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.
Answer key: Probability worksheet — GCSE Foundation
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- (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.
- (b) 12/25 — List every outcome as an ordered pair (first dice, second dice) out of the 25 equally likely outcomes. The pairs with a difference of exactly 2 are (1, 3), (3, 1), (2, 4), (4, 2), (3, 5) and (5, 3), the pairs with a difference of 3 are (1, 4), (4, 1), (2, 5) and (5, 2), and the pairs with a difference of 4 are (1, 5) and (5, 1), giving 6 + 4 + 2 = 12 outcomes with a difference of at least 2, so the probability is 12/25. Choosing 13/25 comes from finding the probability that the two scores differ by LESS than 2 instead, using the remaining 13 outcomes, the opposite of what was asked. Choosing 10/25 comes from forgetting the two outcomes where the difference is 4, (1, 5) and (5, 1), and adding only the difference-2 and difference-3 outcomes, 6 + 4 = 10 out of 25. Choosing 6/25 comes from counting only the pairs with a difference of exactly 2, forgetting that differences of 3 and 4 also count as at least 2.
- (a) £124.80 — On the cake branch, 150 − 100 = 50 cakes were bought by children. Adding the 54 biscuits bought by children gives 50 + 54 = 104 items sold to children in total, and at £1.20 each that raises 104 × £1.20 = £124.80. Writing £60.00 is wrong because 50 × £1.20 = £60.00 only counts the cake sales to children and leaves out the 54 biscuits. Writing £163.20 is wrong because it uses the ADULT sales instead of children's: 100 cake adults plus 90 − 54 = 36 biscuit adults gives 136 × £1.20 = £163.20. Writing £136.80 is wrong because it finds the cake children's number by subtracting the wrong branch (150 − 90 = 60 instead of 150 − 100 = 50), giving 60 + 54 = 114 items and 114 × £1.20 = £136.80. The total raised from sales to children is £124.80.
- (c) 7/8 — Method: 'at least one head' is the opposite of 'no heads at all', so work out the probability of three tails and take it away from 1. Working: a flip that is not a head has probability 1/2, and the flips are independent, so three tails in a row has probability 1/2 × 1/2 × 1/2 = 1/8. Taking this from 8/8 leaves 7/8. Answer: the probability is 7/8. The distractors: 1/8 is the probability of three tails, written down without the final subtraction; 3/8 is the probability of exactly one head, which comes from reading 'at least one' as 'exactly one'; 1/2 comes from giving the probability of a head on a single flip and ignoring that three flips are made.
- (b) She is wrong; 12 rolls is too few to judge — Method: compare the result with what is expected, then ask whether the experiment is long enough for a difference to mean anything. Working: if the dice were fair the expected number of sixes in 12 rolls is 12 × 1 ÷ 6 = 2, so 4 sixes is 2 above what was expected. But over only 12 rolls a result like this turns up often by chance: the relative frequency here is 4/12, which is 1/3, and over so few trials a relative frequency can sit well away from 1/6 with no bias at all. Answer: Erin is wrong, because 12 rolls is far too few to decide; she should roll the dice many more times and see whether the relative frequency settles near 1/6. The distractors: saying she is right because 4 beats the expected 2 uses the correct expected value but treats any difference as proof, which so short an experiment cannot give; saying a fair dice gives each score twice in 12 rolls treats an expected value as a guaranteed one; saying that 4 sixes in 12 rolls cancels down to 1 in 6 mis-cancels the fraction, because 4/12 is 1/3, which is twice 1/6, so that comment reaches the right verdict from arithmetic that is wrong.
- (b) 8/52 — Method: a card cannot be an ace and a king at the same time, so the two events are mutually exclusive and their probabilities are added, keeping the denominator the same. Working: P(ace) = 4/52 and P(king) = 4/52, so P(ace or king) = 4/52 + 4/52, and 4 + 4 = 8 fifty-seconds. Answer: 8/52. The distractors: 4/52 comes from giving the probability of just one of the two events and forgetting to add the other; 16/52 comes from multiplying the two counts, 4 × 4, instead of adding them; 1/52 comes from giving the probability of one particular named card rather than any of the eight.
- (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.
- (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.
- (d) 9 — Method: each of the first spin's 3 outcomes can be paired with each of the second spin's 3 outcomes, so the two counts are multiplied together. Working: 3 × 3 = 9 ordered pairs. Answer: 9. Watch out: writing down 6 comes either from adding the two spins' counts, 3 + 3, instead of multiplying them, or from listing colour pairs without order, treating (red, blue) as the same entry as (blue, red) — both shortcuts miss outcomes that the ordered list contains. Writing down 3 lists only a single spin's outcomes and never pairs the two spins up at all. And writing down 27 treats the spinner as though it had been spun three times, 3 × 3 × 3, one spin too many.
- (a) 1/6 — Method: write the results of the two dice as ordered pairs, count the pairs whose scores add to the total asked for, divide by the number of ordered pairs there are, then cancel the fraction down. Working: there are 6 × 6 = 36 equally likely ordered pairs. The pairs whose scores add to 7 are (1, 6), (2, 5), (3, 4), (4, 3), (5, 2) and (6, 1), which is 6 pairs, so the probability is 6/36. Dividing the top and the bottom by 6 gives 1/6. Answer: the probability is 1/6. The distractors: 7/36 comes from taking the number of favourable pairs to be 7 because 7 is the total asked for, confusing the size of a total with the number of ways of making it; 1/7 comes from using the 21 different combinations of two scores as the equally likely results, finding the 3 combinations 1 and 6, 2 and 5, 3 and 4, and cancelling 3/21; 6/11 comes from counting the 6 favourable pairs correctly but dividing by the 11 possible totals from 2 to 12 rather than by the 36 pairs.
- (d) 3/10 — Method: list the full possibility space as pairs of coin and spinner results, then count how many pairs satisfy both conditions and divide by the size of the whole space. Working: the coin gives 2 outcomes and the spinner gives 5, so the full space has 2 × 5 = 10 equally likely pairs. The pairs with a tail and an odd number are (T,1), (T,3) and (T,5), which is 3 out of 10. Answer: 3/10. Watch out: writing down 1/2 uses only the coin's own chance of a tail and ignores that the spinner also has to land on an odd number. Writing down 3/5 uses only the spinner's chance of landing on an odd number and ignores the coin altogether. And writing down 1/10 counts just one matching outcome, such as (T,1), instead of all three pairs that satisfy both conditions.
- (a) 0.15 — Let P(water) = x, so P(coffee) = 3x. The four outcomes are exhaustive: 0.36 + 0.04 + x + 3x = 1, so 0.4 + 4x = 1, giving 4x = 0.6 and x = 0.15. So P(water) = 0.15. Reporting 3x, the coffee probability, instead of water gives 0.45. Splitting the remaining 0.6 evenly between coffee and water, ignoring the 3:1 ratio, gives 0.30. Stopping once the remaining probability 0.6 is found, without dividing by the four equal shares, gives 0.60.
- (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.
- (b) 40 — n(P ∪ Q) = n(P) + n(Q) − n(P ∩ Q) = 34 + 27 − 11 = 50. The complement is everyone outside both sets: n((P ∪ Q)′) = 90 − 50 = 40. Adding P and Q without subtracting the overlap gives 34 + 27 = 61, so 90 − 61 = 29 double-subtracts the 11 who are in both. Reporting n(P ∪ Q) itself, 50, forgets to take the complement at all. Subtracting only n(P) from the universal set, 90 − 34 = 56, ignores set Q altogether.
- (a) Game B, by £8 — Game A's expected total is 40 × 0.2 × £5 = £40. Game B's expected total is 40 × 0.1 × £12 = £48. Game B is higher, by £48 − £40 = £8. Writing 'Game A, by £8' is wrong because it has the right difference but the wrong game — Game A's total (£40) is actually LOWER than Game B's, not higher. Writing 'Game B, by £48' is wrong because £48 is Game B's whole expected total, not the DIFFERENCE between the two games. Writing 'Game A, by £40' is wrong in the same way, using Game A's whole total as if it were the margin, and naming the wrong game as the winner. Game B gives the higher expected total, by £8.
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