Printable · GCSE Higher · ages 14-16
Probability worksheet — GCSE Higher
Fifteen questions across the probability statements at Higher 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 Higher
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- (a) 1/12 — Method: for draws with nothing put back, multiply the probabilities of the three draws, reducing both the number of red counters and the total each time a red counter is removed. Working: the first counter is red with probability 5/10. One red counter has gone, so the second is red with probability 4/9, and then the third is red with probability 3/8. Multiplying gives 60/720. Answer: the probability is 1/12. The distractors: 1/8 comes from using 5/10 three times, which is what happens only if each counter is put back; 2/9 comes from stopping after two draws and giving 5/10 × 4/9; 3/50 comes from taking one off the red count each time but leaving the total at 10, giving 5/10 × 4/10 × 3/10.
- (d) 21 — Method: 'yoga only' means yoga but not pilates, so subtract the number who do both from the total who do yoga. Working: 32 − 11 = 21. Answer: 21. Watch out: writing down 32, the total who do yoga, answers 'how many do yoga' rather than 'how many do yoga only' — it still includes the 11 who also do pilates. Adding the overlap instead of subtracting it, 32 + 11 = 43, moves in the wrong direction entirely. And writing down 11 gives the number who do both activities, which is the opposite of yoga only.
- (b) 1/3 — Method: the first sweet has already been taken and it was red, so work out the second probability from what is actually left in the bag. Working: one red sweet has gone, so 3 red sweets remain out of 9 sweets altogether, giving 3/9. Dividing the numerator and the denominator by 3 gives 1/3. Answer: the probability is 1/3. The distractors: 2/5 is 4/10, the probability for the first draw used again, which is only right if the first sweet is put back; 3/10 takes one off the red count but leaves the total at 10, updating half of the fraction; 4/9 takes one off the total but leaves the red count at 4, updating the other half of the fraction.
- (b) 500 — Method: when a dice is known to be fair, the theoretical probability is the best thing to work from, and the more trials there are the closer the results tend to it. Working: for a fair dice the probability of a six is 1/6, so the expected number of sixes in 3000 rolls is 3000 × 1 ÷ 6 = 500. The class experiment gave a relative frequency of 14/60, but 60 trials is far too few to overturn a known theoretical value, and the school's 3000 rolls will tend towards 1/6 in any case. Answer: about 500 sixes. The distractors: 700 comes from using the class relative frequency instead of the theory, 3000 × 14 ÷ 60 = 700; 600 comes from splitting the difference between the two, since 1/6 is about 0.167 and 14/60 is about 0.233, whose mean is 0.2, and 3000 × 0.2 = 600; 2500 uses 5/6 instead of 1/6 and counts the rolls expected not to be a six.
- (a) 0.16 — The probability that Kofi fails a single attempt is 1 − 0.6 = 0.4. Since the attempts are independent, the probability he fails both is 0.4 × 0.4 = 0.16. Choosing 0.36 comes from squaring the probability of PASSING instead, 0.6 × 0.6 = 0.36, which is the probability of passing both attempts, not failing both. Choosing 0.4 comes from giving the probability of failing just one attempt, forgetting to combine two attempts. Choosing 0.24 comes from multiplying the fail probability by the pass probability, 0.4 × 0.6 = 0.24, mixing up passing and failing between the two attempts.
- (b) 8/35 — Method: the pupil picked is known to play at least one of the two sports, so first count how many pupils that is, then divide the number who play both by it. Working: 25 play football and 18 play tennis, but the 8 who play both have been counted in each figure, so the number who play at least one sport is 25 + 18 minus 8, which is 35. The pupils who play both give 8/35, which will not cancel. Answer: the probability is 8/35. The distractors: 2/15 is 8/60, dividing by the whole year group instead of by the 35 pupils who play at least one sport; 8/43 uses 25 + 18 as the denominator, forgetting that the 8 pupils who play both have been counted twice; 8/25 conditions on the footballers alone, answering the probability that a footballer also plays tennis rather than using every pupil who plays a sport.
- (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) 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.
- (a) 9/25 — Method: P(badminton | tennis) = n(tennis and badminton) ÷ n(tennis) — restrict to the tennis-players, then find what fraction of them also play badminton. Working: n(tennis and badminton) = 18, n(tennis) = 50, so P(badminton | tennis) = 18/50 = 9/25. Answer: 9/25. Watch out: dividing by 40 (the badminton total) finds P(tennis | badminton) instead of P(badminton | tennis) — the wrong direction. Dividing by 90 (all the members named in the question) ignores that you already know the member plays tennis. And dividing by 72 (50 + 40 − 18, the number who play at least one of the two sports) answers a question about the union, not the condition you were given.
- (a) unlikely — 3/8 = 3 ÷ 8 = 0.375, which is between 0 and 0.5, so the probability is best described as unlikely. Rounding 3/8 loosely to 'about a half' without converting it properly leads to evens, which is wrong since 0.375 is noticeably below 0.5. Reading 3/8 as the complement and using 1 − 3/8 = 5/8 = 0.625 instead gives likely, but the question asks about the probability of 3/8 itself, not its complement. Flipping the fraction to 8/3 ≈ 2.67, which is impossible for a probability, and rounding it towards the top of the scale gives certain.
- (a) 12/25 — P(red then blue) = 4/10 × 6/10 = 24/100. P(blue then red) is the same, 6/10 × 4/10 = 24/100. Since either order counts as different colours, add them: 24/100 + 24/100 = 48/100 = 12/25. Working out only one order, red-then-blue, and forgetting blue-then-red, gives 24/100 = 6/25. Working out the probability that the two counters are the SAME colour instead — 4/10 × 4/10 + 6/10 × 6/10 = 16/100 + 36/100 = 52/100 = 13/25 — answers a different question. Using 6/9 for the second draw, as if the first counter were not replaced, gives 2 × (4/10 × 6/9) = 48/90 = 8/15.
- (c) 5/18 — Method: knowing the total is even cuts the 36 equally likely outcomes down to the even ones, so count those first and then count how many of them give 8. Working: the even totals occur as 2 once, 4 three times, 6 five times, 8 five times, 10 three times and 12 once, which is 18 outcomes. The total is 8 for 2 and 6, 3 and 5, 4 and 4, 5 and 3, and 6 and 2, which is 5 outcomes. The probability is 5/18, which will not cancel. Answer: the probability is 5/18. The distractors: 5/36 keeps the right count of ways to make 8 but divides by all 36 outcomes, ignoring the fact that the odd totals have already been ruled out; 1/6 treats the six even totals 2, 4, 6, 8, 10 and 12 as equally likely and picks one of them, which they are not; 1/11 treats the eleven possible totals from 2 to 12 as equally likely and uses neither the counting nor the condition.
- (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.
- (b) 1/4 — Method: the person picked is known to be aged 30 or over, so the sample space is those 140 people; divide the number of them who had been to the cinema by 140. Working: 35 of the 140 people aged 30 or over had been to the cinema, giving 35/140. Dividing the numerator and the denominator by 35 gives 1/4. Answer: the probability is 1/4. The distractors: 7/20 is 35/100, taking the count from the older group but the total from the under 30s, which is reading across the wrong row; 7/48 is 35/240, dividing by everyone surveyed instead of by the age group named; 3/4 is 105/140, the probability that someone aged 30 or over had NOT been to the cinema, the opposite event inside the correct group.
- (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.
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