Conditional probability - Edexcel GCSE Maths (Calculator) Exam Questions

Practise 37 exam-style questions on Conditional probability 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.

37 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

The table shows information about the members of a gym.
Each member has either standard or premium membership.

A member with premium membership is chosen at random.

Work out the probability that this member is over 40 years old.

Table with columns (row headings), "Under 25", "25 to 40", "Over 40". Standard; Under 25: 46; 25 to 40: 58; Over 40: 31 | Premium; Under 25: 22; 25 to 40: 37; Over 40: 26.

Model answer

Number of premium members = 22 + 37 + 26 = 85
26 of them are over 40.
Probability = \(\dfrac{26}{85}\) (= 0.305...)

Explanation

The member is chosen from the premium members only, so the denominator is the premium total, 85. The numerator is the number of premium members who are over 40. A common mistake is to divide by all 220 members or by the 57 members over 40.

Mark scheme

1 mark
M1 for 22 + 37 + 26 (= 85) used as the denominator
1 mark
A1 for \(\dfrac{26}{85}\) oe
accept awrt 0.306; correct answer with no working scores full marks

Question 2

2 marks

The table shows the percentages of the employees of a company who work in each department, full-time or part-time.

An employee who works part-time is chosen at random.

Work out the probability that this employee works in Production.

Table with columns "Department", "Full-time", "Part-time". Department: Sales; Full-time: 18%; Part-time: 7% | Department: Production; Full-time: 34%; Part-time: 11% | Department: Office; Full-time: 21%; Part-time: 9%.

Model answer

Percentage of employees who are part-time = 7% + 11% + 9% = 27%
Part-time in Production = 11%
Probability = \(\dfrac{11}{27}\) (= 0.407...)

Explanation

You do not need the number of employees: the percentages act like frequencies out of 100. The employee is chosen from the part-time workers, so divide 11 by the part-time total, 27. A common mistake is to give 0.11, which is the probability that a randomly chosen employee is a part-time Production worker.

Mark scheme

1 mark
M1 for 7 + 11 + 9 (= 27) used as the denominator
or 0.07 + 0.11 + 0.09 = 0.27
1 mark
A1 for \(\dfrac{11}{27}\) oe
accept awrt 0.407; correct answer with no working scores full marks

Question 3

3 marks

A bag contains 5 red counters and 7 blue counters.
Two counters are taken at random from the bag without replacement.

The two counters are different colours.

Lily says,
"The probability that the first counter taken was red is \(\dfrac{1}{2}\)."

Is Lily correct?
You must show how you get your answer.

Model answer

P(red, blue) = \(\dfrac{5}{12} \times \dfrac{7}{11} = \dfrac{35}{132}\)
P(blue, red) = \(\dfrac{7}{12} \times \dfrac{5}{11} = \dfrac{35}{132}\)
P(first red given different colours) = \(\dfrac{35}{132} \div \dfrac{70}{132} = \dfrac{1}{2}\)
Yes, Lily is correct.

Explanation

Given the colours are different, the only possible outcomes are red-blue and blue-red. Their probabilities use the same numbers multiplied in a different order, so they are equal and each is half of the total. It may be surprising that there are more blue counters, but the answer is still \(\dfrac{1}{2}\).

Mark scheme

1 mark
M1 for P(red, blue) = \(\dfrac{5}{12} \times \dfrac{7}{11}\) (= \(\dfrac{35}{132}\)) or P(blue, red) = \(\dfrac{7}{12} \times \dfrac{5}{11}\) (= \(\dfrac{35}{132}\))
1 mark
M1 for \(\dfrac{35}{132} \div \left(\dfrac{35}{132} + \dfrac{35}{132}\right)\) oe
1 mark
C1 for Yes, with a correct probability of \(\dfrac{1}{2}\) shown
accept a correct argument that P(red, blue) = P(blue, red)

Question 4

5 marks

150 people were asked whether they live in a house or a flat and whether they have a pet.

92 of the people live in a house.
The probability that a person who lives in a house has a pet is 0.75.
The probability that a person with no pet lives in a flat is \(\dfrac{2}{3}\).

One of the people who live in a flat is chosen at random.
Work out the probability that this person has a pet.

Model answer

Flat = 150 − 92 = 58
House with pet = 92 × 0.75 = 69, house with no pet = 92 − 69 = 23
People with no pet: \(\dfrac{2}{3}\) live in a flat, so the 23 in a house are the other \(\dfrac{1}{3}\).
People with no pet = 3 × 23 = 69, so flat with no pet = 69 − 23 = 46
Flat with pet = 58 − 46 = 12
P(pet given flat) = \(\dfrac{12}{58} = \dfrac{6}{29}\)

Explanation

Build a two-way table (house/flat against pet/no pet) and fill it in one step at a time. Each conditional probability applies to its own group: 0.75 to the people in houses, \(\dfrac{2}{3}\) to the people with no pet. The final answer divides by the 58 people in flats. A common mistake is to apply \(\dfrac{2}{3}\) to all 150 people.

Mark scheme

1 mark
P1 for the number living in a flat, 150 − 92 (= 58)
correct answer with no working scores full marks
1 mark
P1 for house with a pet 92 × 0.75 (= 69) and house with no pet 92 − 69 (= 23)
1 mark
P1 for a correct method to find the number in a flat with no pet, e.g. 23 is \(\dfrac{1}{3}\) of the people with no pet, so 46 in a flat with no pet
or \(\dfrac{f}{f + 23} = \dfrac{2}{3}\) leading to f = 46
1 mark
P1 for flat with a pet = 58 − 46 (= 12)
1 mark
A1 for \(\dfrac{12}{58}\) oe
accept \(\dfrac{6}{29}\) or awrt 0.207

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