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) 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.
- (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) 3,860 — The probability a bulb works correctly is the complement of being defective: 1 − 0.035 = 0.965. Expected number working correctly = 0.965 × 4,000 = 3,860. Using the probability of being defective instead of its complement gives 4,000 × 0.035 = 140, the expected number of DEFECTIVE bulbs, not working ones. Shifting the decimal point in the complement, using 0.0965 instead of 0.965, gives 4,000 × 0.0965 = 386. Assuming every bulb works, ignoring the 0.035 probability altogether, gives the full batch of 4,000.
- (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) 80 — To find the number of attempts needed, divide the target number of successes by the probability of success: 60 ÷ 0.75 = 80. Writing 45 is wrong because 60 × 0.75 = 45 multiplies instead of dividing — that is the number of successes expected from 60 attempts, not the number of attempts needed for 60 successes. Writing 240 is wrong because 60 ÷ 0.25 = 240 uses 0.25, the probability of MISSING, instead of 0.75, the probability of scoring. Writing 90 is wrong because it comes from misremembering 0.75 as 2/3 and dividing by that instead: 60 ÷ (2/3) = 90. She needs to attempt 80 free throws.
- (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.
- (c) 30 — The numbers less than 4 are 1, 2 and 3, so the probability of that event is 3/6, and the expected count in 90 rolls is 90 × 3/6 = 45. The probability of rolling a 6 is 1/6, and the expected count is 90 × 1/6 = 15. The difference between the two expected counts is 45 − 15 = 30. A candidate who answers 45 has given the expected count for 'less than 4' only, forgetting to subtract the other expected count. A candidate who answers 15 has given the expected count for '6' only. A candidate who answers 36 has used a dice with 5 possible numbers instead of 6, giving 90 × 3/5 = 54 and 90 × 1/5 = 18, a difference of 36.
- (d) 3/16 — Method: work out each draw's own probability first, then multiply them together since the two draws are independent. Working: there are 8 tickets in all, 3 of them blue, so P(blue) = 3/8. P(spinner number greater than 2) = 2/4 = 1/2, since 3 and 4 qualify. Multiplying gives 3/8 × 1/2, which comes to 3/16. Answer: 3/16. Watch out: writing down 3/8 stops after the first draw and never brings in the spinner at all. Writing down 1/2 does the opposite, using only the spinner and ignoring the ticket draw. And writing down 5/16 uses 5/8, the probability of a RED ticket, instead of 3/8 for blue — reading the wrong colour off the raffle.
- (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.
- (d) 1/20 — Method: add the counts on the faulty end branches, divide by the total number of items in the experiment, then cancel. Working: the faulty items number 15 + 5 = 20, and 400 items were checked, so the probability is 20/400. Dividing the top and the bottom by 20 gives 1/20. Answer: the probability is 1/20. The distractors: 3/50 is 15/250 and comes from dividing machine A's faults by machine A's output, which is that machine's own fault rate rather than the probability for the whole batch; 1/30 is 5/150 and does the same on machine B's branch; 19/20 is 380/400 and gives the probability that the item picked is not faulty.
- (b) 0.225 — The relative frequency of rain is the number of rainy days out of all days recorded: 9 ÷ 40 = 0.225, which is noticeably less than the forecaster's claimed 0.3. Using the number of dry days, 40 − 9 = 31, as the denominator instead of the total of 40 gives 9 ÷ 31 = 0.29 (2 d.p.). Simply reporting the forecaster's claimed value, 0.3, without calculating anything from the data at all, ignores the recorded results completely. Misplacing the decimal point, treating 9 out of 40 as 9%, gives 0.09 instead of 0.225.
- (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) 100 — There are 3 even numbers on a fair dice (2, 4 and 6), so the probability of landing on an even number is 3/6 = 1/2, and 300 × 1/2 = 150. The probability of landing on a six is 1/6, so 300 × 1/6 = 50. The dice is expected to land on an even number 150 − 50 = 100 more times than on a six. Writing 50 is wrong because that is just the expected number of sixes on its own, without comparing it to the expected number of evens. Writing 150 is wrong because that is just the expected number of evens on its own, without subtracting the sixes. Writing 200 is wrong because it adds the two expected frequencies together (150 + 50 = 200) instead of finding the difference between them. The dice is expected to land on an even number 100 more times than on a six.
- (a) 1200 — Method: take the estimate from the larger sample, because an unbiased relative frequency tends towards the true probability as the sample grows, then multiply by the number of bulbs made in a week. Working: Inspector B tested 500 bulbs, far more than Inspector A's 40, so use B's relative frequency: 30 ÷ 500 = 0.06. A week's production is 4000 × 5 = 20000 bulbs. The expected number of faulty bulbs is 20000 × 0.06 = 1200. Answer: about 1200 faulty bulbs a week. The distractors: 2000 uses Inspector A's estimate, 4 ÷ 40 = 0.1, giving 20000 × 0.1 = 2000, and so rests on a sample of only 40 bulbs; 1600 comes from averaging the two estimates of 0.1 and 0.06 to get 0.08, and 20000 × 0.08 = 1600, which gives the small sample equal weight with the large one; 240 uses the right estimate but stops at a single day, 4000 × 0.06 = 240.
- (b) 8,100 — First find the total number of alerts sent in the month: 1,500 × 30 = 45,000. Then apply the probability of a 'STOP' reply: 45,000 × 0.18 = 8,100. Stopping after finding only one day's expected replies, 1,500 × 0.18 = 270, forgets to scale up to the whole month. Multiplying the number of days by the probability instead of by the daily total of alerts gives 30 × 0.18 = 5.4, which rounds to 5. Shifting the decimal point in the probability, using 0.018 instead of 0.18, gives 45,000 × 0.018 = 810.
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