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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Answer key: Probability worksheet — GCSE Foundation
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- (b) 3 — Method: build the list of possible results systematically, taking the first flip as a head and then as a tail, and keeping the two flips in order so that a head then a tail is a different result from a tail then a head; then count the results that match the description. Working: with a head at the first flip the results are HH and HT, and with a tail at the first flip they are TH and TT, so the list is HH, HT, TH, TT — four results in all. The results containing at least one tail are HT, TH and TT. Answer: 3 of the results in the list contain at least one tail. The distractors: 2 comes from treating a head and a tail as one result however they are ordered, which shortens the list to HH, HT, TT so that only two entries hold a tail; 1 comes from reading 'at least one tail' as a tail at both flips, which is the single result TT; 4 comes from counting the tails written across the list — one in HT, one in TH and two in TT — instead of counting the results that contain a tail.
- (c) 0.37 — Method: pool the two runs into one combined set of results, then find the relative frequency of red across all of the spins together. Working: total reds = 16 + 21 = 37. Total spins = 40 + 60 = 100. Relative frequency = 37 ÷ 100 = 0.37. Answer: 0.37. Watch out: writing down 0.40 uses only the first run, 16 ÷ 40, and throws away the extra evidence from the second 60 spins. Writing down 0.35 uses only the second run, 21 ÷ 60, and throws away the first run instead. And writing down 0.375 averages the two runs' separate rates, (0.40 + 0.35) ÷ 2, which treats a run of 40 spins and a run of 60 spins as equally weighted, when pooling the actual counts gives the larger run its fair share of influence.
- (a) 0.54 — The four colours are exhaustive, so all four probabilities sum to 1: the probability of damson is 1 − 0.18 − 0.22 − 0.24 = 0.36. Amber and damson cannot both happen on one spin, so the probability of amber or damson is 0.18 + 0.36 = 0.54. Stopping after finding the probability of damson alone, without adding the probability of amber, gives 0.36. Adding the three given probabilities together, 0.18 + 0.22 + 0.24 = 0.64, and treating that total as the answer never finds the probability of damson at all. Subtracting only the probability of amber from 1, 1 − 0.18 = 0.82, ignores black, cyan and damson completely.
- (d) 7/15 — Both socks are the same colour either if both are red or if both are blue. The probability both are red is 4/6 × 3/5 = 12/30. The probability both are blue is 2/6 × 1/5 = 2/30. Adding these gives 12/30 + 2/30 = 14/30 = 7/15. Choosing 2/5 comes from only working out the 'both red' path, 12/30, and forgetting the 'both blue' path also counts. Choosing 5/9 comes from treating the draws as if the first sock were replaced, using 4/6 × 4/6 + 2/6 × 2/6 = 20/36 = 5/9, instead of reducing the totals for the second draw. Choosing 7/18 comes from reducing the number of socks removed but not the number left to choose from, using 4/6 × 3/6 + 2/6 × 1/6 = 14/36 = 7/18, instead of 5 remaining socks for the second draw.
- (a) 1/12 — The probability that the first dice shows a 6 is 1/6. The probability that the second dice shows an even number, 2, 4 or 6, is 3/6 = 1/2. Since the two dice are independent, multiply the probabilities: 1/6 × 1/2 = 1/12. A candidate who answers 1/6 has considered only the first dice and forgotten the condition on the second dice. A candidate who answers 1/2 has considered only the second dice and forgotten the condition on the first dice. A candidate who answers 1/36 has treated 'an even number' as a single specific value rather than three possible values, using 1/6 × 1/6.
- (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.
- (c) 47 — Method: fill the first pair of branches of the frequency tree, then the failures on each branch, then add only the failing end branches. Working: 70% of 200 is 140, so 140 cars were more than 3 years old and 200 − 140 = 60 cars were 3 years old or less. One quarter of the older cars failed: 140 ÷ 4 = 35. The newer branch gives 12 failures. Adding the two failing branches gives 35 + 12 = 47. Answer: 47 of the cars failed the test. The distractors: 35 is the older branch on its own, with the 12 newer failures never added; 62 comes from taking 1 in 4 of all the cars, 200 ÷ 4 = 50, and then adding the 12; 153 is 200 − 47 and counts the cars that passed.
- (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.
- (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.
- (c) 120 — 27 of the 45 sweets are orange (45 − 18 = 27), so the probability of taking an orange sweet is 27/45 = 3/5, and 200 × 3/5 = 120. Writing 80 is wrong because 200 × 18/45 = 80 uses the LEMON sweets' fraction instead of orange. Writing 182 is wrong because it takes the 18 lemon sweets away from the 200 repeats (200 − 18 = 182), applying the ‘the rest are orange’ subtraction to the number of goes instead of to the 45 sweets in the bag. Writing 27 is wrong because it is simply the number of orange sweets in the bag — it has not been scaled up to account for the 200 repeats. The expected number of times an orange sweet is taken is 120.
- (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) 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) 120 — Purple's probability is its angle out of the full circle: 144° ÷ 360° = 2/5. Expected number on purple = 2/5 × 300 = 120. Assuming the three colours are equally likely, ignoring their different angles, gives 300 ÷ 3 = 100. Using orange's angle, 90°, instead of purple's 144° gives a probability of 1/4, so 300 × 1/4 = 75. Treating 144 as a percentage instead of finding its fraction of 360° gives 300 × 0.144 = 43.2, which rounds to 43.
- (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) 40% — Method: a fraction becomes a percentage by scaling it so that its denominator is 100, or equivalently by dividing the numerator by the denominator and multiplying the decimal by 100. Working: 5 × 20 = 100, so the numerator is scaled by 20 as well, 2 × 20 = 40, giving 40/100; the same figure comes from 2 ÷ 5 = 0.4 and 0.4 × 100 = 40. Answer: 40%. The distractors: 20% comes from dividing 100 by the denominator alone, 100 ÷ 5 = 20, which is the percentage for 1/5 and never uses the numerator 2; 4% comes from finding the decimal 0.4 correctly and then writing the 4 down as the percentage instead of multiplying by 100; 2% comes from scaling only the denominator up to 100 and leaving the numerator as it was, giving 2/100.
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