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

Practise 28 exam-style questions on Probabilities that sum to 1 for Edexcel GCSE Maths (Non-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 bag contains only red, blue, green and yellow counters.
A counter is taken at random from the bag.
The table shows the probability that the counter is red, blue or green.

Work out the probability that the counter is yellow.

Table with columns "Colour", "Red", "Blue", "Green", "Yellow". Colour: Probability; Red: 0.2; Blue: 0.35; Green: 0.1; Yellow: .

Model answer

P(yellow) = 1 − (0.2 + 0.35 + 0.1) = 1 − 0.65 = 0.35

Explanation

The four colours are all the possible outcomes (an exhaustive set) and a counter cannot be two colours at once, so the four probabilities add up to 1.

Mark scheme

1 mark
M1 for 1 − (0.2 + 0.35 + 0.1)
correct answer with no working scores full marks
1 mark
A1 for 0.35
oe

Question 2

2 marks

A and B are mutually exclusive events.
P(A) = 0.3 and P(B) = 0.45

Work out the probability that neither A nor B happens.

Model answer

A and B cannot both happen, so P(A or B) = 0.3 + 0.45 = 0.75
P(neither) = 1 − 0.75 = 0.25

Explanation

Because A and B are mutually exclusive, their probabilities can be added. 'A', 'B' and 'neither' form an exhaustive set of mutually exclusive events, so their probabilities add up to 1.

Mark scheme

1 mark
M1 for P(A or B) = 0.3 + 0.45 = 0.75
correct answer with no working scores full marks
1 mark
A1 for 0.25
oe

Question 3

3 marks

A bag contains only red, blue and green counters.
A counter is taken at random from the bag.
P(red) : P(blue) : P(green) = 2 : 3 : 5
There are 12 more green counters than red counters.

How many blue counters are in the bag?

Model answer

The probabilities add up to 1 and are in the ratio 2 : 3 : 5, so P(red) = 0.2, P(blue) = 0.3, P(green) = 0.5
The numbers of counters are in the same ratio 2 : 3 : 5.
Green − red = 5 parts − 2 parts = 3 parts = 12, so 1 part = 4
Blue = 3 parts = 12

Explanation

The probability of each colour is proportional to the number of counters of that colour, so the counters are in the same ratio. The difference of 3 parts between green and red tells you the size of one part.

Mark scheme

1 mark
P1 for P(red) = 0.2, P(blue) = 0.3, P(green) = 0.5, or for the numbers of counters being in the ratio 2 : 3 : 5
correct answer with no working scores full marks
1 mark
P1 for 3 parts = 12 (1 part = 4), or 0.3 × total = 12 giving a total of 40
1 mark
A1 for 12
cao

Question 4

4 marks

At a restaurant, each customer chooses exactly one of three desserts: cake, ice cream or fruit.
The probability that a customer chooses cake is 0.35
The probability that a customer chooses ice cream is 0.15 more than the probability that a customer chooses fruit.
200 customers are expected tomorrow.

Work out an estimate for the number of these customers who will choose fruit.

Model answer

P(ice cream) + P(fruit) = 1 − 0.35 = 0.65
Let P(fruit) = \(f\): \(f + f + 0.15 = 0.65\)
\(2f = 0.5\), so \(f = 0.25\)
Expected number choosing fruit = 0.25 × 200 = 50

Explanation

The three desserts are exhaustive and mutually exclusive, so their probabilities add up to 1. Form an equation for the two unknown probabilities, solve it, then multiply P(fruit) by the number of customers.

Mark scheme

1 mark
P1 for P(ice cream) + P(fruit) = 1 − 0.35 = 0.65
correct answer with no working scores full marks
1 mark
P1 for \(f + (f + 0.15) = 0.65\) oe
1 mark
P1 for P(fruit) = 0.25
1 mark
A1 for 50
cao

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