IB DP Mathematics: Analysis and Approaches – Probability, discrete and continuous distributions, binomial and normal 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 – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

This practice set covers the probability and distributions unit of IB Diploma Programme Mathematics: Analysis and Approaches. It is aligned to the IB Mathematics: analysis and approaches guide, first assessment 2021, syllabus sections 4.5–4.9 and 4.11–4.12 (SL and HL) and 4.13–4.14 (HL only). Questions 10 and 11 are HL only. It follows the IB guide for first assessment 2021, which remains the examined syllabus until the new course is first assessed in May 2029, so it applies to the May and November 2026, 2027 and 2028 sessions.

Paper 1 allows no technology; Paper 2 (and HL Paper 3) require it. Questions are labelled “(calculator-free)” or “(calculator allowed)” to match. Give calculator-allowed answers exactly or to 3 significant figures.

Learn the methods first in the probability and distributions study guide and the revision notes. The IB DP Maths AA course hub and the printable syllabus checklist show where this unit sits.

Questions

1. (calculator-free) The probability that a seed germinates is 0.85.

(a) Write down the probability that a seed does not germinate. [1] (b) 240 seeds are planted. Find the expected number that do not germinate. [1]

2. (calculator-free) Events A and B satisfy P(A) = 0.5, P(B) = 0.4 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. Justify your answer. [2]

3. (calculator-free) A box contains 4 green and 6 yellow pens. Two pens are taken at random without replacement.

(a) Find the probability that both pens are green. [2] (b) Find the probability that at least one pen is yellow. [1] (c) Given that both pens are the same colour, find the probability that they are both yellow. [3]

4. (calculator-free) The discrete random variable X has P(X = x) = k(x² + 1) for x ∈ {0, 1, 2, 3}.

(a) Find the value of k. [2] (b) Find E(X). [2] (c) Find P(X ≥ 2). [1]

5. (calculator-free) A player pays c points to spin a spinner once. The player receives 12 points with probability 1/8, 4 points with probability 1/4, and nothing otherwise. Find the value of c for which the game is fair. [3]

6. (calculator allowed) 12% of the emails an office receives are spam. A random sample of 25 emails is taken, and X is the number that are spam.

(a) Find P(X = 3). [2] (b) Find P(X ≥ 5). [2] (c) Find E(X) and Var(X). [2]

7. (calculator allowed) The masses of apples from an orchard are normally distributed with mean 180 g and standard deviation 12 g.

(a) Find the probability that an apple has mass between 170 g and 200 g. [2] (b) A crate holds 150 apples. Find the expected number with mass over 200 g. [2] (c) 10% of apples are heavier than m grams. Find m. [2]

8. (calculator allowed) The time T minutes that runners take to finish a course is normally distributed with mean μ and standard deviation σ. 15% of runners take less than 40 minutes and 10% take more than 55 minutes. Find μ and σ. [5]

9. (calculator allowed) The length L mm of a bolt is normally distributed with mean 50 and standard deviation 0.4. A bolt is accepted if 49.3 < L < 50.7; otherwise it is rejected.

(a) Find the probability that a bolt is accepted. [2] (b) Find the probability that an accepted bolt is longer than 50.5 mm. [3] (c) 20 bolts are made. Find the probability that at most one is rejected. [3]

10. (calculator-free, HL only) A shop buys phone chargers from three suppliers: 45% from A, 35% from B and 20% from C. The probability that a charger is faulty is 0.04 from A, 0.02 from B and 0.06 from C.

(a) Find the probability that a charger chosen at random is faulty. [2] (b) A charger is faulty. Find the probability that it came from C. [2] (c) A charger is not faulty. Find the probability that it came from A. [3]

11. (calculator-free, HL only) The continuous random variable X has probability density function

f(x) = kx,             0 ≤ x ≤ 2
f(x) = k(6 − x)/2,     2 < x ≤ 6
f(x) = 0,              otherwise

(a) Show that k = 1/6. [2] (b) Write down the mode of X. [1] (c) Find the exact median of X. [3] (d) Find E(X). [2] (e) Find Var(X). [2] (f) Find E(3X − 2) and Var(3X − 2). [2]

Answers

1. (a) 1 − 0.85 = 0.15 [1] (b) 240 × 0.15 = 36 [1] Examiner insight: The expected number is n × p; it does not need to be a whole number, so never round it unless the question asks.

2. (a) 0.7 = 0.5 + 0.4 − P(A ∩ B) [1], so P(A ∩ B) = 0.2 [1] (b) P(A | B) = 0.2/0.4 = 0.5 [1] (c) P(A)P(B) = 0.5 × 0.4 = 0.2 [1]. This equals P(A ∩ B), so A and B are independent [1] Examiner insight: “Independent” with no numerical comparison earns nothing; the reasoning mark needs P(A)P(B) = P(A ∩ B), or P(A | B) = P(A), written out.

3. (a) (4/10)(3/9) [1] = 2/15 [1] (b) 1 − 2/15 = 13/15 [1] (c) P(both yellow) = (6/10)(5/9) = 1/3 [1]. P(same colour) = 2/15 + 1/3 = 7/15 [1]. P(yellow | same) = (1/3)/(7/15) = 5/7 [1] Examiner insight: Using 4/10 on both branches is a with-replacement model and loses the method mark as well as the answer.

4. (a) k(1 + 2 + 5 + 10) = 18k = 1 [1], so k = 1/18 [1] (b) E(X) = (0 × 1 + 1 × 2 + 2 × 5 + 3 × 10)/18 [1] = 42/18 = 7/3 [1] (c) (5 + 10)/18 = 5/6 [1] Examiner insight: On a calculator-free paper give exact fractions; a decimal such as 2.33 for 7/3 can lose the accuracy mark.

5. Expected gain = 12 × 1/8 + 4 × 1/4 − c [1]. For a fair game the expected gain is 0 [1], so 1.5 + 1 − c = 0 and c = 2.5 [1] Examiner insight: Set the expected gain to 0, not the expected receipt; forgetting to subtract the entry cost is the usual error.

6. (a) X ~ B(25, 0.12) [1]; P(X = 3) = 0.239 [1] (b) P(X ≥ 5) = 1 − P(X ≤ 4) [1] = 0.173 [1] (c) E(X) = 25 × 0.12 = 3 [1]; Var(X) = 25 × 0.12 × 0.88 = 2.64 [1] Examiner insight: Writing 1 − P(X ≤ 5) for P(X ≥ 5) is a common method error; stating the distribution first secures the method mark even if a later value slips.

7. (a) M ~ N(180, 12²); P(170 < M < 200) [1] = 0.750 [1] (b) P(M > 200) = 0.04779 [1]; 150 × 0.04779 = 7.17 [1] (c) P(M < m) = 0.9 [1], so m = 195 g [1] Examiner insight: In (c) the inverse normal needs the area to the left, 0.9; entering 0.1 gives 165 g and scores only if the tail is set to “right”.

8. P(T < 40) = 0.15 gives z = −1.0364 [1]. P(T < 55) = 0.9 gives z = 1.2816 [1]. So 40 = μ − 1.0364σ and 55 = μ + 1.2816σ [1]. Subtracting, 15 = 2.3180σ, so σ = 6.47 [1] and μ = 46.7 [1] Examiner insight: The z-values must come from N(0, 1); keep four or more figures in z and σ until the end, or μ can drift in the third figure.

9. (a) P(49.3 < L < 50.7) [1] = 0.920 [1] (b) P(L > 50.5 | accepted) = P(50.5 < L < 50.7)/P(accepted) [1] = 0.06559/0.91988 [1] = 0.0713 [1] (c) P(rejected) = 1 − 0.91988 = 0.08012, and R ~ B(20, 0.08012) [1]. P(R ≤ 1) [1] = 0.516 [1] Examiner insight: The numerator in (b) is the intersection, 50.5 < L < 50.7, not P(L > 50.5); the method mark depends on showing that range.

10. (a) P(F) = 0.45 × 0.04 + 0.35 × 0.02 + 0.2 × 0.06 [1] = 0.037 [1] (b) P(C | F) = 0.012/0.037 [1] = 12/37 [1] (c) P(A ∩ F′) = 0.45 × 0.96 = 0.432 [1]. P(F′) = 1 − 0.037 = 0.963 [1]. P(A | F′) = 0.432/0.963 = 48/107 [1] Examiner insight: Each Bayes answer earns its method mark for showing the branch product over the total; a bare fraction with no working can score zero if it is wrong.

11. (a) ∫₀² kx dx = 2k [1] and ∫₂⁶ k(6 − x)/2 dx = 4k, so 6k = 1 and k = 1/6 [1] (b) f has its maximum at x = 2, so the mode is 2 [1] (c) ∫₀² x/6 dx = 1/3 < 1/2, so the median lies in (2, 6] [1]. ∫ₘ⁶ (6 − x)/12 dx = (6 − m)²/24 = 1/2 [1], so (6 − m)² = 12 and, since m < 6, m = 6 − 2√3 [1] (d) E(X) = ∫₀² x²/6 dx + ∫₂⁶ x(6 − x)/12 dx = 4/9 + 20/9 [1] = 8/3 [1] (e) E(X²) = ∫₀² x³/6 dx + ∫₂⁶ x²(6 − x)/12 dx = 2/3 + 8 = 26/3 [1]; Var(X) = 26/3 − 64/9 = 14/9 [1] (f) E(3X − 2) = 3 × 8/3 − 2 = 6 [1]; Var(3X − 2) = 9 × 14/9 = 14 [1] Examiner insight: In a “show that” such as (a), both integrals must be shown before the conclusion; writing only “6k = 1” loses the first mark.

Where marks are usually lost

• Subtracting the overlap twice, or not at all, when using P(A ∪ B).
• Declaring independence without a numerical test.
• Keeping the same denominator on the second draw of a without-replacement question.
• Converting “≥” and “<” wrongly for a binomial variable, for example P(X ≥ 5) = 1 − P(X ≤ 5).
• Giving GDC answers with no distribution or bounds written, so no method mark is possible.
• Entering the variance where the GDC asks for the standard deviation.
• Using the right-tail area in inverse normal without setting the tail, as in Question 7(c).
• Rounding z-values early when solving for μ and σ together.
• (HL) Integrating only one piece of a piecewise pdf, or searching for the median in the wrong piece.
• (HL) Adding b to the variance under a linear transformation.

Next steps

• Recap the key facts in the probability and distributions revision notes.
• Re-learn any weak topic in the probability and distributions study guide.
• Revise integration for Question 11 with the AA calculus study guide.
• See the rest of the course on the IB DP Maths AA course hub and tick off the printable syllabus checklist.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

International Baccalaureate Organization, Diploma Programme, Mathematics: analysis and approaches guide, first assessment 2021 (published February 2019, updated November 2020).