Practise Conditional probability. 2 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Build a paper on this topicWritten for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy3 marks
The table shows information about 105 students invited on a school trip. One student is chosen at random.
[object Object]
(a) Find the probability that the student went on the trip.[1]
(b) Given that the student was in Year 12, find the probability that the student went on the trip.[2]
Show the answer and mark scheme
(a) Answer: \(\frac{22}{35} \approx 0.629\)
- B1 for \(\frac{22}{35}\) (or awrt 0.629)
(b) Answer: \(\frac{10}{17} \approx 0.588\)
- M1 for \(\dfrac{40}{68}\)
- A1 for \(\frac{10}{17}\) (or awrt 0.588)
Question 2Medium5 marks
The table shows information about 75 orders placed with a retailer. One order is chosen at random. \(C\) is the event that the order was placed with the city centre branch and \(T\) is the event that the order was delivered on time.
[object Object]
(a) Given that the order was delivered late, find the probability that the order was placed with the retail park branch.[2]
(b) Determine whether the events \(C\) and \(T\) are independent. Show your working.[3]
Show the answer and mark scheme
(a) Answer: \(\frac{17}{27} \approx 0.630\)
- M1 for \(\dfrac{17}{27}\)
- A1 for \(\frac{17}{27}\) (or awrt 0.630)
(b) Answer: \(P(T \mid C) = \dfrac{25}{35} = \frac{5}{7}\) and \(P(T) = \dfrac{48}{75} = \frac{16}{25}\), which are not equal, so \(C\) and \(T\) are not independent.
- M1 for a correct test, e.g. comparing \(P(T \mid C)\) with \(P(T)\), or \(P(C \cap T) = \frac{1}{3}\) with \(P(C) \times P(T) = \frac{7}{15} \times \frac{16}{25}\)
- A1 for correct values (\(\frac{5}{7}\) and \(\frac{16}{25}\), or equivalent)
- A1 for the correct conclusion (not independent) with a reason
Question 3Hard10 marks
A screening test for a disease is used on a population in which 2% of people have the disease. If a person has the disease, the probability that the test is positive is 0.95. If a person does not have the disease, the probability that the test is positive is 0.10.
(a) Find the probability that a randomly chosen person tests positive.[2]
(b) Given that a person tests positive, find the probability that they have the disease.[2]
(c) A newspaper states: “The test is 95% accurate, so if you test positive you almost certainly have the disease.”
Use your answer to part (b) to comment on this statement.[2]
(d) People who test positive are tested a second time. Assuming that, for each person, the results of the two tests are independent, find the probability that a person who tests positive twice has the disease.[3]
(e) Explain why the independence assumption in part (d) may not be valid.[1]
Show the answer and mark scheme
(a) Answer: 0.117
- M1 for \(0.02 \times 0.95 + 0.98 \times 0.10\)
- A1 for 0.117
Worked solution: \(P(+) = 0.02 \times 0.95 + 0.98 \times 0.10 = 0.019 + 0.098 = 0.117\).
(b) Answer: 0.162 (3 s.f.)
- M1 for \(\frac{0.019}{0.117}\)
- A1 for awrt 0.162
Worked solution: \(P(D \mid +) = \dfrac{0.019}{0.117} = 0.162\).
(c) Answer: Only about 16% of people who test positive have the disease. Because the disease is rare, the false positives from the large healthy group outnumber the true positives.
- B1 for the statement is wrong: only about 16% of positives have the disease
- B1 for the reason: the disease is rare (2%), so most positive results come from the 98% who do not have it (false positives)
Worked solution: The statement is misleading: only about 16% of people who test positive actually have the disease. Although the test detects 95% of cases, the disease is rare, so the 10% false-positive rate among the 98% without the disease produces many more positives (0.098) than the true positives (0.019).
(d) Answer: 0.648 (3 s.f.)
- M1 for the numerator \(0.02 \times 0.95^2\)
- M1 for the denominator \(0.02 \times 0.95^2 + 0.98 \times 0.10^2\)
- A1 for awrt 0.648
Worked solution: \(P(D \mid ++) = \dfrac{0.02 \times 0.95^2}{0.02 \times 0.95^2 + 0.98 \times 0.1^2} = \dfrac{0.01805}{0.02785} = 0.648\).
(e) Answer: Whatever caused a false positive in one person (e.g. a similar condition) is likely to cause another false positive, so the two results for the same person are likely to be related.
- B1 for a reason in context, e.g. a person who gives a false positive once is more likely to do so again (the cause is a feature of that person), so the results are not independent
Worked solution: A false positive is often caused by something about the person (such as a related condition), which would also affect the second test. So a second positive is more likely after a first false positive than the model assumes, and the tests are not independent.