TMUA Paper 1: Statistics and probability, set 1

• Answer all questions. No calculator.
• Each question has exactly one correct answer.

1. 11 mark

   A two-way table records, for 120 sixth-formers, which year group they are in (Year 12 or Year 13) and whether they hold a driving licence. The Year 12 total is 70 and the Year 13 total is 50. Across both year groups, 45 students hold a licence in total, and the number of Year 13 students with a licence is exactly twice the number of Year 12 students with a licence.

   Work out how many Year 12 students do NOT hold a driving licence.

   1. A 15
   2. B 55
   3. C 40
   4. D 35
2. 21 mark

   A histogram shows the time, in minutes, that 18 customers spent in a shop, using unequal class widths. The bar for the interval 20 <= t < 50 has a frequency of 18.

   The bar for the interval 50 <= t < 70 is drawn exactly twice as tall (that is, its frequency density is exactly twice that of the first bar).

   Work out the frequency represented by the bar for 50 <= t < 70.

   1. A 36
   2. B 18
   3. C 12
   4. D 24
3. 31 mark

   Six numbers have a mean of 15. A seventh number is included with them, and the mean of all seven numbers is 14.

   Work out the value of the seventh number.

   1. A 8
   2. B 98
   3. C 14
   4. D 90
4. 41 mark

   A researcher records, for a sample of house fires, the number of fire engines sent to each fire and the amount of damage caused, in pounds. A scatter graph of the two variables shows strong positive correlation, and the researcher concludes: 'Sending more fire engines to a fire causes more damage to be done.'

   Which of the following is the best comment on this conclusion?

   1. A The strong positive correlation proves that sending more fire engines directly causes more damage at the scene.
   2. B The correlation must run the other way round: it is the amount of damage that determines how many fire engines get sent, so damage causes the number of engines, and not the reverse.
   3. C The correlation is better explained by a third factor, such as the size of the fire, which independently increases both the number of engines sent and the amount of damage caused.
   4. D Before any conclusion can be drawn, the trend shown must first be extrapolated well beyond the range of the fires actually recorded.
5. 51 mark

   A frequency tree records data for 200 sixth-formers. First, it splits them by whether they drive to school: 30% of the 200 drive, and the rest do not.

   Then, of those who do NOT drive, exactly 42 use the bus to get to school.

   Work out how many of the 200 sixth-formers neither drive to school nor use the bus.

   1. A 140
   2. B 42
   3. C 60
   4. D 98
6. 61 mark

   A biased coin is twice as likely to land on heads as it is to land on tails.

   If the coin is flipped 300 times, how many times would you expect it to land on heads?

   1. A 150
   2. B 200
   3. C 100
   4. D 600
7. 71 mark

   A fair six-sided dice is rolled 60 times, and lands on 6 a total of 15 times, giving a relative frequency of 15/60 = 1/4. The theoretical probability of rolling a 6 on a fair dice is 1/6.

   Which of the following is the best comparison between this relative frequency and the theoretical probability?

   1. A The relative frequency of 1/4 is higher than the theoretical probability of 1/6; a gap like this over only 60 rolls is expected, and it would typically shrink if the dice were rolled many more times.
   2. B Since the dice is fair, the relative frequency must equal the theoretical probability of 1/6 exactly, so a result of 15 sixes in 60 rolls could not really have happened.
   3. C Theoretical probability only describes what happens once an experiment has actually been run a large number of times; before that, every one of the six faces is equally likely regardless of how the dice is loaded.
   4. D Because 15 out of 60 rolls is more than the theoretical probability predicts, this result proves that the dice is not actually fair.
8. 81 mark

   A biased four-sided spinner can land on 1, 2, 3 or 4. These are the only possible outcomes, so their probabilities are exhaustive and mutually exclusive.

   P(1) = 0.15, P(3) = x, P(2) = 2x, and P(4) = 0.25.

   Work out P(2).

   1. A 0.2
   2. B 0.6
   3. C 0.3
   4. D 0.4
9. 91 mark

   In a class of 30 students, 18 study French, 15 study Spanish, and 5 study neither language. Some students study both languages.

   By reasoning systematically about the class as a Venn diagram of two overlapping sets, work out how many students study both French and Spanish.

   1. A 3
   2. B 33
   3. C 8
   4. D 25
10. 101 mark

    A fair 4-sided dice, numbered 1 to 4, and a fair 6-sided dice, numbered 1 to 6, are rolled together, and their scores are added.

    By using a possibility space diagram of all 24 equally likely outcomes, find the probability that the total is 7.

    1. A 1/6
    2. B 1/9
    3. C 1/4
    4. D 5/24
11. 111 mark

    A box contains 4 red pens and 6 blue pens, 10 pens in total. Two pens are removed at random, one after the other, without the first pen being replaced.

    Using a tree diagram, find the probability that the two pens removed are different colours.

    1. A 4/15
    2. B 8/15
    3. C 6/25
    4. D 7/15
12. 121 mark

    In a group of 50 students, 20 study Physics, 25 study Chemistry, and 10 study both Physics and Chemistry.

    A student is chosen at random from those who study Chemistry. Find the probability that this student also studies Physics.

    1. A 1/5
    2. B 1/2
    3. C 2/5
    4. D 3/5

Worked solutions

Every question below carries the reasoning, not just the answer. The official material for this test publishes a correct option letter and nothing else.

1. Question 1Answer: B

   1. Let x be the number of Year 12 students who hold a licence. Since the Year 13 count is twice this, Year 13's licence holders number 2x.
   2. The two groups together account for all 45 licence holders, so x + 2x = 45, which is 3x = 45.
   3. Dividing by 3 gives x = 15, so 15 Year 12 students hold a licence.
   4. The Year 12 total is 70, so the number who do NOT hold a licence is 70 - 15 = 55.
   5. The answer is B.
   • Why not A: This is the number of Year 12 students who DO hold a licence, found correctly along the way, but it answers the wrong question: the table asks for the number who do NOT hold one.
   • Why not C: This swaps which year group's count is doubled. Taking Year 12's licence count as twice Year 13's instead of the other way round gives a different split of the 45 licence holders, and 70 minus that (wrong) Year 12 figure lands on 40.
   • Why not D: This ignores the 2:1 ratio given in the question entirely and instead assumes an even 50:50 split within Year 12 itself, treating half of the 70 Year 12 students (35) as holding a licence and half as not, without ever using the 45 figure.

2. Question 2Answer: D

   1. On a histogram with unequal class widths, frequency = class width x frequency density, so density = frequency / width.
   2. The first bar has width 50 - 20 = 30 and frequency 18, so its frequency density is 18 / 30 = 0.6.
   3. The second bar is twice as tall, so its frequency density is 2 x 0.6 = 1.2.
   4. The second bar's width is 70 - 50 = 20, so its frequency is 20 x 1.2 = 24.
   5. The answer is D.
   • Why not A: This doubles the frequency directly (18 x 2 = 36) because the bar is 'twice as tall', without accounting for the fact that the second bar also has a different (narrower) width from the first.
   • Why not B: This assumes the frequency (the area of the bar) stays the same because the change in height and the change in width happen to compensate for each other, when in fact the question states it is the DENSITY, not the area, that doubles.
   • Why not C: This finds the correct width (20) for the second bar but uses the first bar's frequency density (0.6) without doubling it, missing the 'twice as tall' condition altogether.

3. Question 3Answer: A

   1. The mean of the first six numbers is 15, so their total is 6 x 15 = 90.
   2. Once the seventh number is included, the mean of all seven becomes 14, so the new total is 7 x 14 = 98.
   3. The seventh number itself is the difference between the new total and the original total: 98 - 90 = 8.
   4. The answer is A.
   • Why not B: This is the total of all seven numbers (7 x 14 = 98), not the seventh number by itself. It stops one step early, without subtracting the original six numbers' total.
   • Why not C: This assumes the seventh number is simply equal to the new overall mean, but a single number that merely equalled the new mean would not be able to pull the mean down from 15 to 14 on its own; it needs to be far enough below 15 to do that.
   • Why not D: This is the total of the original six numbers (6 x 15 = 90), which is a step in the working, not the answer to what the seventh number is.

4. Question 4Answer: C

   1. A strong correlation between two variables shows only that they tend to rise and fall together, not that one directly causes the other.
   2. Here, a bigger fire would independently need more engines sent to it AND cause more damage, without either of those two recorded variables causing the other.
   3. This kind of unmeasured factor, driving both variables at once, is called a confounding variable, and it is the standard reason correlation does not indicate causation.
   4. So the fire engines are not causing the damage; the size of the fire is the more likely explanation for both.
   5. The answer is C.
   • Why not A: This treats correlation as proof of causation in the direction stated, which is exactly the fallacy the question is testing: correlation between two variables does not by itself show that either one causes the other.
   • Why not B: This still assumes one of the two recorded variables directly causes the other, merely reversing which way the arrow points, rather than considering that a separate factor might cause both.
   • Why not D: This confuses two different ideas: the danger of extrapolating a line of best fit beyond the data is a separate issue from whether correlation implies causation, and it does not address the researcher's causal claim at all.

5. Question 5Answer: D

   1. 30% of the 200 sixth-formers drive to school: 30% of 200 = 60.
   2. So the number who do not drive is 200 - 60 = 140.
   3. Of these 140, exactly 42 use the bus, so the number who neither drive nor use the bus is 140 - 42.
   4. 140 - 42 = 98.
   5. The answer is D.
   • Why not A: This is the total number who do NOT drive (140), stopping after the first split of the tree and forgetting to then subtract those among them who use the bus.
   • Why not B: This is simply the number who use the bus (42), which answers a different question from the one asked; it does not give the number who avoid BOTH travel methods.
   • Why not C: This is the number who DO drive (30% of 200 = 60), the wrong branch of the tree entirely for a question about those who neither drive nor bus.

6. Question 6Answer: B

   1. Let P(tails) = p, so P(heads) = 2p, since heads is twice as likely.
   2. These are the only two outcomes, so they must sum to 1: p + 2p = 1, which is 3p = 1, so p = 1/3.
   3. This makes P(heads) = 2p = 2/3.
   4. The expected number of heads in 300 flips is 300 x 2/3 = 200.
   5. The answer is B.
   • Why not A: This assumes the coin is fair, using a probability of 1/2 for heads, when the question states the coin is biased so that heads is twice as likely as tails.
   • Why not C: This swaps which outcome is twice as likely, using a probability of 1/3 for heads (as though tails were the more likely outcome), rather than the 2/3 that belongs to heads.
   • Why not D: This takes the ratio '2' from '2:1' and multiplies it directly by 300, rather than first converting the 2:1 ratio into the fraction 2/3 that a probability must be.

7. Question 7Answer: A

   1. The theoretical probability of rolling a 6 on a fair dice is 1/6, fixed by the dice having six equally likely faces.
   2. The relative frequency observed here, 15/60 = 1/4, is only an estimate based on one particular run of 60 rolls, and estimates from a limited number of trials do not have to match the theoretical value exactly.
   3. As the number of rolls increases, the relative frequency would be expected to settle closer and closer to the theoretical probability of 1/6.
   4. So the gap between 1/4 and 1/6 here reflects normal chance variation over a small sample, not proof that the dice is unfair.
   5. The answer is A.
   • Why not B: This wrongly treats theoretical probability as a guarantee for any fixed number of trials, rather than a long-run value that relative frequency only tends towards as the number of trials grows.
   • Why not C: This gets the relationship backwards: theoretical probability (1/6 for a fair dice) can be stated before any rolling happens, from the dice's structure alone, and does not depend on having already run the experiment.
   • Why not D: This over-interprets a single experiment's natural variation as definite proof of bias, when a difference of this size over only 60 rolls is well within the range expected from chance alone.

8. Question 8Answer: D

   1. Since the four outcomes are exhaustive and mutually exclusive, their probabilities must sum to 1: 0.15 + 2x + x + 0.25 = 1.
   2. Collecting terms gives 0.4 + 3x = 1, so 3x = 0.6.
   3. Dividing by 3 gives x = 0.2.
   4. P(2) is defined as 2x, so P(2) = 2 x 0.2 = 0.4.
   5. The answer is D.
   • Why not A: This correctly solves for x, but then reports x itself rather than P(2), which the question defines as 2x, not x.
   • Why not B: This correctly reaches 3x = 0.6 but stops there, reporting the combined total of the two unknown terms rather than dividing by 3 to isolate x first.
   • Why not C: This wrongly treats P(2) and P(3) as equal to each other, splitting the remaining probability 0.6 evenly between them (0.3 each), rather than using the fact that P(2) is exactly twice P(3).

9. Question 9Answer: C

   1. 5 of the 30 students study neither language, so 30 - 5 = 25 students study at least one of French or Spanish.
   2. If French and Spanish did not overlap at all, the two subject totals would add to 18 + 15 = 33 students.
   3. Since only 25 students actually study at least one language, the excess of 33 - 25 = 8 must be students who were counted twice, once in each subject.
   4. A student is counted twice exactly when they study both languages, so 8 students study both French and Spanish.
   5. The answer is C.
   • Why not A: This uses the whole class of 30 as the 'studies at least one language' total, forgetting to first remove the 5 students who study neither language.
   • Why not B: This simply adds the French and Spanish totals (18 + 15 = 33) without allowing for any overlap between the two sets at all, which is only valid if no student studies both.
   • Why not D: This correctly finds that 25 students study at least one language, but then reports this figure directly as the answer, rather than using it to work out the overlap between the French and Spanish sets.

10. Question 10Answer: A

    1. The possibility space has 4 x 6 = 24 equally likely outcomes, since the two dice have 4 and 6 sides respectively.
    2. For the total to be 7, if the 4-sided dice shows a, the 6-sided dice must show 7 - a, and 7 - a must be between 1 and 6.
    3. Checking a = 1, 2, 3, 4 in turn gives 7 - a = 6, 5, 4, 3, all of which are valid scores on the 6-sided dice, so all four values of a work.
    4. This gives 4 favourable outcomes out of 24, so the probability is 4/24, which simplifies to 1/6.
    5. The answer is A.
    • Why not B: This treats the possibility space as 36 outcomes, as if both dice had 6 sides, instead of the 24 outcomes that come from one dice having only 4 sides.
    • Why not C: This treats the possibility space as 16 outcomes, as if both dice had 4 sides, instead of allowing for the second dice having 6 sides.
    • Why not D: This miscounts the favourable outcomes as 5, likely by including a pairing such as (5, 2) that is impossible, since the 4-sided dice cannot show a 5.

11. Question 11Answer: B

    1. There are two ways to get pens of different colours: red then blue, or blue then red, and their probabilities must be added.
    2. P(red then blue) = (4/10) x (6/9) = 24/90, since after removing a red pen, 9 pens remain of which 6 are blue.
    3. P(blue then red) = (6/10) x (4/9) = 24/90, since after removing a blue pen, 9 pens remain of which 4 are red.
    4. Adding these gives 24/90 + 24/90 = 48/90, which simplifies (dividing by 6) to 8/15.
    5. The answer is B.
    • Why not A: This finds the probability of only one order (red then blue) and forgets that the pens could also come out in the other order (blue then red), which must be added on as well.
    • Why not C: This wrongly treats the two draws as independent, as though the first pen were replaced before the second draw, using (4/10) x (6/10) rather than accounting for only 9 pens remaining on the second draw.
    • Why not D: This computes the probability that the two pens are the SAME colour, which is the complementary event to the one asked about, rather than the probability that they are different colours.

12. Question 12Answer: C

    1. The question asks for a conditional probability: given that a student studies Chemistry, what is the probability they also study Physics?
    2. This restricts the denominator to the 25 students who study Chemistry, rather than the whole group of 50.
    3. The students who study Chemistry AND also study Physics number 10, which is the numerator.
    4. The conditional probability is therefore 10/25, which simplifies to 2/5.
    5. The answer is C.
    • Why not A: This uses the whole group of 50 as the denominator (10/50), rather than restricting attention to the 25 Chemistry students the question specifies as the group to choose from.
    • Why not B: This uses the Physics total (20) as the denominator (10/20), conditioning on the wrong subject; the question asks for a probability given that the student studies Chemistry.
    • Why not D: This uses the number who study Chemistry only, 25 - 10 = 15, as the numerator (15/25), rather than the number who study both subjects, which is what the question actually asks for.