IB DP Mathematics: Applications and Interpretation – Probability, discrete random variables, binomial and normal distributions 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.

These practice questions cover probability, discrete random variables and the binomial and normal distributions for IB Diploma Programme Mathematics: Applications and Interpretation. They are aligned to the IB Mathematics: applications and interpretation guide, first assessment 2021, syllabus sections 4.5–4.9, which are common content for SL and HL. They follow the IB guide for first assessment 2021, which remains the examined syllabus until the new course is first assessed in May 2029, so they apply to the May and November 2026, 2027 and 2028 sessions.

The guide lists every paper in this course as “technology required”, so every question here is labelled (calculator allowed). Give answers exactly or to 3 s.f. unless told otherwise.

Links: course hub · printable checklist · study guide · revision notes · statistics and probability overview

Questions

1. (calculator allowed) A drawing pin is thrown 400 times and lands point up 148 times.

(a) Estimate the probability that it lands point up. [1] (b) Find the expected number of point-up landings in 250 further throws. [1]

2. (calculator allowed) P(A) = 0.45, P(B) = 0.3 and P(A ∪ B) = 0.6.

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

3. (calculator allowed) The discrete random variable X has this distribution.

x	0	1	2	3	4
P(X = x)	0.1	0.2	k	2k	0.25

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

4. (calculator allowed) Two fair four-sided dice, each numbered 1 to 4, are rolled. The score is the larger of the two numbers (or the common number if they are equal).

(a) Draw a sample space diagram showing the score for every outcome. [2] (b) Find P(score = 3). [1] (c) Given that the two numbers are different, find the probability that the score is 4. [2]

5. (calculator allowed) Of 120 visitors to a museum, 70 visited the gallery (G), 45 visited the café (C) and 20 visited neither.

(a) Find the number who visited both. [2] (b) Find P(C | G). [2] (c) Show that G and C are not independent. [2]

6. (calculator allowed) A drawer holds 7 black socks and 5 grey socks. Two socks are taken at random without replacement.

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

7. (calculator allowed) A quiz has 15 multiple-choice questions, each with 4 options. Ravi guesses every answer at random. Let X be the number he gets right.

(a) Find P(X = 5). [2] (b) Find P(X ≥ 6). [2] (c) Find E(X) and the standard deviation of X. [2]

8. (calculator allowed) Bus travel times, T minutes, are modelled by T ~ N(34, 4.5²).

(a) Find P(T > 40). [2] (b) Find P(30 < T < 38). [2] (c) 95% of bus trips take less than t minutes. Find t. [2] (d) Without a GDC, write down an interval containing about 95% of travel times. [1]

9. (calculator allowed) The masses of bags of flour, M grams, are modelled by M ~ N(1005, 6²). Bags lighter than 995 g are rejected.

(a) Find the probability that a bag is rejected. [2] (b) Find the expected number of rejected bags in a day’s output of 2000 bags. [2] (c) The heaviest 3% of bags are checked by hand. Find the least mass of a checked bag, to the nearest gram. [2] (d) A box holds 12 bags chosen independently. Find the probability that at least one bag in the box is rejected. [3]

10. (calculator allowed) At a charity stall a player pays 5 euros and rolls two fair six-sided dice. A double six pays out 50 euros, any other double pays out 10 euros, a total of 7 pays out 5 euros, and anything else pays nothing. Let X be the player’s gain in euros.

(a) Find the probability distribution of X. [4] (b) Find E(X). [2] (c) State, with a reason, whether the game is fair. [1] (d) The game is played 300 times. Find the charity’s expected profit. [2]

11. (calculator allowed) Leila scores each free throw with probability 0.72, independently. She takes 20 free throws. Let X be the number she scores.

(a) Find P(X = 15). [2] (b) Find P(X ≥ 12). [2] (c) Find P(13 < X ≤ 17). [2] (d) Find the mean and variance of X. [2] (e) Find the least number of throws n for which the probability that she misses at least once exceeds 0.999. [3]

Answers

1. (a) 148/400 = 0.37 [1] (b) 250 × 0.37 = 92.5 [1] Examiner insight: an expected number is left as 92.5; rounding it to 92 or 93 when not asked can cost the accuracy mark.

2. (a) 0.6 = 0.45 + 0.3 − P(A ∩ B) [1], so P(A ∩ B) = 0.15 [1] (b) P(A)P(B) = 0.45 × 0.3 = 0.135 [1]; 0.135 ≠ 0.15, so not independent [1] Examiner insight: the reasoning mark needs both numbers written and compared; “not independent” alone scores nothing.

3. (a) 0.1 + 0.2 + k + 2k + 0.25 = 1 [1], so 3k = 0.45, k = 0.15 [1] (b) E(X) = 0(0.1) + 1(0.2) + 2(0.15) + 3(0.3) + 4(0.25) [1] = 2.4 [1] (c) 0.15 + 0.3 + 0.25 = 0.7 [1] Examiner insight: a wrong k carries through to (b) and (c) as follow-through, so show the substitution even if you are unsure of k.

4. (a) A 4 × 4 grid with all 16 outcomes [1] and correct scores [1]:

	1	2	3	4
1	1	2	3	4
2	2	2	3	4
3	3	3	3	4
4	4	4	4	4

(b) 5 of 16 outcomes, 5/16 [1] (c) 12 outcomes have different numbers, and 6 of these have score 4 [1], so 6/12 = 1/2 [1] Examiner insight: in (c) the denominator is the 12 “different” outcomes, not 16; the method mark is for the reduced sample space.

5. (a) 120 − 20 = 100 visited at least one [1]; 70 + 45 − 100 = 15 [1] (b) P(C | G) = 15/70 [1] = 3/14 ≈ 0.214 [1] (c) P(G ∩ C) = 15/120 = 0.125 [1]; P(G)P(C) = (70/120)(45/120) = 0.219 ≠ 0.125, so not independent [1] Examiner insight: “show that” needs both values written out; an exact fraction or 3 s.f. value is fine, but a conclusion with no numbers earns nothing.

6. (a) P(BB) = (7/12)(6/11) = 42/132 [1]; P(GG) = (5/12)(4/11) = 20/132 [1]; total = 62/132 = 31/66 ≈ 0.470 [1] (b) (20/132) ÷ (62/132) [1] = 10/31 ≈ 0.323 [1] Examiner insight: using 7/12 × 7/12 (with replacement) loses the method mark on the first branch and every accuracy mark after it.

7. (a) X ~ B(15, 0.25) [1]; P(X = 5) = 0.165 [1] (b) P(X ≥ 6) = 1 − P(X ≤ 5) [1] = 0.148 [1] (c) E(X) = 15 × 0.25 = 3.75 [1]; Var(X) = 3.75 × 0.75 = 2.8125, so σ = 1.68 [1] Examiner insight: writing the distribution and “1 − P(X ≤ 5)” earns the method mark even if the GDC entry is wrong; a bare 0.148 risks both marks.

8. (a) P(T > 40), normal cdf with lower 40, μ = 34, σ = 4.5 [1] = 0.0912 [1] (b) Lower 30, upper 38 [1], = 0.626 [1] (c) P(T < t) = 0.95 [1]; t = 41.4 minutes [1] (d) 34 ± 2(4.5): 25 to 43 minutes [1] Examiner insight: entering 4.5² = 20.25 as the standard deviation gives wrong answers and no accuracy marks in (a) to (c).

9. (a) P(M < 995) [1] = 0.0478 [1] (b) 2000 × 0.04779… [1] = 95.6 bags [1] (c) P(M < m) = 0.97 [1]; m = 1016.28…, so 1016 g [1] (d) Let Y ~ B(12, 0.04779…) [1]; P(Y ≥ 1) = 1 − P(Y = 0) = 1 − (0.95220…)¹² [1] = 0.444 [1] Examiner insight: (d) needs the switch from a normal to a binomial model stated; the complement 1 − P(Y = 0) is the method mark.

10. (a) Gains 45, 5, 0, −5 [1]; P(X = 45) = 1/36 and P(X = 5) = 5/36 [1]; P(X = 0) = 6/36 [1]; P(X = −5) = 24/36 [1]

x	45	5	0	−5
P(X = x)	1/36	5/36	6/36	24/36

(b) E(X) = (45 + 25 + 0 − 120)/36 [1] = −25/18 ≈ −1.39 euros [1] (c) E(X) ≠ 0, so not fair; the player loses on average [1] (d) Charity gains 25/18 per game [1]; 300 × 25/18 = 416.67 ≈ 417 euros [1] Examiner insight: using payouts (50, 10, 5, 0) instead of gains loses the first mark in (a); follow-through then applies to later parts.

11. (a) X ~ B(20, 0.72) [1]; P(X = 15) = 0.193 [1] (b) 1 − P(X ≤ 11) [1] = 0.922 [1] (c) P(X ≤ 17) − P(X ≤ 13) [1] = 0.631 [1] (d) Mean = 20 × 0.72 = 14.4 [1]; variance = 14.4 × 0.28 = 4.032 [1] (e) 1 − 0.72ⁿ > 0.999 [1]; GDC table: n = 21 gives 0.99899…, n = 22 gives 0.99927… [1]; n = 22 [1] Examiner insight: in (e), show the values either side of 0.999; an answer of 22 with no inequality or table earns at most the final mark.

Where marks are usually lost

• Not stating X ~ B(n, p) or X ~ N(μ, σ²) before a GDC answer, so no method mark is available.
• Converting P(X > r) to 1 − P(X ≤ r − 1) for a binomial, which shifts the boundary by one.
• Entering the variance as σ in normal cdf or inverse normal.
• Using the right-tail area in inverse normal when the GDC wants the left-tail area.
• Using the whole sample space as the denominator in a conditional probability.
• Treating draws without replacement as independent.
• Working from payouts instead of gains in a fair-game question.
• Rounding an intermediate probability early, then using it in a binomial calculation.
• Answering a “show that” or “determine whether” part with a conclusion and no comparison.

Next steps

• Revise from the revision notes.
• Re-read the study guide for any question you missed.
• Return to the course hub.
• Tick off sections on the printable checklist.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

International Baccalaureate Organization, Diploma Programme, Mathematics: applications and interpretation guide, first assessment 2021.