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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- (c) 26 — The number who play football or basketball is 27 + 21 − 9 = 39, subtracting the 9 who play both so they are not counted twice. The number who play neither is then 65 − 39 = 26. Giving 39 stops after finding the number who play football or basketball, without taking the complement within the 65 members. Subtracting all three given numbers from 65, 65 − 27 − 21 − 9 = 8, treats the 9 who play both as a separate group rather than a correction for double-counting. Adding 27 and 21 without removing the double-counted 9, then subtracting that total from 65, gives 65 − (27 + 21) = 17.
- (d) No, because 3/8 + 5/12 + 1/6 = 23/24 — Using a common denominator of 24: 3/8 = 9/24, 5/12 = 10/24 and 1/6 = 4/24. Adding these numerators gives 9 + 10 + 4 = 23, so the three probabilities sum to 23/24, which is less than 1 — Zara is not correct. Adding the original numerators (3 + 5 + 1 = 9) over a denominator of 12 instead of converting each fraction properly gives 9/12 = 3/4, still less than 1 but the wrong fraction. Converting 1/6 to 5/24 instead of 4/24 (using the wrong scaling) makes the total 9/24 + 10/24 + 5/24 = 24/24 = 1, wrongly suggesting the probabilities are valid. Judging validity from the fact that each individual fraction lies between 0 and 1 ignores that an exhaustive set must sum to exactly 1, not merely contain valid individual values.
- (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.
- (d) 1/3 — List the outcomes for the two spinners systematically in a 3 × 3 grid: 1-1, 1-2, 1-3, 2-1, 2-2, 2-3, 3-1, 3-2, 3-3, where the first number is the score on spinner A and the second is the score on spinner B — 9 equally likely outcomes in total. The pairs where the two numbers are the same are 1-1, 2-2 and 3-3, so there are 3 favourable outcomes. P(same number) = 3/9 = 1/3. 2/3 comes from working out the probability that the two numbers are different and then forgetting to take the complement the right way round, so the probability of "different" is given instead of the probability of "same". 1/2 comes from listing only the 6 unordered pairs 1-1, 2-2, 3-3, 1-2, 1-3, 2-3 instead of all 9 ordered outcomes in the grid, then taking 3 out of that 6. 1/9 comes from spotting only one of the three matching pairs, such as 1-1, and missing 2-2 and 3-3.
- (c) 2 — Set A, the multiples of 3, is {3, 6, 9}. Removing the numbers that are also in set B, the even numbers {2, 4, 6, 8, 10}, leaves {3, 9} — 2 numbers are in A but not B. Giving 3, the whole size of set A, forgets to remove the number 6, which is also even. Giving 7 counts every number in A or B combined, {2, 3, 4, 6, 8, 9, 10}, rather than only those in A but not B. Giving 4 counts the numbers in set B but not set A, {2, 4, 8, 10}, the wrong way round.
- (d) unlikely and evens — The marked points are impossible (0), unlikely (0.25), evens (0.5) and certain (1). Since 0.25 < 0.3 < 0.5, the probability 0.3 lies between unlikely and evens. Choosing 'impossible and unlikely' treats 0.3 as below 0.25, which it is not. Choosing 'evens and certain' treats 0.3 as above 0.5, which it is not. Choosing 'impossible and evens' skips over the unlikely marker altogether, even though 0.3 is above it.
- (d) 1/8 — The multiples of 3 from 1 to 8 are 3 and 6, so the probability of that event is 2/8, which simplifies to 1/4. The probability of the coin landing on heads is 1/2. Since the spin and the toss are independent, multiply the two probabilities: 1/4 × 1/2 = 1/8. A candidate who answers 1/4 has considered only the spinner and forgotten to combine it with the coin toss. A candidate who answers 1/2 has considered only the coin and forgotten the spinner condition entirely. A candidate who answers 1/16 has counted only one number, 6, as a multiple of 3 instead of two, giving 1/8 × 1/2.
- (a) 80 — Method: list the outcomes that count as a success, write the probability from them, then multiply by the number of rolls. Working: the scores of 3 or more are 3, 4, 5 and 6, which is 4 of the 6 equally likely scores, so the probability is 4/6, which cancels to 2/3. Over 120 rolls the expected number is 120 × 2 ÷ 3 = 80. Answer: about 80 of the rolls would be expected to give 3 or more. The distractors: 60 comes from reading a score of 3 or more as a score above 3 and counting only 4, 5 and 6, giving 120 × 3 ÷ 6 = 60; 40 is the expected number of rolls that are not 3 or more, 120 × 2 ÷ 6 = 40; 20 is 120 ÷ 6 and is the expected count for one single score.
- (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.
- (c) 5/8 — Method: the sections are all the same size, so every section is equally likely; count the sections that are not red and write that count over the total number of sections. Working: 8 − 3 = 5 sections are not red, and there are 8 sections altogether. Answer: 5/8, a value between 0 and 1 and a little above the halfway point of the scale. The distractors: 3/8 comes from giving the probability that the spinner does land on red; 5/11 comes from adding the 3 red sections to the 8 sections to make a total of 11 instead of using the 8 sections that exist; 1/2 comes from assuming that 'red' and 'not red' must be equally likely because there are only two possibilities.
- (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.
- (b) 11/36 — Method: 'at least one' is the opposite of 'none at all', so work out the probability of no 5 on either roll and take it away from 1. Working: a roll that is not a 5 has probability 5/6, and the rolls are independent, so no 5 at all has probability 5/6 × 5/6 = 25/36. Taking this from 36/36 leaves 11/36. Answer: the probability is 11/36. The distractors: 25/36 is the probability of no 5 at all, written down without the final subtraction; 12/36 comes from counting the 6 pairs with a 5 on the first roll and the 6 pairs with a 5 on the second and adding them, which counts the pair (5, 5) twice; 30/36 comes from working out 1 − 1/6 as though only one roll were made.
- (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.
- (d) 15/64 — Method: because the first marble is put back, the second draw is made from the same 8 marbles, so the two draws are independent and their probabilities are multiplied. Working: the bag holds 8 marbles both times, so yellow first has probability 5/8 and blue second has probability 3/8. Multiplying fractions gives 5 × 3 = 15 on the top and 8 × 8 = 64 on the bottom. Answer: the probability is 15/64. The distractors: 9/64 comes from working out the probability that BOTH marbles are blue, 3/8 × 3/8, instead of one of each colour in the stated order; 15/16 comes from multiplying the numerators but adding the denominators; 5/8 comes from writing down the first draw only and never combining it with the second.
- (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.
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