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.Two fair coins are thrown together 400 times. Work out how many of the 400 throws you would expect to give one head and one tail.
- 2.The probability that a component is faulty is 1/10. Two components are tested independently. Work out the probability that at least one of the two components is faulty.
- 3.A grower knows that the probability that one of their seeds germinates is 0.6. The grower wants to expect 300 of the seeds to germinate. Work out how many seeds the grower should plant.
- 4.A fair spinner is divided into 5 equal sections, 2 labelled win and 3 labelled lose. Zara spins it twice, and the two spins are independent. Work out the probability that she wins on the first spin and loses on the second spin.
- 5.A garden centre tests 250 packets of seeds for germination, recording the results on a frequency tree. 68% of the 250 packets germinated successfully, and the rest did not. Work out how many packets germinated successfully.
- 6.The probability that a seed fails to germinate is 1/5. Three seeds are planted independently. Work out the probability that at least one of the three seeds germinates.
- 7.A train company runs 25 trains a day, every day. The probability that any one train is late is 0.08. Work out how many late trains the company should expect over a period of 4 weeks.
- 8.The probability that Oliver's train is delayed on any morning is 0.3. Work out the probability that his train is not delayed.
- 9.A two-way table records 200 members of a gym. 80 of the members are women and the rest are men. 60 of the women attend yoga classes. Work out the percentage of the women who attend yoga classes.
- 10.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.
- 11.Five balls numbered 1, 2, 3, 4 and 5 are in a bag. Two of them are taken out together at random. Work out the probability that the two numbers on them add up to more than 7.
- 12.A market research company found that, out of 120 shoppers surveyed, 42 preferred paper bags to plastic bags. A supermarket expects 900 shoppers this week. Using this relative frequency, work out how many of the 900 shoppers would be expected to prefer paper bags.
- 13.A fair spinner has 8 equal sections: 5 are red and 3 are blue. Freya spins it twice, and each spin is independent of the other. Work out the probability that she gets one red and one blue in the two spins, in either order.
- 14.On the probability scale from 0 to 1, which word best describes an event with probability 0.9?
- 15.At a summer fair, the probability of winning at the hoopla stall is 6/10 and the probability of winning at the coconut shy is 2/10. Work out how many times as likely a player is to win at the hoopla stall as at the coconut shy.
Answer key
- (b) 200 — Method: list the equally likely outcomes for the two coins before writing any probability, then multiply by the number of throws. Working: the equally likely outcomes are head then head, head then tail, tail then head, and tail then tail, so there are 4 of them. Two of those 4 give one head and one tail, so the probability is 2/4, which is 1/2. Over 400 throws the expected number is 400 × 1 ÷ 2 = 200. Answer: about 200 of the throws would be expected to give one head and one tail. The distractors: 133 comes from treating two heads, two tails and one of each as three equally likely results and working out 400 ÷ 3 = 133.3, then rounding; 100 comes from counting only head then tail as a success, giving 400 × 1 ÷ 4 = 100; 300 is the expected number of throws that do not give two heads, 400 × 3 ÷ 4 = 300.
- (d) 19/100 — It is easier to first find the probability that NEITHER component is faulty, then subtract from 1. The probability a component is not faulty is 9/10, so the probability neither is faulty is 9/10 × 9/10 = 81/100. So the probability at least one is faulty is 1 − 81/100 = 19/100. Choosing 1/5 comes from adding the two probabilities of a fault instead, 1/10 + 1/10 = 1/5, which double-counts the case where both are faulty. Choosing 1/10 comes from giving the probability for just one component being faulty. Choosing 1/100 comes from squaring the probability of a fault directly, 1/10 × 1/10 = 1/100, which is actually the probability that BOTH are faulty, not at least one.
- (c) 500 — Method: the expected number of successes is the number of trials multiplied by the probability, so to find the number of trials, divide the expected number by the probability. Working: let n be the number of seeds planted. Then n multiplied by 0.6 must come to 300, so n = 300 ÷ 0.6 = 500. Answer: the grower should plant 500 seeds. The distractors: 180 comes from multiplying instead of dividing, 300 × 0.6 = 180, which answers how many of 300 seeds would germinate; 750 comes from dividing by the probability of not germinating, 300 ÷ 0.4 = 750; 120 comes from multiplying by that same 0.4, 300 × 0.4 = 120.
- (b) 6/25 — The probability of winning on any one spin is 2/5, and the probability of losing on any one spin is 3/5. Since the spins are independent, multiply the probability of winning on the first spin by the probability of losing on the second spin: 2/5 × 3/5 = 6/25. A candidate who answers 4/25 has used the winning probability for both spins, 2/5 × 2/5. A candidate who answers 9/25 has used the losing probability for both spins, 3/5 × 3/5. A candidate who answers 3/10 has treated the spins as if they were dependent, reducing the second spin's denominator to 4.
- (c) 170 — 68% of the 250 packets germinated: 250 × 0.68 = 170. Using the complement, the 32% that did NOT germinate, gives 250 × 0.32 = 80. Shifting the decimal point, using 0.068 instead of 0.68, gives 250 × 0.068 = 17. Rounding 68% up to 70% before multiplying gives 250 × 0.70 = 175.
- (d) 124/125 — The probability that a seed germinates is 1 − 1/5 = 4/5, so the probability that all three seeds fail to germinate is 1/5 × 1/5 × 1/5 = 1/125. The probability that at least one germinates is 1 − 1/125 = 124/125. Choosing 4/5 comes from giving the probability that a single seed germinates, forgetting to combine all three seeds. Choosing 64/125 comes from working out the probability that ALL three seeds germinate, 4/5 × 4/5 × 4/5 = 64/125, instead of at least one. Choosing 12/125 comes from working out the probability that EXACTLY one seed germinates, 3 × 4/5 × 1/5 × 1/5 = 12/125, instead of at least one.
- (d) 56 — Method: count the trials over the whole period first, then multiply the number of trials by the probability. Working: 4 weeks is 4 × 7 = 28 days, and at 25 trains a day that is 25 × 28 = 700 trains. The expected number of late trains is 700 × 0.08 = 56. Answer: about 56 late trains over the 4 weeks. The distractors: 2 is the expected number for a single day, 25 × 0.08 = 2, with the 28 days never brought in; 14 uses one week instead of four, 25 × 7 × 0.08 = 14; 644 is 700 − 56 and counts the trains expected to be on time.
- (d) 0.7 — Method: delayed and not delayed are the only two outcomes, so the two probabilities form an exhaustive set and add to 1; subtract the given probability from 1. Working: 1 − 0.3, written as 1.0 − 0.3 with the decimal points lined up. Answer: 0.7. The distractors: 0.3 comes from giving back the probability that the train is delayed instead of its complement; 0.97 comes from lining the decimal points up wrongly and working out 1.00 − 0.03; 0.5 comes from assuming that delayed and not delayed must be equally likely because there are only two outcomes.
- (a) 75% — 'Percentage of the women' restricts the group to the 80 women, of whom 60 attend yoga: 60/80 = 0.75 = 75%. Dividing by the number of men (200 − 80 = 120) instead of the number of women gives 60/120 = 0.5 = 50%. Dividing by all 200 members instead of just the 80 women gives 60/200 = 0.3 = 30%. Using the 20 women who do NOT attend yoga (80 − 60) as the numerator instead of the 60 who do gives 20/80 = 0.25 = 25%.
- (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.
- (d) 1/5 — Method: list the pairs systematically, work out each total, then count the pairs that meet the condition and compare that count with the length of the list. Working: the pairs and their totals are 1 and 2 giving 3, 1 and 3 giving 4, 1 and 4 giving 5, 1 and 5 giving 6, 2 and 3 giving 5, 2 and 4 giving 6, 2 and 5 giving 7, 3 and 4 giving 7, 3 and 5 giving 8, and 4 and 5 giving 9. That is 10 pairs, of which 2 have a total of more than 7. Answer: the probability is 1/5. The distractors: 2/5 comes from counting the totals of exactly 7 as well, reading 'more than 7' as '7 or more'; 1/10 comes from finding only the pair 4 and 5 and missing that 3 and 5 also beat 7; 4/5 comes from counting the pairs on the wrong side of the condition, the 8 pairs whose total is 7 or less.
- (d) 315 — The relative frequency from the survey is 42 ÷ 120 = 0.35, so the expected number who prefer paper bags among 900 shoppers is 0.35 × 900 = 315. Giving 42 as the answer reuses the original survey count without scaling it up to 900 shoppers at all. Finding the expected number who prefer PLASTIC bags instead of paper, using the relative frequency 78 ÷ 120 = 0.65, gives 0.65 × 900 = 585. Using 1000 shoppers instead of the 900 actually stated gives 0.35 × 1000 = 350.
- (b) 15/32 — Method: there are two ways to get one of each colour, red then blue and blue then red. Work out the probability of each path by multiplying, then add the two paths. Working: red then blue is 5/8 × 3/8 = 15/64, and blue then red is 3/8 × 5/8 = 15/64. Adding the two paths gives 30/64. Answer: the probability is 15/32. The distractors: 15/64 comes from working out red then blue only and forgetting that blue then red also gives one of each; 15/28 comes from doubling correctly but reducing the total to 7 for the second spin, which is what happens to a bag when an item is kept out, not to a spinner; 39/64 comes from working from the opposite event and subtracting only the two-red case, 1 − 25/64, leaving the two-blue case inside the answer.
- (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) 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.
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