Pearson Edexcel International Advanced Level Mathematics: Probability, and correlation and regression (YMA01) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions follow Unit S1 sections 3 and 4 (probability, then correlation and regression) in Pearson’s International Advanced Subsidiary/Advanced Level specification (YMA01), Issue 3, April 2019. Per the specification, Unit S1 is optional, both for IAS and for IAL, in Mathematics and in Further Mathematics. Teaching is in the study guide and the revision notes. Mark totals are Marlbridge’s own allocations.

Questions

1. A letter is chosen at random from the word STATISTICS. Find (a) P(the letter is S or T) [1], (b) P(the letter is a consonant, given that it is not S) [2].

2. For events A and B, P(A) = 0.5, P(B) = 0.35 and P(A ∪ B) = 0.7.

(a) Find P(A ∩ B). [2] (b) Find P(A | B). [1] (c) Determine whether A and B are independent. [1]

3. Events A and B are independent with P(A) = 0.3 and P(A ∪ B) = 0.65.

(a) Find P(B). [2] (b) Find the probability that exactly one of A and B occurs. [2]

4. Machine M makes 60% of a factory’s items and machine N makes the rest. Of M’s items, 3% are defective. Of N’s items, 5% are defective.

(a) Find the probability that an item chosen at random is defective. [2] (b) Given that it is defective, find the probability that it came from N. [2] (c) Two items are chosen at random from N’s output. Find the probability that exactly one is defective. [2]

5. A box holds 4 yellow and 3 green counters. Two are taken at random without replacement.

(a) Find the probability that they are the same colour. [3] (b) Given that they are the same colour, find the probability that both are green. [2]

6. Of 40 students, 18 take art, 16 take music and 14 take drama. Also 6 take art and music, 5 take art and drama, 4 take music and drama, and 2 take all three.

(a) Find the number of students in each of the eight regions of a three-set Venn diagram. [3] (b) Find the probability that a randomly chosen student takes exactly one of the three subjects. [2] (c) Find P(art | drama) and state, with a reason, whether taking art and taking drama are independent. [2]

7. On nine days the mean outdoor temperature T (°C) and the gas used G (units) were recorded. Σ T = 76, Σ G = 380, Σ T² = 796, Σ G² = 17 616, Σ TG = 2721. Find the regression line of G on T, giving the constants to 3 significant figures. [4]

8. In question 7 the temperatures lay between 2 °C and 15 °C, and the line is G = 68.9 − 3.16T.

(a) Which is the explanatory variable? [1] (b) Interpret the gradient in context. [1] (c) Predict G when T = 9. [1] (d) Explain why the line should not be used to predict G when T = 25. [2] (e) Comment on the meaning of the constant 68.9. [1]

9. A regression line for cars of mass m kg (between 800 and 1600) and fuel use E litres per 100 km is q = 2.1 + 0.35p, where p = (m − 1000)/100 and q = E − 4.

(a) Express E in terms of m. [3] (b) Estimate E when m = 1400. [1] (c) The correlation coefficient between p and q is 0.94. State the coefficient between m and E, with a reason. [1]

10. (a) For a data set, Sxx = 85, Syy = 120 and Sxy = −78. Calculate r. [2] (b) Interpret your value. [1] (c) Nine points have x = 1, 2, …, 9 and y = (x − 5)². Show that r = 0 and comment. [2]

11. A shop records weekly advertising spend x (£100) and sales y (tens of items): (2, 14), (3, 19), (5, 22), (6, 31), (8, 35), (10, 47).

(a) Find Sxx, Sxy and Syy. [3] (b) Find the regression line of y on x. [3] (c) Find r. [2] (d) Predict y when x = 7, and comment on a prediction for x = 20. [2]

12. A candidate has up to three attempts at a test. The probability of passing is 0.6 on the first attempt, 0.7 on the second and 0.8 on the third. Attempts stop after a pass.

(a) Find the probability that the candidate eventually passes. [3] (b) Given that the candidate passes, find the probability that it was on the third attempt. [2] (c) Two candidates take the test independently. Find P(both eventually pass) [1] and P(at least one passes on the first attempt) [2].

Answers

1. (a) S and T appear 3 times each out of 10: P = 6/10 = 3/5 [1] (b) Not S leaves 7 letters, of which T, T, T, C are consonants [1]. P = 4/7 [1] Examiner insight: in a conditional question, the denominator shrinks to the given condition, so state it (“7 letters”) to show the method.

2. (a) P(A ∩ B) = 0.5 + 0.35 − 0.7 [1] = 0.15 [1] (b) P(A | B) = 0.15/0.35 = 3/7 [1] (c) P(A) P(B) = 0.175 ≠ 0.15 [1], so not independent. Examiner insight: a conclusion needs both values quoted and a verdict. Writing only “no” gives little to credit.

3. (a) 0.3 + p − 0.3p = 0.65 [1] so 0.7p = 0.35 and p = 0.5 [1] (b) (0.3)(0.5) + (0.7)(0.5) [1] = 0.5 [1] Examiner insight: “exactly one” needs two routes added. A common slip is giving only P(A′ ∩ B), the route that was easiest to find.

4. (a) (0.6)(0.03) + (0.4)(0.05) [1] = 0.018 + 0.02 = 0.038 [1] (b) 0.02 / 0.038 [1] = 10/19 = 0.526 [1] (c) 2 × (0.05)(0.95) [1] = 0.095 [1] Examiner insight: writing the route products on the tree before adding keeps part (b) easy, because the numerator is already there.

5. (a) P(YY) = (4/7)(3/6) = 2/7 [1]; P(GG) = (3/7)(2/6) = 1/7 [1]; total 3/7 [1] (b) P(GG | same) = (1/7)/(3/7) [1] = 1/3 [1] Examiner insight: using 4/7 twice, as with replacement, is the usual error here, and it changes every later line.

6. (a) Three-way region: 2. Two-only regions: art and music 6 − 2 = 4, art and drama 5 − 2 = 3, music and drama 4 − 2 = 2 [1]. Only art 18 − 4 − 3 − 2 = 9, only music 16 − 4 − 2 − 2 = 8, only drama 14 − 3 − 2 − 2 = 7 [1]. Union = 9 + 8 + 7 + 4 + 3 + 2 + 2 = 35, so none = 5 [1] (b) (9 + 8 + 7)/40 [1] = 24/40 = 3/5 [1] (c) P(art | drama) = 5/14 [1]. P(art) P(drama) = (18/40)(14/40) = 0.1575 ≠ 5/40 = 0.125, so not independent [1] Examiner insight: checking that all eight regions sum to 40 is a quick test that catches an early subtraction slip.

7. Sxx = 796 − 76²/9 = 154.2 [1]. Sxy = 2721 − (76)(380)/9 = −487.9 [1]. b = −487.9/154.2 = −3.16 [1]. a = 380/9 − (−3.1635)(76/9) = 68.9, so G = 68.9 − 3.16T [1] Examiner insight: keep extra decimals until the end. Rounding b to −3.16 before finding a shifts the third significant figure of a.

8. (a) Temperature [1] (b) Each 1 °C rise in temperature reduces the predicted gas use by about 3.16 units [1] (c) 68.94 − 3.1635(9) = 40.5 [1] (d) 25 °C is outside the observed range 2 to 15 °C, so this is extrapolation [1]. The line gives a negative value (about −10), which is impossible [1] (e) It is the prediction at 0 °C, which is outside the data (2 to 15 °C), so it is an extrapolated estimate [1] Examiner insight: a comment gains credit when it refers to the actual range or value in the question, rather than “extrapolation is unreliable” alone.

9. (a) E − 4 = 2.1 + 0.35(m − 1000)/100 [1]. E = 6.1 + 0.0035(m − 1000) [1]. E = 2.6 + 0.0035m [1] (b) 2.6 + 0.0035(1400) = 7.5 [1] (c) 0.94, because both codings multiply by positive constants so r is unchanged [1] Examiner insight: substituting both codings, not just one, is the main step to show. Check by converting one point from the coded line.

10. (a) r = −78 / √(85 × 120) [1] = −78/101.0 = −0.772 [1] (b) Fairly strong negative linear correlation [1] (c) Σx = 45, Σy = 60, Σxy = 300, so Sxy = 300 − (45)(60)/9 = 0 [1], hence r = 0. The points fit a curve exactly, so r = 0 does not mean no relationship, only no linear one [1] Examiner insight: “no correlation” and “no relationship” are not the same claim, and answers that blur them lose the point.

11. (a) Σx = 34, Σy = 168, Σx² = 238, Σy² = 5436, Σxy = 1131 [1]. Sxx = 238 − 34²/6 = 45.33 [1], Sxy = 1131 − (34)(168)/6 = 179 and Syy = 5436 − 168²/6 = 732 [1] (b) b = 179/45.33 [1] = 3.95 (3 s.f.). x̄ = 34/6, ȳ = 28, so a = 28 − 3.9485(34/6) = 5.625 [1]. y = 5.63 + 3.95x [1] (c) r = 179 / √(45.33 × 732) [1] = 179/182.2 = 0.983 [1] (d) y = 5.625 + 3.9485(7) = 33.3 [1]. x = 20 is far outside the data (2 to 10), so extrapolation: unreliable [1] Examiner insight: showing the sums and S-values separately lets a slip in one be seen and not repeated across the rest.

12. (a) P(pass) = 0.6 + (0.4)(0.7) + (0.4)(0.3)(0.8) [1] = 0.6 + 0.28 + 0.096 [1] = 0.976 [1] (b) 0.096/0.976 [1] = 6/61 = 0.0984 [1] (c) 0.976² = 0.953 [1]. P(at least one first-attempt pass) = 1 − (0.4)² [1] = 0.84 [1] Examiner insight: the later probabilities here are conditional on earlier failures, so draw the tree and multiply along every branch used.

Where marks are usually lost

• Using the unconditional probability where a conditional one is needed, or swapping P(A | B) and P(B | A).
• Not stating the condition’s total as the denominator in a conditional probability.
• Calling events independent because they look unrelated, rather than testing P(A ∩ B) = P(A) P(B).
• Keeping the same denominators when items are not replaced.
• Adding where the product law applies, or multiplying where routes should be added.
• Leaving out an arithmetic step in Sxx or Sxy, so one slip changes everything after it.
• Rounding a and b early, then predicting with the rounded values.
• Commenting on a prediction without referring to the data range given.
• Saying “no correlation” when the pattern is a curve.

Next steps

• Revise from the revision notes.
• Reread the study guide for any question that went wrong.
• Return to the Edexcel IAL Mathematics hub and the checklist.
• Find gaps with the diagnostics.
• Take a free trial class with our teachers.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and Pure Mathematics specification, Issue 3, April 2019, published by Pearson Education Limited. Unit S1: Statistics 1, sections 3 and 4.