 people passed. The course takers who passed give 120/170, and dividing the numerator and the denominator by 10 gives 12/17. Answer: the probability is 12/17. The distractors: 4/5 is 120/150, the probability that someone passed given that they took the course, which reverses the condition and the event; 12/25 is 120/250, dividing by everyone who sat the test rather than by the 170 who passed; 17/25 is 170/250, the probability that a person picked from everyone sitting the test passed, which answers a different question altogether.
- (c) 26 — The number who play football or basketball is 27 + 21 − 9 = 39, subtracting the 9 who play both so they are not counted twice. The number who play neither is then 65 − 39 = 26. Giving 39 stops after finding the number who play football or basketball, without taking the complement within the 65 members. Subtracting all three given numbers from 65, 65 − 27 − 21 − 9 = 8, treats the 9 who play both as a separate group rather than a correction for double-counting. Adding 27 and 21 without removing the double-counted 9, then subtracting that total from 65, gives 65 − (27 + 21) = 17.
- (a) 50 — To find the number of shots needed for an expected 12 hits, divide the number of hits wanted by the probability of a hit: 12 ÷ 0.24 = 50. Multiplying the number of hits by the probability instead of dividing gives 12 × 0.24 = 2.88, which rounds to 3 shots. Rounding 0.24 to 0.25 before dividing gives 12 ÷ 0.25 = 48. Using the probability of missing, 1 − 0.24 = 0.76, instead of the probability of hitting, gives 12 ÷ 0.76 = 15.79, which rounds to 16.
- (b) 1/12 — Method: the coin does not affect the dice, so the two events are independent and the probability that both happen is the product of their probabilities. Working: heads has probability 1/2 and a 6 on an ordinary dice has probability 1/6. Multiplying gives 1 on the top and 2 × 6 = 12 on the bottom. Answer: the probability is 1/12. The distractors: 2/3 comes from adding 1/2 and 1/6 instead of multiplying them; 1/6 comes from using the dice alone and ignoring the condition on the coin; 1/8 comes from counting the possible results as 6 + 2 = 8 and treating the winning result as one of those eight.
- (a) 1/3 — Method: for two independent spinners, multiply the probability of each separate outcome, but first work out each spinner's own probability correctly, using how many of its equal sections actually carry that result. Working: Spinner A has 2 even numbers, 2 and 4, out of 4 sections, so P(even) = 2/4 = 1/2. Spinner B has 2 red sections out of 3, so P(red) = 2/3. Multiplying gives 1/2 × 2/3, which cancels down to 1/3. Answer: 1/3. Watch out: writing down 1/4 treats Spinner B's two colours as equally likely and uses 1/2 for red, when in fact 2 of its 3 sections are red — the sections are not split evenly between the two colours. Writing down 1/6 undercounts Spinner A's even numbers as just one out of four instead of two. And writing down 5/6 applies the 'at least one' formula, P(A) + P(B) − P(A)×P(B), which answers a different question about EITHER spinner landing the right way, not both together.
- (a) 4/9 — There are 3 × 6 = 18 equally likely outcomes. The product is a multiple of 5 whenever Spinner C shows 5, whatever Spinner D shows, which is 6 outcomes, or whenever Spinner D shows 5 and Spinner C shows 2 or 3, which is 2 more outcomes. This gives 6 + 2 = 8 outcomes, so the probability is 8/18 = 4/9. Choosing 1/3 comes from only counting the 6 outcomes where Spinner C shows 5 and forgetting the 2 outcomes where Spinner D shows 5 instead. Choosing 1/9 comes from only counting the 2 outcomes where Spinner D shows 5 and forgetting the 6 outcomes where Spinner C shows 5. Choosing 1/2 comes from counting 9 outcomes instead of 8, by listing the case where Spinner C shows 5 and Spinner D shows 5 twice over.
- (b) 3 times as likely — Method: to say how many times as likely one event is as another, divide the larger probability by the smaller one; subtracting them gives the gap between the two probabilities, not the multiple. Working: both probabilities are counted in tenths, so 6/10 ÷ 2/10 compares 6 tenths with 2 tenths, and 6 ÷ 2 = 3. Answer: winning at the hoopla stall is 3 times as likely, which is why 6/10 sits three times as far along the 0 to 1 scale as 2/10. The distractors: 4 times as likely comes from subtracting the two counts, 6 − 2, instead of dividing them, which measures the gap rather than the multiple; 6 times as likely comes from reading the larger probability's 6 tenths straight off as the multiple without ever comparing it with the 2 tenths at the other stall; 12 times as likely comes from multiplying the two counts, 6 × 2, instead of dividing one by the other.
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