IB · Mathematics: Analysis & Approaches SL & HL · Statistics & probability

Discrete random variables: IB Mathematics: Analysis & Approaches questions with answers

Free Discrete random variables questions for IB Mathematics: Analysis & Approaches (SL & HL), part of Statistics & probability. Each question has the answer and the marking points an examiner would look for.

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Discrete random variables practice questions

Question 1Structured · 13 marks

Two fair 4-sided dice, each numbered 1 to 4, are rolled. Let M be the larger of the two scores (if the scores are equal, M is that score).

(a)

Show that P(M = 4) = 716.

[2]

(b)

Complete the probability distribution of M in a table.

m	1	2	3	4
P(M = m)

Table 1.1

[3]

(c)

Find E(M).

[2]

(d)

In a game, a player pays $c to roll the dice and receives $M². Find the value of c for which the game is fair.

[3]

(e)

The dice are rolled twice. Find the probability that M ≤ 3 on both occasions.

[3]

Show answers and mark scheme

Part	Answer and marking points	Marks
(a)	shown P(M ≤ 3) = (34)² M1 1 − 916 = 716 AG A1	2
(b)	P(M = 1) = 116, P(M = 2) = 316, P(M = 3) = 516, P(M = 4) = 716 P(M = m) = 2m − 116 or listing outcomes (M1) at least two further probabilities correct A1 all correct A1	3
(c)	258 Σ_m=1⁴m × 2m − 116 (M1) 258 A1	2
(d)	$858 fair game: E(winnings) = cost (M1) E(M²) = 858 A1 c = 858 (= 10.63) A1	3
(e)	81256 P(M ≤ 3) = (34)² = 916 (M1)(A1) 81256 A1	3

Question 2Structured · 6 marks

The probability distribution of a discrete random variable X is given by P(X = x) = kx, for x ∈ {1, 2, 3, 4}.

(a)

Find the value of k.

[2]

(b)

Find E(X).

[2]

(c)

Find P(X > E(X)).

[2]

Show answers and mark scheme

Part	Answer and marking points	Marks
(a)	110 k + 2k + 3k + 4k = 1 (M1) k = 110 A1	2
(b)	3 ΣxP(X = x) = 1 × k + 2 × 2k + 3 × 3k + 4 × 4k (M1) 3 A1	2
(c)	25 values of x greater than 3 (M1) 25 A1	2

Question 3Structured · 13 marks

Two fair 4-sided dice, each numbered 1 to 4, are rolled. Let M be the larger of the two scores (if the scores are equal, M is that score).

(a)

Show that P(M = 4) = 716.

[2]

(b)

Complete the probability distribution of M in a table.

m	1	2	3	4
P(M = m)

Table 3.1

[3]

(c)

Find E(M).

[2]

(d)

In a game, a player pays $c to roll the dice and receives $M². Find the value of c for which the game is fair.

[3]

(e)

The dice are rolled twice. Find the probability that M ≤ 2 on both occasions.

[3]

Show answers and mark scheme

Part	Answer and marking points	Marks
(a)	shown P(M ≤ 3) = (34)² M1 1 − 916 = 716 AG A1	2
(b)	P(M = 1) = 116, P(M = 2) = 316, P(M = 3) = 516, P(M = 4) = 716 P(M = m) = 2m − 116 or listing outcomes (M1) at least two further probabilities correct A1 all correct A1	3
(c)	258 Σ_m=1⁴m × 2m − 116 (M1) 258 A1	2
(d)	$858 fair game: E(winnings) = cost (M1) E(M²) = 858 A1 c = 858 (= 10.63) A1	3
(e)	116 P(M ≤ 2) = (24)² = 14 (M1)(A1) 116 A1	3

Question 4Structured · 6 marks

The probability distribution of a discrete random variable X is given by P(X = x) = k(5 − x), for x ∈ {1, 2, 3, 4}.

(a)

Find the value of k.

[2]

(b)

Find E(X).

[2]

(c)

Find P(X > E(X)).

[2]

Show answers and mark scheme

Part	Answer and marking points	Marks
(a)	110 4k + 3k + 2k + k = 1 (M1) k = 110 A1	2
(b)	2 ΣxP(X = x) = 1 × 4k + 2 × 3k + 3 × 2k + 4 × k (M1) 2 A1	2
(c)	310 values of x greater than 2 (M1) 310 A1	2

Question 5Structured · 13 marks

Two fair 6-sided dice, each numbered 1 to 6, are rolled. Let M be the larger of the two scores (if the scores are equal, M is that score).

(a)

Show that P(M = 6) = 1136.

[2]

(b)

Complete the probability distribution of M in a table.

m	1	2	3	4	5	6
P(M = m)

Table 5.1

[3]

(c)

Find E(M).

[2]

(d)

In a game, a player pays $c to roll the dice and receives $M². Find the value of c for which the game is fair.

[3]

(e)

The dice are rolled twice. Find the probability that M ≤ 2 on both occasions.

[3]

Show answers and mark scheme

Part	Answer and marking points	Marks
(a)	shown P(M ≤ 5) = (56)² M1 1 − 2536 = 1136 AG A1	2
(b)	P(M = 1) = 136, P(M = 2) = 112, P(M = 3) = 536, P(M = 4) = 736, P(M = 5) = 14, P(M = 6) = 1136 P(M = m) = 2m − 136 or listing outcomes (M1) at least two further probabilities correct A1 all correct A1	3
(c)	16136 Σ_m=1⁶m × 2m − 136 (M1) 16136 A1	2
(d)	$79136 fair game: E(winnings) = cost (M1) E(M²) = 79136 A1 c = 79136 (= 21.97) A1	3
(e)	181 P(M ≤ 2) = (26)² = 19 (M1)(A1) 181 A1	3

Question 6Structured · 6 marks

The probability distribution of a discrete random variable X is given by P(X = x) = kx², for x ∈ {1, 2, 3, 4}.

(a)

Find the value of k.

[2]

(b)

Find E(X).

[2]

(c)

Find P(X > E(X)).

[2]

Show answers and mark scheme

Part	Answer and marking points	Marks
(a)	130 k + 4k + 9k + 16k = 1 (M1) k = 130 A1	2
(b)	103 ΣxP(X = x) = 1 × k + 2 × 4k + 3 × 9k + 4 × 16k (M1) 103 A1	2
(c)	815 values of x greater than 103 (M1) 815 A1	2

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Original IB-style practice questions written for mentorriyaz.in. These are not IB past paper questions, and this site is not affiliated with or endorsed by the International Baccalaureate Organization. Updated 2026-11-04. Paper generator · mentorriyaz.in