IGCSE · Mathematics 0580 · Probability

Introduction to probability: IGCSE Mathematics questions with answers

Free Introduction to probability questions for IGCSE Mathematics (0580), part of Probability. Each question has the answer and the marking points an examiner would look for.

Generate a full Introduction to Probability worksheet →All Probability questions

Introduction to probability practice questions

Question 1Structured · 12 marksA* challenge

Three fair six-sided dice are thrown.
The three scores can sometimes be arranged in an order that makes an arithmetic sequence (a sequence with a constant difference, which may be zero), for example 5, 1, 3 → 1, 3, 5.

(a)

Find the probability that the three scores can be arranged to make an arithmetic sequence.

[4]
(b)

Find the probability that the scores, in the order the dice are numbered, already make an arithmetic sequence.

[3]
(c)

Given that the three scores can be arranged to make an arithmetic sequence, find the probability that the three scores are all different.

[2]
(d)

Five fair dice are thrown. Find the probability that the five scores can be arranged to make an arithmetic sequence.

[3]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)736
all three equal: 6 outcomes B1
distinct sets {1,2,3}, {2,3,4}, {3,4,5}, {4,5,6}, {1,3,5}, {2,4,6} B1
6 sets × 6 orders = 36 M1
42216 = 736 A1
4
(b)112
6 constant sequences B1
each of the 6 sets in increasing or decreasing order: 12 M1
18216 = 112 A1
3
(c)67
3642 M1
67 A1
2
(d)411296
only difference 0 or 1 is possible (difference 2 would need 9) M1
6 + 2 × 5! = 246 outcomes M1
2467776 = 411296 A1
3
Question 2Structured · 6 marksA* challenge

A bag contains x red sweets and (x + 3) yellow sweets. Two sweets are taken at random without replacement. The probability that both sweets are red is 655.

(a)

Show that 31x2 − 115x − 36 = 0.

[3]
(b)

Solve the equation to find the number of sweets in the bag.

[3]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)x2x + 3 × x − 12x + 2 = 655 leads to 31x2 − 115x − 36 = 0
x2x + 3 × x − 12x + 2 = 655 M1
cross-multiplied and expanded correctly M1
31x2 − 115x − 36 = 0 with no errors A1
3
(b)x = 4, so 11 sweets
factorise or use the formula M1
x = 4 (other root rejected) A1
11 A1
3
Question 3Structured · 3 marks

A bag contains 3 red, 2 blue and 1 green counters. One counter is taken at random.

(a)

Find the probability that it is red.

[1]
(b)

Find the probability that it is not green.

[1]
(c)

Find the probability that it is yellow.

[1]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)12
12 B1
1
(b)56
56 B1
1
(c)0
0 B1
1
Question 4Structured · 3 marks

The table shows the probabilities of a biased spinner landing on each colour.

ColourRedBlueGreenYellow
Probability0.150.300.30x
Table 4.1
(a)

Find x.

[2]
(b)

The spinner is spun 200 times. Find the expected number of times it lands on blue.

[1]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)0.25
1 − sum M1
0.25 A1
2
(b)60
60 B1
1
Question 5Structured · 3 marks

A biased spinner can land on red, blue, green or yellow. The table shows some of the probabilities.

ColourRedBlueGreenYellow
Probability0.240.270.24
Table 5.1
(a)

Find the probability that the spinner lands on yellow.

[2]
(b)

The spinner is spun 50 times. Find the expected number of times it lands on red.

[1]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)0.25
1 − (0.24 + 0.27 + 0.24) M1
0.25 A1
2
(b)12
12 B1
1
Question 6Structured · 6 marksA* challenge

A bag contains x red sweets and (x + 2) yellow sweets. Two sweets are taken at random without replacement. The probability that both sweets are red is 1895.

(a)

Show that 23x2 − 203x − 36 = 0.

[3]
(b)

Solve the equation to find the number of sweets in the bag.

[3]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)x2x + 2 × x − 12x + 1 = 1895 leads to 23x2 − 203x − 36 = 0
x2x + 2 × x − 12x + 1 = 1895 M1
cross-multiplied and expanded correctly M1
23x2 − 203x − 36 = 0 with no errors A1
3
(b)x = 9, so 20 sweets
factorise or use the formula M1
x = 9 (other root rejected) A1
20 A1
3

More Probability questions

Relative & expected frequencyCombined events & tree diagrams

Want a mentor, not just a paper?

Riyaz Shaikh has mentored IGCSE, IB and A Level students for 15+ years, one-on-one, online and in Mumbai. The hardest questions on this site are the ones he uses to find a student's weak points.

Meet Riyaz →

Original Cambridge-style practice questions written for mentorriyaz.in. These are not Cambridge past paper questions, and this site is not affiliated with or endorsed by Cambridge University Press & Assessment. Updated 2026-11-04. Paper generator · mentorriyaz.in