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.
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Probability worksheet — GCSE Foundation
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- 1.On the probability scale from 0 to 1, which word best describes an event with probability 0.9?
- 2.Two fair coins are thrown together 400 times. Work out how many of the 400 throws you would expect to give one head and one tail.
- 3.Jack rolls two ordinary fair dice and adds the two scores together. Work out the probability that the total is 6.
- 4.A market stall sells umbrellas. Over the last 250 days, it rained on 70 of them. Using this as an estimate of the probability of rain, work out how many rainy days would be expected in the next 365 days.
- 5.The probability that Priya scores with a netball shot is 3/4. She takes two shots, and the shots are independent. Work out the probability that she scores with her first shot but misses her second shot.
- 6.A café offers 3 types of sandwich, cheese, ham and egg, and 4 types of drink, tea, coffee, juice and water. A customer chooses one sandwich and one drink at random. Work out the probability that the customer chooses egg and water.
- 7.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.
- 8.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.
- 9.A small pack contains 5 cards, numbered 1 to 5. One card is drawn at random from the pack, and a fair coin is flipped. Every outcome is listed as a pair, such as (3, H). Work out how many equally likely outcomes there are in total.
- 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.Spinner P and Spinner Q are each fair and have 5 equal sections, numbered 1 to 5. Both spinners are spun once, and every outcome is listed as an ordered pair, Spinner P then Spinner Q, such as (2, 4). Work out how many of the outcomes have the score on Spinner P less than the score on Spinner Q.
- 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.Kwame flips a fair coin three times and writes down what it lands on each time. Work out the probability that it lands on heads all three times.
- 14.A spinner has equal sections, and the probability that it lands on blue in a single spin is 1/4. It is spun twice, and the two spins are independent. Work out the probability that it lands on blue on both spins.
- 15.Priya and Ben each want to estimate the probability that a coin lands on heads. Priya flips the coin 50 times and works out her relative frequency. Ben flips the same coin 500 times and works out his relative frequency. Whose relative frequency is more likely to be close to the true probability? Give a reason for your answer.
Answer key
- (b) likely — On the probability scale, 'certain' is reserved for a probability of exactly 1, and 'evens' describes a probability of exactly 0.5. A probability of 0.9 is high but not equal to 1, so the correct word is 'likely'. Choosing 'certain' treats a probability close to 1 as if it were exactly 1, which it is not. Choosing 'evens' misjudges 0.9 as being close to the midpoint of the scale, when it is close to the 'certain' end instead. Choosing 'unlikely' reads the scale the wrong way round, as if a high probability meant a low chance of happening.
- (b) 200 — Method: list the equally likely outcomes for the two coins before writing any probability, then multiply by the number of throws. Working: the equally likely outcomes are head then head, head then tail, tail then head, and tail then tail, so there are 4 of them. Two of those 4 give one head and one tail, so the probability is 2/4, which is 1/2. Over 400 throws the expected number is 400 × 1 ÷ 2 = 200. Answer: about 200 of the throws would be expected to give one head and one tail. The distractors: 133 comes from treating two heads, two tails and one of each as three equally likely results and working out 400 ÷ 3 = 133.3, then rounding; 100 comes from counting only head then tail as a success, giving 400 × 1 ÷ 4 = 100; 300 is the expected number of throws that do not give two heads, 400 × 3 ÷ 4 = 300.
- (a) 5/36 — Method: list every result of the two dice as an ordered pair, first score then second score, count the pairs that give the total asked for and divide by how many pairs the list holds. Working: each dice can show 6 scores, so there are 6 × 6 = 36 equally likely ordered pairs. The pairs whose scores add to 6 are (1, 5), (2, 4), (3, 3), (4, 2) and (5, 1), which is 5 pairs out of the 36. Answer: the probability is 5/36. The distractors: 4/36 comes from listing (1, 5), (5, 1), (2, 4) and (4, 2) and leaving (3, 3) out, because a double does not look like a pair that can be turned round; 5/12 comes from finding the 5 pairs but taking the number of possible results to be 6 + 6 = 12, adding the two dice instead of multiplying them; 1/11 comes from treating the eleven possible totals 2, 3, 4 and so on up to 12 as equally likely, so that a total of 6 is one result out of eleven.
- (c) 102 — Method: turn the past record into a relative frequency, then use it as an estimate of the probability of rain and multiply by the number of days being predicted for. Working: relative frequency of rain = 70 ÷ 250 = 0.28. Expected rainy days in 365 days = 365 × 0.28 = 102.2, which rounds to about 102 days. Answer: about 102 days. Watch out: writing down 48 swaps which number is the sample and which is the target, working out 70 ÷ 365 × 250 instead of 70 ÷ 250 × 365. Writing down 70 just repeats the original count of rainy days without scaling it up to the new, longer period at all. And writing down 110 comes from rounding the relative frequency to 0.3 before multiplying, 365 × 0.3 = 109.5, when 70 ÷ 250 is exactly 0.28 and needs no rounding at all.
- (c) 3/16 — The probability that Priya misses a shot is 1 − 3/4 = 1/4. The two shots are independent, so multiply the probability of scoring the first shot by the probability of missing the second shot: 3/4 × 1/4 = 3/16. Choosing 9/16 comes from using the probability of scoring, 3/4, for both shots instead of switching to the miss probability for the second shot, 3/4 × 3/4 = 9/16, which is the probability that she scores with both shots. Choosing 1/2 comes from subtracting the two probabilities instead of multiplying them, 3/4 − 1/4 = 1/2. Choosing 3/4 comes from giving only the probability that she scores with the first shot, forgetting to combine it with what happens on the second shot.
- (d) 1/12 — Method: list the full possibility space of sandwich-and-drink pairs, then divide the one matching pair by the size of the whole space. Working: there are 3 × 4 = 12 equally likely sandwich-and-drink pairs, and exactly one of them is egg and water. Answer: 1/12. Watch out: writing down 1/7 comes from adding the two counts, 3 + 4 = 7, instead of multiplying them to build the possibility space. Writing down 1/3 uses only the chance of choosing egg out of 3 sandwiches and ignores the drink altogether. And writing down 1/4 uses only the chance of choosing water out of 4 drinks and ignores the sandwich altogether.
- (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.
- (c) 10 — There are 5 possible cards and 2 possible coin results, so listing every pair gives 5 × 2 = 10 equally likely outcomes. Choosing 5 comes from listing only the card outcomes and forgetting the coin flip adds a second stage to each one. Choosing 7 comes from adding the two stages instead of combining them, 5 + 2 = 7, rather than pairing every card with every coin result. Choosing 20 comes from counting each coin result twice for every card, 5 × 2 × 2 = 20, effectively pairing every card with the coin twice over.
- (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.
- (b) 10 — There are 25 equally likely ordered pairs in total. Listing the pairs where the Spinner P score is less than the Spinner Q score gives 10 outcomes, running from (1, 2) up to (4, 5). Choosing 15 comes from also including the 5 pairs where the two scores are equal, such as (1, 1) and (2, 2), which do not satisfy 'less than'. Choosing 4 comes from counting only the outcomes where the scores differ by exactly 1, such as (1, 2) and (2, 3), and missing the pairs that differ by 2, 3 or 4. Choosing 20 comes from correctly excluding the 5 outcomes where the two scores are equal, 25 − 5 = 20, but forgetting to also exclude the outcomes where Spinner P is bigger than Spinner Q, rather than smaller.
- (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.
- (d) 1/8 — Method: a run of flips of a fair coin gives equally likely sequences of heads and tails, so count the sequences that match and divide by how many sequences there are. Working: each flip lands two ways and no flip affects another, so three flips give 2 × 2 × 2 = 8 equally likely sequences: HHH, HHT, HTH, HTT, THH, THT, TTH and TTT. Only HHH has a head at every flip, so 1 sequence of the 8 matches. Answer: the probability is 1/8. The distractors: 1/4 comes from treating 'three heads', 'two heads', 'one head' and 'no heads' as four equally likely results, which they are not, since one sequence gives three heads and three sequences give two; 1/6 comes from taking the number of sequences to be 2 + 2 + 2 = 6, adding the two ways each flip can land instead of multiplying them; 1/2 comes from reading the first flip only and giving the probability of a head on one flip, without combining it with the other two.
- (a) 1/16 — The two spins are independent, so multiply the probability of blue on each spin: 1/4 × 1/4 = 1/16. Choosing 1/2 comes from adding the two probabilities instead of multiplying, 1/4 + 1/4 = 1/2. Choosing 1/8 comes from wrongly treating the second spin as having a 1/2 chance of blue instead of the given 1/4, giving 1/4 × 1/2 = 1/8. Choosing 1/4 comes from giving the probability for a single spin and forgetting that the spinner is spun twice.
- (a) Ben, because a larger sample is closer to the theory — Method: a relative frequency is an estimate of a probability, and for an unbiased experiment that estimate tends towards the theoretical value as the sample grows. Working: Priya's estimate rests on 50 results, so a few unexpected heads move it a long way; one extra head shifts her relative frequency by 1 ÷ 50 = 0.02. Ben's estimate rests on 500 results, where one extra head shifts his relative frequency by only 1 ÷ 500 = 0.002. The larger sample therefore swings far less around the true value. Answer: Ben's relative frequency is the one more likely to be close, because a larger unbiased sample tends closer to the theoretical probability. The distractors: saying a small sample is less affected by luck reverses the result, since it is the small sample that swings most; saying every flip is a separate random event is true of the flips themselves but says nothing about the estimates, and is often used to argue wrongly that the number of trials does not matter; saying 500 flips must give exactly 250 heads confuses an expected value with a guaranteed one, and 500 flips very rarely give exactly 250 heads.
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