Practise Venn diagrams and set notation. 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 Venn diagram shows the sets ξ, \(A\) and \(B\).
ξ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18}
[object Object]
(a) List the members of the set \(A \cap B\).[1]
(b) One of the numbers in ξ is chosen at random.
Find \(P(A \cup B)\).[2]
Show the answer and mark scheme
(a) Answer: 2, 3
- B1 for 2, 3 (and no others)
(b) Answer: \(\frac{13}{18}\)
- M1 for a fraction with numerator 13 or with denominator 18
- A1 for \(\frac{13}{18}\) oe
Worked solution: The members of \(A \cup B\) are 1, 2, 3, 4, 5, 6, 7, 9, 11, 12, 13, 17, 18, so \(P(A \cup B) = \frac{13}{18}\).
Question 2Medium4 marks
There are 30 students in a class.
18 of the students play football, 15 play tennis and 5 play neither.
(a) Explain why some students must play both football and tennis.[1]
(b) A student is chosen at random from the class.
Work out the probability that this student plays tennis but not football.[3]
Show the answer and mark scheme
(a) Answer: \(18 + 15 + 5 = 38\), which is more than 30, so 8 students have been counted twice.
- C1 for \(18 + 15 + 5 = 38 \gt 30\) (or \(18 + 15 = 33 \gt 25\)), so some students are in both groups
Worked solution: Only 25 students play at least one sport, but \(18 + 15 = 33\). The extra 8 must be students counted in both groups.
(b) Answer: \(\frac{7}{30}\)
- M1 for both = \(38 - 30 = 8\)
- M1 for \(15 - 8 = 7\) playing tennis but not football (ft their 8)
- A1 for \(\frac{7}{30}\) oe
Worked solution: Both: \(33 - 25 = 8\). Tennis only: \(15 - 8 = 7\). Probability \(= \frac{7}{30}\).
Question 3Hard4 marks
\(A\) and \(B\) are two events.
\(P(A) = 0.4\), \(P(B) = 0.5\) and \(P(A \cup B) = 0.7\).
(a) Work out \(P(A \cap B)\).[2]
(b) Are the events \(A\) and \(B\) independent? Give a reason for your answer.[1]
(c) Find \(P(A' \cap B)\).[1]
Show the answer and mark scheme
(a) Answer: 0.2
- M1 for \(0.4 + 0.5 - 0.7\) or a correct Venn diagram method
- A1 for 0.2
Worked solution: \(P(A \cap B) = P(A) + P(B) - P(A \cup B) = 0.4 + 0.5 - 0.7 = 0.2\)
(b) Answer: Yes: \(P(A) \times P(B) = 0.4 \times 0.5 = 0.2 = P(A \cap B)\).
- C1 for 'yes' with \(0.4 \times 0.5 = 0.2\) compared with \(P(A \cap B)\) (ft their part (a))
Worked solution: For independent events \(P(A \cap B) = P(A) \times P(B)\). Here \(0.4 \times 0.5 = 0.2\), which equals \(P(A \cap B)\), so they are independent.
(c) Answer: 0.3
- B1 for 0.3 (ft \(0.5 -\) their part (a))
Worked solution: \(P(A' \cap B) = P(B) - P(A \cap B) = 0.5 - 0.2 = 0.3\)