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.
Probability worksheet — GCSE Foundation
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- 1.80 people are asked whether they drink tea and whether they drink coffee, and the results are recorded on a frequency tree. The first pair of branches splits the 80 people into 50 tea drinkers and 30 non-tea-drinkers. Of the 50 tea drinkers, 32 also drink coffee. Of the 30 non-tea-drinkers, 6 drink coffee. Work out n(T ∩ C), the number of people who drink both tea and coffee.
- 2.Two ordinary fair dice are rolled. Work out the probability that both dice show the same score.
- 3.A library recorded loans for 250 books over a week, using a frequency tree. The first branch splits the books into 160 fiction and 90 non-fiction. Of the fiction books, 112 were returned on time and the rest were returned late. All the non-fiction books were returned on time. Work out the probability that a book, chosen at random from the 250, was returned late. Give your answer as a fraction in its simplest form.
- 4.A red dice and a blue dice are rolled. Both dice are ordinary and fair. Work out the probability of getting a 1 on the red dice and a 2 on the blue dice.
- 5.In Manchester the probability of rain on any day in November is taken to be 0.3, and whether it rains on one day is independent of whether it rains on the next. Work out the probability that it rains on both the 10th and the 11th of November.
- 6.Spinner A has 4 equal sections, numbered 1, 2, 3 and 4. Spinner B has 3 equal sections, coloured red, red and blue. Both spinners are spun once. Work out the probability of getting an even number on Spinner A and red on Spinner B.
- 7.A charity raffle sells 250 tickets. A group of friends buy 40 of the tickets between them. Three prizes are awarded, each to a different randomly drawn ticket, and no ticket can win more than one prize. Work out the expected number of prizes won by the group. Give your answer as a fraction in its simplest form.
- 8.A bag contains 5 letter tiles spelling the word MATHS, with one letter on each tile. Two tiles are drawn at random, one after the other, without the first tile being put back. Work out the probability that both tiles drawn are consonants.
- 9.A spinner has equal sections, and the probability that it lands on blue in a single spin is 1/4. It is spun twice, and the two spins are independent. Work out the probability that it lands on blue on both spins.
- 10.One card is taken at random from an ordinary pack of 52 playing cards. 26 of the cards are red. Work out the probability that the card is red. Give your answer as a percentage.
- 11.Amir has 3 different coloured pens, red, blue and green, and 2 different types of paper, lined and plain. He wants to choose one pen and one type of paper, and draws a grid to list every combination systematically. Work out how many combinations are in his grid.
- 12.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.
- 13.A survey asked 80 people to name their favourite drink. The table shows the results: Tea 20 people, Coffee 32 people, Juice 16 people, Water 12 people. Work out the percentage of people whose favourite drink was Coffee.
- 14.150 pupils are asked about swimming and cycling, and the results are recorded on a frequency tree. The first pair of branches splits the 150 pupils into 90 who swim and 60 who do not swim. Of the 90 who swim, 54 also cycle. Of the 60 who do not swim, 18 cycle. Work out P(S ∩ C), the probability that a pupil, chosen at random, swims and cycles. Give your answer as a fraction in its simplest form.
- 15.A biased six-sided dice is rolled once. The probability that it lands on 6 is 0.25. The other five scores are all equally likely. Work out the probability that it lands on 3.
Answer key
- (d) 32 — The frequency tree already shows the tea-and-coffee branch directly: of the 50 tea drinkers, 32 also drink coffee, so n(T ∩ C) = 32. Adding both coffee branches together, 32 + 6 = 38, gives n(C), the total number of coffee drinkers, not just those who also drink tea. Using the non-tea branch's figure, 6, describes people who drink coffee but NOT tea. Subtracting to get 50 − 32 = 18 finds the tea drinkers who do NOT drink coffee, the opposite region to the one asked for.
- (b) 1/6 — Method: list the ordered pairs where the two scores match, and divide by the 36 equally likely pairs. Working: the matching pairs are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6), which is 6 pairs out of 36, cancelling down to 1/6. Answer: 1/6. Watch out: writing down 1/36 finds the probability of one particular double, such as (6, 6), rather than any double at all. Guessing 1/2 treats 'same' and 'different' as equally likely events, when there are only 6 matching pairs against 30 non-matching ones. And writing down 1/3 comes from listing each double twice, once for each order of the two dice, giving 12 pairs out of 36 — but (1, 1) is a single outcome, and swapping the two dice over does not produce a second one.
- (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.
- (c) 1/36 — Method: the two dice do not affect each other, so the probability that both results happen is the product of the two separate probabilities. Working: a 1 on the red dice has probability 1/6 and a 2 on the blue dice has probability 1/6. Multiplying gives 1 × 1 = 1 on the top and 6 × 6 = 36 on the bottom. Answer: the probability is 1/36. The distractors: 1/18 comes from allowing either order, counting a 2 on the red dice with a 1 on the blue dice as well, although the stem names which dice shows which score; 1/3 comes from adding 1/6 and 1/6 instead of multiplying; 1/6 comes from giving the probability for one dice only and treating the other as making no difference.
- (a) 0.09 — Method: two independent events that must both happen are combined by multiplying their probabilities. Working: the same probability 0.3 applies to each day, so the calculation is 0.3 × 0.3. Written as fractions this is 3/10 × 3/10 = 9/100. Answer: the probability is 0.09. The distractors: 0.6 comes from adding 0.3 and 0.3 instead of multiplying them; 0.3 comes from quoting the single-day probability, as though the second day added no further condition; 0.9 comes from multiplying 3 by 3 correctly but keeping only one decimal place in the product instead of two.
- (a) 1/3 — Method: for two independent spinners, multiply the probability of each separate outcome, but first work out each spinner's own probability correctly, using how many of its equal sections actually carry that result. Working: Spinner A has 2 even numbers, 2 and 4, out of 4 sections, so P(even) = 2/4 = 1/2. Spinner B has 2 red sections out of 3, so P(red) = 2/3. Multiplying gives 1/2 × 2/3, which cancels down to 1/3. Answer: 1/3. Watch out: writing down 1/4 treats Spinner B's two colours as equally likely and uses 1/2 for red, when in fact 2 of its 3 sections are red — the sections are not split evenly between the two colours. Writing down 1/6 undercounts Spinner A's even numbers as just one out of four instead of two. And writing down 5/6 applies the 'at least one' formula, P(A) + P(B) − P(A)×P(B), which answers a different question about EITHER spinner landing the right way, not both together.
- (d) 12/25 — The group holds 40 of the 250 tickets, so for any one prize the probability the group wins it is 40/250 = 4/25. There are 3 prizes and the group has the same chance at each one, so the expected number won is 3 × 4/25 = 12/25. Writing 4/25 is wrong because it is the chance of winning just ONE prize, without multiplying by the 3 prizes available. Writing 4/75 is wrong because it divides by the 3 prizes instead of multiplying (4/25 ÷ 3 = 4/75), which would mean the group did worse the more prizes were on offer. Writing 64/15625 is wrong because it multiplies the single-prize probability by itself three times, (4/25)³, as though all three prizes had to be won together, instead of adding up the expected number across the three separate prizes. The expected number of prizes won by the group is 12/25.
- (d) 3/5 — The first tile drawn is a consonant with probability 4/5, since 4 of the 5 letters in MATHS are consonants. Since it is not replaced, 3 consonants remain out of the 4 tiles left, so the second tile is a consonant with probability 3/4. Multiplying gives 4/5 × 3/4 = 3/5. Choosing 4/5 comes from only using the probability for the first draw and forgetting the second draw changes once a tile is removed. Choosing 3/4 comes from only using the probability for the second draw on its own. Choosing 16/25 comes from treating the two draws as independent, using 4/5 × 4/5 = 16/25, which would only be correct if the first tile were replaced before the second draw.
- (a) 1/16 — The two spins are independent, so multiply the probability of blue on each spin: 1/4 × 1/4 = 1/16. Choosing 1/2 comes from adding the two probabilities instead of multiplying, 1/4 + 1/4 = 1/2. Choosing 1/8 comes from wrongly treating the second spin as having a 1/2 chance of blue instead of the given 1/4, giving 1/4 × 1/2 = 1/8. Choosing 1/4 comes from giving the probability for a single spin and forgetting that the spinner is spun twice.
- (d) 50% — Method: write the number of red cards over the total number of cards, then turn that fraction into a percentage by dividing and multiplying by 100. Working: the probability is 26/52; 26 ÷ 52 = 0.5 and 0.5 × 100 = 50. Answer: 50%, the exact middle of the 0 to 1 probability scale, which matches the fact that half the pack is red. The distractors: 26% comes from writing the number of red cards straight down as the percentage without dividing by the total; 25% comes from taking one of the four suits as the red ones and giving a quarter; 2% comes from dividing the wrong way round, 52 ÷ 26 = 2, and writing that as a percentage.
- (c) 6 — Method: to list every combination of one item from a group of 3 and one item from a group of 2 systematically, multiply the two numbers of choices together. Working: 3 × 2 = 6. Answer: 6. Watch out: adding the two totals instead of multiplying, 3 + 2 = 5, misses combinations that a full grid would show — a grid with 3 rows and 2 columns has 6 cells, not 5. Writing down 3 counts only the pens, and writing down 2 counts only the types of paper — neither one pairs every pen with every type of paper.
- (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.
- (b) 40% — 32 out of the 80 people chose Coffee, so the percentage is (32 ÷ 80) × 100 = 40%. Choosing 32% comes from treating 32 as if it were already a percentage out of 100, instead of dividing by the actual total of 80 people surveyed. Choosing 25% comes from using the Tea row instead of the Coffee row, (20 ÷ 80) × 100 = 25%. Choosing 20% comes from using the Juice row instead of the Coffee row, (16 ÷ 80) × 100 = 20%.
- (c) 9/25 — The frequency tree already shows the swim-and-cycle branch directly: 54 out of 150, which simplifies to 9/25. Adding both cycling branches together, 54 + 18 = 72, gives the total number of cyclists, so 72/150 = 12/25, not just those who also swim. Using the non-swimmers' cycling figure, 18, gives 18/150 = 3/25, the probability of cycling WITHOUT swimming. Using the swimmers' total of 90, before splitting by cycling, gives 90/150 = 3/5, the probability of swimming on its own.
- (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.
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