Probabilities that sum to 1 - Edexcel GCSE Maths (Calculator) Exam Questions

Practise 28 exam-style questions on Probabilities that sum to 1 for Edexcel GCSE Maths (Calculator). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.

28 questions
1–5 marks each
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20 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

2 marks

A biased five-sided spinner is numbered 1 to 5.
The table shows the probability that the spinner lands on each of the numbers 1, 2, 3 and 4.
The probability for 5 is missing from the table.

The spinner is spun once.
Work out the probability that it lands on an odd number.

Table with columns "Number", "1", "2", "3", "4", "5". Number: Probability; 1: 0.18; 2: 0.24; 3: 0.09; 4: 0.31; 5: .

Model answer

P(5) = 1 − (0.18 + 0.24 + 0.09 + 0.31) = 1 − 0.82 = 0.18
P(odd) = P(1) + P(3) + P(5) = 0.18 + 0.09 + 0.18 = 0.45

Explanation

The five outcomes are mutually exclusive and exhaustive, so their probabilities add up to 1 - this gives P(5). The odd numbers are 1, 3 and 5, and because the outcomes are mutually exclusive you can add their probabilities.

Mark scheme

1 mark
M1 for 1 − (0.18 + 0.24 + 0.09 + 0.31) (= 0.18) for P(5)
oe; correct answer with no working scores full marks
1 mark
A1 for 0.45
cao; P(odd) = 0.18 + 0.09 + 0.18

Question 2

3 marks

The table shows the probabilities that a student chosen at random gets each result in a music exam.
The probability for Fail is missing from the table.

500 students take the exam.
Work out an estimate for the number of these students who fail.

Table with columns "Result", "Distinction", "Merit", "Pass", "Fail". Result: Probability; Distinction: 0.18; Merit: 0.36; Pass: 0.29; Fail: .

Model answer

P(fail) = 1 − (0.18 + 0.36 + 0.29) = 1 − 0.83 = 0.17
Estimate = 0.17 × 500 = 85 students

Explanation

Every student gets exactly one result, so the four probabilities add up to 1. Subtract the three given probabilities from 1 to find P(Fail), then multiply by the number of students to get the expected number.

Mark scheme

1 mark
M1 for 1 − (0.18 + 0.36 + 0.29) (= 0.17)
oe
1 mark
M1 for their 0.17 × 500
their 0.17 must come from subtracting from 1
1 mark
A1 for 85
cao; correct answer with no working scores full marks

Question 3

3 marks

A, B and C are three mutually exclusive events.
P(A) = 0.3
P(B) = 2x
P(C) = x + 0.1

Find the range of possible values of x.
You must show your working.

Model answer

Mutually exclusive events cannot have probabilities that total more than 1:
0.3 + 2x + (x + 0.1) ≤ 1
3x + 0.4 ≤ 1
3x ≤ 0.6
x ≤ 0.2
Also, P(B) = 2x cannot be negative, so x ≥ 0 (and then x + 0.1 ≥ 0 too).
So 0 ≤ x ≤ 0.2

Explanation

For mutually exclusive events, P(A or B or C) is the sum of their probabilities, and this cannot be more than 1 - it equals 1 only if the events are also exhaustive. Each probability must also be at least 0, which gives the lower limit. Both limits are needed for a full range.

Mark scheme

1 mark
M1 for 0.3 + 2x + x + 0.1 ≤ 1 or 3x + 0.4 ≤ 1
accept = for this mark
1 mark
A1 for x ≤ 0.2
cao
1 mark
B1 for 0 ≤ x ≤ 0.2
accept 0 < x ≤ 0.2; needs a reason for the lower limit, e.g. P(B) = 2x cannot be negative

Question 4

5 marks

A biased six-sided dice is numbered 1 to 6.
The table shows the probability of each score.

The probability of rolling an even number is 0.53.

The dice is rolled 400 times.
Work out an estimate for the number of times the dice lands on 4 or 5.

Table with columns "Score", "1", "2", "3", "4", "5", "6". Score: Probability; 1: x; 2: x; 3: x; 4: y; 5: y; 6: 0.25.

Model answer

The probabilities add up to 1:
3x + 2y + 0.25 = 1, so 3x + 2y = 0.75
P(even) = P(2) + P(4) + P(6) = x + y + 0.25 = 0.53, so x + y = 0.28
Multiply the second equation by 2: 2x + 2y = 0.56
Subtract: x = 0.75 − 0.56 = 0.19
y = 0.28 − 0.19 = 0.09
P(4 or 5) = 2 × 0.09 = 0.18
Estimate = 0.18 × 400 = 72

Explanation

Two facts give two equations: all six probabilities sum to 1, and the even scores (2, 4 and 6) have probabilities x, y and 0.25. Solve the simultaneous equations, then multiply P(4 or 5) = 2y by 400. Check: 3(0.19) + 2(0.09) + 0.25 = 1.

Mark scheme

1 mark
P1 for 3x + 2y + 0.25 = 1 (or 3x + 2y = 0.75)
oe; correct answer with no working scores full marks
1 mark
P1 for x + y + 0.25 = 0.53 (or x + y = 0.28)
P(even) = P(2) + P(4) + P(6)
1 mark
P1 for a correct method to solve their simultaneous equations, e.g. 3x + 2y − 2(x + y) = 0.75 − 0.56
1 mark
A1 for x = 0.19 and y = 0.09
cao
1 mark
A1 for 72
cao; from 2 × 0.09 × 400

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