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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- (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.
- (c) 11/20 — 'Red or white' combines two mutually exclusive events, so add their probabilities: 3/10 = 6/20 and 1/4 = 5/20, giving 6/20 + 5/20 = 11/20. Multiplying the two probabilities instead of adding them, 3/10 × 1/4, gives 3/40, which would be the probability of red and white together, not red or white — and a bead can't be both colours. Subtracting the sum from 1, 1 − 11/20 = 9/20, gives the probability of the bead being black instead of red or white. Converting 1/4 as 4/20 instead of 5/20 (dividing 20 by 4 but forgetting to scale the numerator) gives 6/20 + 4/20 = 1/2.
- (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.
- (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.
- (b) 4 — With 150 rolls and probability 1/6 for each number, the expected count is 150 ÷ 6 = 25. Comparing each actual count with 25: 1 is 22 (3 below), 2 is 27 (2 above), 3 is 24 (1 below), 4 is 34 (9 above), 5 is 21 (4 below) and 6 is 22 (3 below). Number 4 is furthest above its expected count, so it is the most over-represented. Number 2 is also above its expected count, but by only 2, far less than 4's 9. Number 3's count of 24 is below the expected 25, so it is under-represented, not over. Number 6's count of 22 is also below the expected 25, so it too is under-represented.
- (b) 1/6 — There are 2 outcomes for the coin and 3 outcomes for the spinner, giving 2 × 3 = 6 equally likely pairs: (H,X), (H,Y), (H,Z), (T,X), (T,Y), (T,Z). Only one of these, (H,Z), matches both conditions, so the probability is 1/6. A candidate who answers 1/3 has listed only the spinner's 3 outcomes and ignored the coin. A candidate who answers 1/2 has considered only the coin and ignored the spinner. A candidate who answers 1/5 has missed one pair when listing the possibility space, treating it as 5 outcomes instead of 6.
- (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) 24/125 — On the fiction branch, 160 − 112 = 48 books were returned late. None of the non-fiction books were late, so the total number of late books is 48, out of 250 books altogether: 48/250 = 24/125. Writing 3/10 is wrong because it divides the 48 late fiction books by the fiction total (160) instead of the whole library total (250). Writing 56/125 is wrong because 112/250 simplifies to 56/125, and 112 is the number of fiction books returned ON TIME, not late. Writing 9/25 is wrong because 90/250 simplifies to 9/25, and 90 is simply the number of non-fiction books, which has nothing to do with late returns. The probability is 24/125.
- (b) 11/12 — Method: raining and not raining are the only two outcomes, so their probabilities add to 1; subtract the given probability from 1. Working: writing 1 as 12/12 gives 12/12 − 1/12, and only the numerators are subtracted, 12 − 1 = 11. Answer: 11/12, close to the right-hand end of the 0 to 1 scale because rain is unlikely. The distractors: 1/12 comes from giving back the probability that it does rain; 1/11 comes from subtracting the 1 from the denominator instead of subtracting the fraction from 1; 11/11 comes from subtracting 1 from the numerator and from the denominator of 12/12 rather than from the numerator alone.
- (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.
- (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.
- (d) 15 — Method: each of Spinner A's outcomes can be paired with each of Spinner B's outcomes, so the two counts are combined by multiplying, not adding. Working: Spinner A has 3 outcomes and Spinner B has 5, so there are 3 × 5 = 15 equally likely outcomes in total. Answer: 15. Watch out: adding the two counts, 3 + 5 = 8, undercounts drastically — every one of Spinner A's 3 outcomes pairs with all 5 of Spinner B's outcomes, not just one each. Writing down 3 counts Spinner A alone and leaves Spinner B out completely. And writing down 25 comes from squaring Spinner B's own outcome count, 5 × 5, and forgetting Spinner A altogether.
- (d) 24 — There are 4 choices for the first digit. Once that digit is used, 3 digits remain for the second position, and then 2 digits remain for the third position: 4 × 3 × 2 = 24 codes. Choosing 64 comes from allowing a digit to be reused at every position, 4 × 4 × 4 = 64, which is not allowed here since no digit repeats. Choosing 12 comes from multiplying only the first two positions, 4 × 3 = 12, and forgetting that a third digit is also chosen from the digits that remain. Choosing 6 comes from counting only the arrangements of one single set of three digits, 3 × 2 × 1 = 6, and forgetting that there are 4 different sets of three digits that can be chosen from 2, 3, 4 and 5.
- (d) 0.420 — Method: use the relative frequency worked out from the larger number of trials as the best estimate of the probability, since a bigger sample tends to sit closer to the true probability. Working: 84 out of 200 drops land open end up, so the relative frequency from the total is 84 ÷ 200 = 0.420. Answer: 0.420. Watch out: writing down 0.650 comes from 13 ÷ 20, using only the first, much smaller sample instead of the total. Writing down 0.535 comes from averaging 0.650 and 0.420, treating the 20-drop run and the 200-drop run as equally reliable instead of using the larger sample on its own. And writing down 0.580 comes from 1 − 0.420, working out the probability that the bottle top lands the other way up instead of open end up.
- (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.
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