Revision Libraryv1.6.8
Choose a subject →
Search
WorksheetsBooksFreePracticeHow it worksPricingFreeAboutFAQ
Choose a subject →

Admissions tests / ESAT / Maths 1 / Statistics and probability

Foundation. 15 questions, 15 marks, about 22 minutes.

ESAT Mathematics 1: Statistics and probability, set 1

Tables, charts and diagrams, averages and spread, sampling, probability of single and combined events, tree and Venn diagrams and conditional probability.

Download the questions (PDF) Download with worked solutions (PDF)

  • Answer all questions. No calculator.
  • Each question has exactly one correct answer.
  1. 11 mark

    A survey asked 150 pupils, in Year 9 and Year 10, whether they prefer Maths or Art.

    The results were: Year 9 - Maths 35, Art 45 (Year 9 total 80). Year 10 - Maths 40, Art 30 (Year 10 total 70). Overall totals - Maths 75, Art 75 (grand total 150).

    A pupil is chosen at random from Year 10 only. What is the probability that this pupil prefers Art?

    1. A 4/7
    2. B 3/7
    3. C 1/5
    4. D 2/5
  2. 21 mark

    A pie chart shows the favourite hobby of 90 people, split between Reading, Sport, Gaming and Other.

    The sector for Reading has an angle of 100 degrees at the centre of the circle.

    Work out how many of the 90 people chose Reading as their favourite hobby.

    1. A 50
    2. B 25
    3. C 100
    4. D 65
  3. 31 mark

    A histogram shows the time, in minutes, taken by a group of runners to complete a course.

    One bar covers the class interval 10 <= t < 25 (a class width of 15 minutes) and has a frequency density of 2.4.

    Work out the frequency (the number of runners) represented by this bar.

    1. A 36
    2. B 17.4
    3. C 6.25
    4. D 60
  4. 41 mark

    The cumulative frequency graph for the finishing times, in minutes, of 100 runners in a race passes through the points (10,0), (20,10), (30,30), (40,70), (50,90) and (60,100), where each point is (time, cumulative frequency).

    Use the graph to estimate the median finishing time.

    1. A 40
    2. B 35
    3. C 30
    4. D 60
  5. 51 mark

    The table shows the number of pets owned by each of 20 pupils.

    Number of pets: 0, 1, 2, 3. Frequency: 6, 5, 7, 2.

    Work out the mean number of pets per pupil.

    1. A 2
    2. B 1.5
    3. C 6.25
    4. D 1.25
  6. 61 mark

    The table shows the heights, in cm, of 30 plants, grouped into class intervals.

    Class interval: 0 <= h < 10, frequency 5. Class interval: 10 <= h < 20, frequency 10. Class interval: 20 <= h < 30, frequency 10. Class interval: 30 <= h < 40, frequency 5.

    Work out an estimate for the mean height of the plants.

    1. A 25
    2. B 15
    3. C 20
    4. D 150
  7. 71 mark

    Two classes sat the same test. Class A's scores had a median of 62 and an interquartile range (IQR) of 18. Class B's scores had a median of 58 and an IQR of 9.

    Which statement correctly compares the two classes, using these summary statistics?

    1. A Class B had the higher median, and Class B's scores were more spread out.
    2. B Class A had the higher median, but Class B's scores were more spread out.
    3. C Class A had the higher median, and Class A's scores were more spread out.
    4. D Class A had the higher median, but the two classes' scores were equally spread out.
  8. 81 mark

    A scatter graph plots exam score (y) against hours of revision (x) for a group of students. The line of best fit has equation y = 3x + 40, and this line is only intended to be used for x-values within the range of the data collected.

    Using this line of best fit, estimate the exam score for a student who revised for 5 hours.

    1. A 55
    2. B 15
    3. C 135
    4. D 43
  9. 91 mark

    A frequency tree records data for 200 students at a school. Of these, 120 are right-handed and the rest are left-handed. Of the right-handed students, 45 play a musical instrument. Of the left-handed students, 32 play a musical instrument.

    Work out the total number of students who play a musical instrument.

    1. A 200
    2. B 123
    3. C 45
    4. D 77
  10. 101 mark

    A fair six-sided dice is rolled 150 times.

    How many times would you expect the dice to show a number greater than 4?

    1. A 25
    2. B 100
    3. C 75
    4. D 50
  11. 111 mark

    A spinner is spun 80 times, and it lands on red 24 times.

    Based on this experiment, estimate the probability that the spinner lands on red on the next spin.

    1. A 10/3
    2. B 3/10
    3. C 3/7
    4. D 7/10
  12. 121 mark

    A bag contains only red, blue and green counters. A counter is picked at random. The probability it is red is 0.35, and the probability it is blue is 0.4.

    Work out the probability that the counter is green.

    1. A 0.75
    2. B 0.65
    3. C 0.25
    4. D 0.6
  13. 131 mark

    A padlock code uses two different digits chosen from 1, 2, 3 and 4. The two digits must be different, and the order matters, so the code 12 is different from the code 21.

    By systematically listing the possibilities, work out how many different codes there are.

    1. A 12
    2. B 6
    3. C 16
    4. D 7
  14. 141 mark

    Two fair, ordinary six-sided dice are rolled, and their scores are added together.

    By using a possibility space diagram of all 36 equally likely outcomes, find the probability that the total is 9.

    1. A 1/9
    2. B 1/12
    3. C 1/11
    4. D 5/36
  15. 151 mark

    A drawer contains 5 blue socks and 3 red socks, 8 socks in total. Two socks are taken out at random, one after the other, without the first sock being replaced.

    Work out the probability that both socks are blue.

    1. A 25/64
    2. B 5/16
    3. C 5/14
    4. D 5/8

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. The question restricts attention to Year 10 pupils only, so the denominator must be the Year 10 total, which is 70.
    2. Within Year 10, the number who prefer Art is 30 (since Year 10 totals 70 and Maths accounts for 40 of those, 70 - 40 = 30).
    3. The probability is therefore 30 out of 70, written as the fraction 30/70.
    4. Dividing numerator and denominator by 10 gives 3/7, which does not simplify further since 3 and 7 share no common factor.
    5. So the answer is 3/7.
    • Why not A: This uses the Year 10 Maths figure (40) instead of the Year 10 Art figure (30) as the numerator, effectively answering for the wrong subject.
    • Why not C: This is 30/150, using the grand total of 150 pupils as the denominator instead of the Year 10 total of 70; the question restricts the choice to Year 10 only.
    • Why not D: This is 30/75, using the overall Art total (75) as the denominator instead of the Year 10 total (70); 75 is a column total, not the row total for Year 10.

  2. Question 2Answer: B

    1. A full pie chart represents all 90 people spread over 360 degrees, so one person corresponds to 360 / 90 = 4 degrees.
    2. The Reading sector has an angle of 100 degrees, and each person is worth 4 degrees, so the number of people is 100 / 4.
    3. 100 / 4 = 25.
    4. So 25 of the 90 people chose Reading.
    • Why not A: This comes from treating a full circle as 180 degrees instead of 360 degrees: 100/180 x 90 = 50. A full turn is 360 degrees, not 180.
    • Why not C: This takes the angle itself (100) as the number of people, but the angle measures degrees, not people; it must be converted using the total of 90 people over 360 degrees.
    • Why not D: This is 90 - 25, the number of people who did NOT choose Reading (everyone in the other three sectors), rather than the number who did choose Reading.

  3. Question 3Answer: A

    1. On a histogram, frequency density is defined so that frequency = class width x frequency density.
    2. The class interval is 10 <= t < 25, so its width is 25 - 10 = 15 minutes.
    3. Multiplying the width by the frequency density gives 15 x 2.4.
    4. 15 x 2.4 = 15 x 2 + 15 x 0.4 = 30 + 6 = 36.
    5. So the bar represents 36 runners.
    • Why not B: This adds the class width and the frequency density (15 + 2.4 = 17.4) instead of multiplying them; frequency is width times density, not their sum.
    • Why not C: This divides the class width by the frequency density (15 / 2.4 = 6.25) instead of multiplying; the correct relationship is frequency = width x density.
    • Why not D: This uses a class width of 25, as if the interval ran from 0 to 25, instead of the true width of 15 (from 10 to 25); the width must be measured between the interval's own boundaries.

  4. Question 4Answer: B

    1. With 100 runners, the median is the value at position n/2 = 50 on the cumulative frequency scale.
    2. The point (30,30) and the point (40,70) show that the cumulative frequency rises from 30 to 70 (an increase of 40) as time rises from 30 to 40 minutes (an increase of 10).
    3. The target cumulative frequency of 50 is 50 - 30 = 20 above the value at t = 30, which is 20/40 = 1/2 of the way through that rise.
    4. Moving 1/2 of the way through the 10-minute interval from t = 30 gives 30 + (1/2 x 10) = 30 + 5 = 35.
    5. So the estimated median finishing time is 35 minutes.
    • Why not A: This reads off the upper boundary (40) of the class interval that contains the median, rather than estimating a value inside that interval by interpolation.
    • Why not C: This reads off the lower boundary (30) of the class interval that contains the median, rather than estimating a value inside that interval by interpolation.
    • Why not D: This looks for the point where the cumulative frequency equals the total number of runners (100), rather than half of that total (50), which is the position of the median.

  5. Question 5Answer: D

    1. The mean of grouped data is found from sum(frequency x value) divided by the total frequency.
    2. Multiplying each value by its frequency: 0 x 6 = 0, 1 x 5 = 5, 2 x 7 = 14, 3 x 2 = 6.
    3. Adding these gives 0 + 5 + 14 + 6 = 25.
    4. The total number of pupils is 6 + 5 + 7 + 2 = 20.
    5. The mean is 25 / 20 = 1.25 pets per pupil.
    • Why not A: This gives the mode (the most common number of pets, which is 2 since it has the highest frequency of 7), not the mean.
    • Why not B: This averages the four pet-count values themselves, (0 + 1 + 2 + 3) / 4 = 1.5, ignoring how many pupils reported each value; the frequencies must be used as weights.
    • Why not C: This divides the total of (frequency x value) by the number of different pet-count categories (4) instead of by the total number of pupils (20).

  6. Question 6Answer: C

    1. For grouped data, each class interval is represented by its midpoint: the midpoints here are 5, 15, 25 and 35.
    2. Multiplying each midpoint by its frequency: 5 x 5 = 25, 15 x 10 = 150, 25 x 10 = 250, 35 x 5 = 175.
    3. Adding these gives 25 + 150 + 250 + 175 = 600.
    4. The total number of plants is 5 + 10 + 10 + 5 = 30.
    5. The estimated mean is 600 / 30 = 20 cm.
    • Why not A: This uses each interval's upper class boundary (10, 20, 30, 40) instead of its midpoint (5, 15, 25, 35); an estimated mean for grouped data must use midpoints.
    • Why not B: This uses each interval's lower class boundary (0, 10, 20, 30) instead of its midpoint (5, 15, 25, 35); an estimated mean for grouped data must use midpoints.
    • Why not D: This divides the total of (frequency x midpoint) by the number of class intervals (4) instead of by the total number of plants (30).

  7. Question 7Answer: C

    1. The median gives a typical, central value: Class A's median of 62 is higher than Class B's median of 58, so Class A typically scored higher.
    2. The interquartile range measures spread: Class A's IQR of 18 is larger than Class B's IQR of 9.
    3. A larger IQR means the middle half of the scores is spread over a wider range, so Class A's scores were more spread out than Class B's.
    4. Putting the two comparisons together: Class A had the higher median and the more spread out scores, which is statement C.
    • Why not A: This reverses which class scored higher: Class A's median of 62 is higher than Class B's median of 58, not the other way round.
    • Why not B: This reverses the spread comparison: Class A's IQR of 18 is larger than Class B's IQR of 9, so Class A's scores were more spread out, not Class B's.
    • Why not D: This ignores the actual IQR values given: an IQR of 18 for Class A against 9 for Class B shows the two classes did not have equal spread.

  8. Question 8Answer: A

    1. The line of best fit gives an estimated y-value from y = 3x + 40, where x is the number of hours of revision.
    2. Substituting x = 5 gives y = 3 x 5 + 40.
    3. 3 x 5 = 15, and 15 + 40 = 55.
    4. So the estimated exam score is 55.
    5. This is an interpolation, since x = 5 hours falls within the range of the data collected; the line should not be trusted for x-values far outside that range.
    • Why not B: This calculates only the gradient term, 3 x 5 = 15, and leaves out the y-intercept of 40 entirely.
    • Why not C: This adds x and the intercept before multiplying by the gradient, 3 x (5 + 40) = 135, instead of multiplying the gradient by x first and then adding the intercept.
    • Why not D: This adds the gradient (3) to the intercept (40) to get 43, without multiplying the gradient by the number of hours (5) first.

  9. Question 9Answer: D

    1. The tree splits 200 students into 120 right-handed and 200 - 120 = 80 left-handed.
    2. Of the 120 right-handed students, 45 play an instrument.
    3. Of the 80 left-handed students, 32 play an instrument.
    4. The total number who play an instrument is the sum of both branches: 45 + 32 = 77.
    • Why not A: This adds every branch of the tree together, which gives the whole sample of 200 students rather than just those who play an instrument.
    • Why not B: This is the total number of students who do NOT play an instrument (75 right-handed non-players plus 48 left-handed non-players), rather than the number who do.
    • Why not C: This uses only the right-handed branch of the tree (45) and ignores the left-handed students who also play an instrument.

  10. Question 10Answer: D

    1. On a fair dice, the numbers greater than 4 are 5 and 6, which is 2 out of the 6 equally likely outcomes.
    2. The probability of rolling a number greater than 4 is therefore 2/6, which simplifies to 1/3.
    3. The expected number of successes in 150 rolls is 150 x 1/3.
    4. 150 x 1/3 = 50.
    5. So you would expect a number greater than 4 about 50 times.
    • Why not A: This uses a probability of 1/6, as if only the number 6 satisfies 'greater than 4', but 5 also satisfies it.
    • Why not B: This uses a probability of 4/6, treating 3, 4, 5 and 6 as all satisfying 'greater than 4'; only 5 and 6 actually do.
    • Why not C: This uses a probability of 3/6, treating 4, 5 and 6 as all satisfying 'greater than 4'; 4 itself is not greater than 4.

  11. Question 11Answer: B

    1. An estimated probability from an experiment is the relative frequency: the number of successes divided by the total number of trials.
    2. Here, red occurred on 24 of the 80 spins, so the relative frequency is 24/80.
    3. Dividing both numbers by 8 gives 24/80 = 3/10.
    4. So the estimated probability of landing on red is 3/10.
    • Why not A: This inverts the fraction, dividing the total number of spins by the number of reds (80/24), which cannot be a probability since it is greater than 1.
    • Why not C: This uses the number of spins that were NOT red (80 - 24 = 56) as the denominator, giving 24/56, instead of using the total number of spins (80).
    • Why not D: This is 56/80, the relative frequency of NOT landing on red, rather than the relative frequency of landing on red.

  12. Question 12Answer: C

    1. Red, blue and green are the only possible outcomes, so their probabilities form an exhaustive set and must sum to 1.
    2. P(green) = 1 - P(red) - P(blue).
    3. P(green) = 1 - 0.35 - 0.4.
    4. 1 - 0.35 = 0.65, and 0.65 - 0.4 = 0.25.
    5. So the probability the counter is green is 0.25.
    • Why not A: This adds the two given probabilities (0.35 + 0.4 = 0.75), but P(green) is found by subtracting their sum from 1, not by adding them together.
    • Why not B: This subtracts only P(red) from 1 (1 - 0.35 = 0.65), forgetting to also subtract P(blue).
    • Why not D: This subtracts only P(blue) from 1 (1 - 0.4 = 0.6), forgetting to also subtract P(red).

  13. Question 13Answer: A

    1. There are 4 choices for the first digit of the code.
    2. Since the second digit must be different from the first, there are only 3 remaining choices for the second digit.
    3. Because order matters, the total number of codes is found by multiplying the choices for each position: 4 x 3.
    4. 4 x 3 = 12.
    5. So there are 12 different codes, which can also be checked by listing them systematically: 12, 13, 14, 21, 23, 24, 31, 32, 34, 41, 42, 43.
    • Why not B: This counts each pair of digits only once, treating 12 and 21 as the same code, but the problem states that order matters, so they are different codes.
    • Why not C: This allows the same digit to be used twice (4 choices for each of the two positions, 4 x 4 = 16), but the problem requires the two digits to be different.
    • Why not D: This adds the number of choices for the first digit and the second digit (4 + 3 = 7) instead of multiplying them.

  14. Question 14Answer: A

    1. When two dice are rolled, there are 6 x 6 = 36 equally likely outcomes, since each die has 6 faces.
    2. The pairs of scores that add to 9 are (3,6), (4,5), (5,4) and (6,3), which is 4 outcomes out of the 36.
    3. The probability is therefore 4/36.
    4. Dividing numerator and denominator by 4 gives 4/36 = 1/9.
    5. So the probability that the total is 9 is 1/9.
    • Why not B: This finds only 3 of the 4 pairs of dice scores that add to 9 (missing one of (3,6), (4,5), (5,4) and (6,3)), giving 3/36 = 1/12 instead of 4/36.
    • Why not C: This treats the 11 possible totals, from 2 to 12, as if they were equally likely outcomes, giving 1/11; the 36 individual dice combinations are equally likely, but the totals themselves are not.
    • Why not D: This counts 5 pairs of dice scores as adding to 9, mistakenly counting one of the four genuine pairs twice.

  15. Question 15Answer: C

    1. For the first sock, there are 5 blue socks out of 8 in total, so P(first blue) = 5/8.
    2. Since the first sock is not replaced, only 7 socks remain for the second draw, of which 4 are blue (one blue sock has already been removed).
    3. So P(second blue, given the first was blue) = 4/7.
    4. Because the events happen one after the other, the probabilities are multiplied: P(both blue) = 5/8 x 4/7.
    5. 5/8 x 4/7 = 20/56, which simplifies (dividing by 4) to 5/14.
    • Why not A: This treats the two draws as independent, using a probability of 5/8 for each draw, as if the first sock were replaced before the second draw; the problem states there is no replacement.
    • Why not B: This correctly reduces the number of blue socks to 4 for the second draw, but forgets to also reduce the total number of socks remaining to 7, using 4/8 instead of 4/7.
    • Why not D: This only finds the probability for the first sock being blue (5/8) and stops there, without accounting for the second draw at all.

More on this strand

More free ESAT practice

Every strand of the published ESAT specification, with worked solutions throughout.

Choose a subject →