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.A spinner is spun 50 times. It lands on green 18 times. Work out the relative frequency of landing on green.
- 2.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.
- 3.A cinema sold 150 tickets for a Saturday screening. A frequency tree records the sales: 90 of the tickets were bought by adults and the rest by children. Of the adult tickets, 54 were for the 3D showing and the rest were for the standard showing. Work out how many adult tickets were for the standard showing.
- 4.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.
- 5.Four bags each contain counters, some of which are winning counters. The probability of taking a winning counter is 5/12 from the red bag, 3/8 from the blue bag, 7/24 from the green bag and 1/3 from the yellow bag. Work out which bag has the smallest probability of giving a winning counter.
- 6.Two games at a school fete are played independently. The probability of winning game X is 3/8. The probability of winning game Y is 0.4. Which statement correctly compares the two games?
- 7.A machine produces bolts. In a sample of 250 bolts, 15 are faulty. Work out the relative frequency of a bolt being faulty, giving your answer as a fraction in its simplest form.
- 8.One card is taken at random from an ordinary pack of 52 playing cards. 26 of the cards are red. Work out the probability that the card is red. Give your answer as a percentage.
- 9.A fair four-sided dice, numbered 1 to 4, is rolled twice. Every outcome is listed as an ordered pair, first roll then second roll, such as (2, 3). Work out how many different outcomes are in the list.
- 10.A fair coin is flipped and an ordinary fair dice is rolled. Every possible result is listed in a table. Work out how many different results the table contains altogether.
- 11.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?
- 12.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.
- 13.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.
- 14.A leisure centre has 150 members. 80 of the members are male and the rest are female. Every member uses either the pool or the gym, but not both. 66 members use the pool, and 35 of those pool users are male. Work out how many female members use the gym.
- 15.An ordinary pack of 52 playing cards contains 13 hearts. One card is taken at random from the pack. Work out the probability that the card is not a heart. Give your answer as a percentage.
Answer key
- (a) 0.36 — Relative frequency is the number of times the event happened divided by the total number of trials: 18 ÷ 50 = 0.36. Dividing by 100 instead of the actual 50 spins gives 18 ÷ 100 = 0.18. Finding the relative frequency of NOT landing on green, using 50 − 18 = 32 spins, gives 32 ÷ 50 = 0.64. Misplacing the decimal point in the division, so that 18 ÷ 50 is carried out as 18 ÷ 500, gives 0.036 — a tenth of the correct value.
- (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.
- (c) 36 — On the adult branch there are 90 tickets in total, and 54 of them are for the 3D showing, so the standard-showing branch is 90 − 54 = 36. Subtracting 54 from the overall total of 150 gives 96, but 150 is the total for ALL tickets, not just the adult branch, so 96 is wrong. Writing 60 is wrong because that is the number of CHILD tickets (150 − 90 = 60), not adult standard tickets. Writing 54 again is wrong because that is the number of adult 3D tickets, not the number of adult standard tickets. The adult standard-showing branch is 36.
- (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) The green bag (7/24) — Method: fractions can only be ordered once they share a denominator, so rewrite all four over the lowest common denominator and compare the numerators. Working: the lowest common denominator of 12, 8, 24 and 3 is 24, and scaling gives 5/12 = 10/24, 3/8 = 9/24, 7/24 stays as it is, and 1/3 = 8/24; the numerators are then 10, 9, 7 and 8. Answer: the smallest numerator is 7, so the green bag, with 7/24, is the least likely and sits furthest to the left on the 0 to 1 scale. The distractors: the red bag (5/12) comes from finding the largest of the four probabilities instead of the smallest; the blue bag (3/8) comes from scaling 3/8 by changing only the denominator to 24, which turns it into 3/24 and makes it look the smallest; the yellow bag (1/3) comes from comparing numerators alone and assuming the fraction with the numerator 1 must be the smallest.
- (c) Game Y — 0.4 is greater than 3/8. — Converting 3/8 to a decimal gives 3 ÷ 8 = 0.375, and 0.4 is greater than 0.375, so game Y gives the better chance of winning: 'Game Y — 0.4 is greater than 3/8.' Comparing the numerator 3 to the digit 0 in front of the decimal point in 0.4 compares two unrelated numbers and gives 'Game X — 3 is greater than 0, the units digit of 0.4.' Treating a larger denominator as meaning more chances, rather than smaller equal shares, gives 'Game X — 3/8 has the larger denominator.' Rounding 0.375 to 0.4 to one decimal place makes the two values look equal, but they are not exactly equal, so 'Equal — 3/8 rounds to 0.4 to one decimal place' mistakes a rounded match for a true one.
- (d) 3/50 — Relative frequency is the number of faulty bolts divided by the total sample size: 15/250, which simplifies to 3/50 by dividing both the numerator and the denominator by 5. Using 235, the number of bolts that were NOT faulty, as the denominator instead of the total 250 gives 15/235, which simplifies to 3/47. Inverting the fraction, dividing the total by the number of faulty bolts instead of the other way round, gives 250/15, which simplifies to 50/3 — a value greater than 1, which cannot be a probability. Simplifying by dividing the numerator and the denominator by different numbers, 15 ÷ 15 = 1 and 250 ÷ 25 = 10, gives 1/10.
- (d) 50% — Method: write the number of red cards over the total number of cards, then turn that fraction into a percentage by dividing and multiplying by 100. Working: the probability is 26/52; 26 ÷ 52 = 0.5 and 0.5 × 100 = 50. Answer: 50%, the exact middle of the 0 to 1 probability scale, which matches the fact that half the pack is red. The distractors: 26% comes from writing the number of red cards straight down as the percentage without dividing by the total; 25% comes from taking one of the four suits as the red ones and giving a quarter; 2% comes from dividing the wrong way round, 52 ÷ 26 = 2, and writing that as a percentage.
- (a) 16 — Method: keep the two rolls in order, first roll then second roll, and count every ordered pair the dice can land on, so a (2, 3) result is counted separately from a (3, 2) result. Working: the first roll can land on any of 4 numbers, and for each of those the second roll can also land on any of 4 numbers, giving 4 × 4 = 16 ordered pairs. Answer: 16. Watch out: writing down 8 comes from adding the two rolls' counts, 4 + 4, instead of multiplying them. Writing down 10 comes from listing unordered pairs, so (2, 3) and (3, 2) are counted as the same entry — that gives the 4 doubles plus 6 mixed pairs, 10 in total, instead of 16 ordered pairs. And writing down 4 comes from counting only one roll's outcomes and never pairing the two rolls together at all.
- (d) 12 — Method: each result of the first experiment can be paired with every result of the second, so the two counts are combined by multiplying. Working: the coin lands in 2 ways and the dice lands in 6 ways. Each of the 6 dice scores can appear with a head or with a tail, so the table has 6 × 2 rows. Answer: there are 12 different results. The distractors: 8 comes from adding the two counts, 6 + 2, instead of multiplying them; 6 comes from listing the dice scores only and treating the coin as making no difference to the table; 36 comes from working out 6 × 6, counting the coin as though it too had six equally likely results.
- (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.
- (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.
- (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) 39 — Method: put the counts into a two-way table and fill each missing cell by subtracting along a row or down a column. Working: the number of female members is 150 − 80 = 70. The pool column holds 66 members and 35 of them are male, so the number of female pool users is 66 − 35 = 31. Subtracting along the female row, 70 − 31 = 39 female members use the gym. Answer: 39 female members use the gym. The distractors: 45 comes from subtracting along the male row instead, 80 − 35 = 45, which counts male gym users; 31 is the female pool cell, written down one step before the gym cell; 84 is 150 − 66 and counts every gym user, male and female together.
- (d) 75% — Method: count how many cards are not hearts, write that count over the total number of cards, cancel the fraction down and then turn it into a percentage. Working: 52 − 13 = 39 cards are not hearts, so the probability is 39/52; dividing the numerator and the denominator by 13 gives 3/4, and 3/4 = 0.75, so 0.75 × 100 = 75. Answer: 75%, three quarters of the way along the 0 to 1 scale. The distractors: 25% comes from giving the probability that the card is a heart, 13 out of 52, which cancels to 1/4; 50% comes from reading 'not a heart' as 'not a red card' and halving the pack; 39% comes from writing the count of 39 cards straight down as the percentage without comparing it with the 52 cards in the pack.
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