Exam questions · Maths · Probability
Venn Diagrams and Set Notation
- 6 exam questions
- 19 marks
- 9 quick checks
-
1 Write down [2 marks]
\(\xi = \{1, 2, 3, \ldots, 12\}\). \(A\) is the set of factors of 12 and \(B\) is the set of odd numbers. (a) List the numbers that are in both \(A\) and \(B\). [1 mark] (b) How many of the numbers in \(\xi\) are in neither \(A\) nor \(B\)? [1 mark]
Show answerHide answer
Model answer
(a) \(A = \{1, 2, 3, 4, 6, 12\}\) and \(B = \{1, 3, 5, 7, 9, 11\}\), so the numbers in both are 1 and 3. (b) The numbers in neither set are 8 and 10, so there are 2 of them.
Mark scheme
- (a) 1 and 3 — B1
- (b) 2 — B1
-
2 Calculate [4 marks]
There are 40 students in a class. 19 have a cat and 15 have a dog. The incomplete Venn diagram shows 4 students with both and 10 with neither. (a) Complete the Venn diagram. [2 marks] (b) A student is chosen at random. Calculate the probability that the student has a dog but not a cat. [2 marks]
Show answerHide answer
Model answer
(a) Cats only is \(19 - 4 = 15\) and dogs only is \(15 - 4 = 11\). Check: \(15 + 4 + 11 + 10 = 40\). (b) \(\dfrac{11}{40}\).
Mark scheme
- (a) 15 — B1
- (a) 11 — B1
- (b) \(\dfrac{11}{40}\) with the correct denominator 40 — M1
- (b) \(\dfrac{11}{40}\) — A1
-
3 Calculate [3 marks]
In a survey of 80 people, 45 read newspaper \(N\), 30 read magazine \(M\) and 15 read both. Calculate the number of people who read neither. [3 marks]
Show answerHide answer
Model answer
At least one: \(45 + 30 - 15 = 60\). Neither: \(80 - 60 = 20\).
Mark scheme
- \(45 + 30 - 15\) — M1
- 60 — A1
- 20 — A1
-
4 Calculate [3 marks]
\(A\) and \(B\) are mutually exclusive events. \(P(A) = 0.5\) and \(P(B) = 0.35\). (a) Calculate \(P(A \text{ or } B)\). [1 mark] (b) Calculate the probability that neither event happens. [1 mark] (c) Write down \(P(A \text{ and } B)\). [1 mark]
Show answerHide answer
Model answer
(a) \(0.5 + 0.35 = 0.85\). (b) \(1 - 0.85 = 0.15\). (c) The events cannot both happen, so \(P(A \text{ and } B) = 0\).
Mark scheme
- (a) \(0.85\) — B1
- (b) \(0.15\) — B1
- (c) 0 — B1
-
5 Calculate [4 marks]
In a Venn diagram of 68 people, \(x + 5\) are in set \(A\) only, \(2x\) are in both sets, \(x\) are in set \(B\) only and \(3x\) are in neither set. (a) Calculate the value of \(x\). [2 marks] (b) A person is chosen at random. Calculate the probability that the person is in both sets. [2 marks]
Show answerHide answer
Model answer
(a) \(x + 5 + 2x + x + 3x = 68\), so \(7x + 5 = 68\) and \(x = 9\). (b) Both sets: \(2x = 18\), so \(\dfrac{18}{68} = \dfrac{9}{34}\).
Mark scheme
- (a) \(7x + 5 = 68\) — M1
- (a) \(x = 9\) — A1
- (b) \(\dfrac{18}{68}\) — M1
- (b) \(\dfrac{9}{34}\) — A1
-
6 Calculate [3 marks]
\(\xi = \{1, 2, 3, \ldots, 20\}\). \(A\) is the set of multiples of 4 and \(B\) is the set of square numbers. (a) List the numbers that are in both \(A\) and \(B\). [1 mark] (b) Calculate how many numbers are in \(A\) or \(B\) or both. [2 marks]
Show answerHide answer
Model answer
(a) \(A = \{4, 8, 12, 16, 20\}\) and \(B = \{1, 4, 9, 16\}\), so the numbers in both are 4 and 16. (b) \(5 + 4 - 2 = 7\) numbers.
Mark scheme
- (a) 4 and 16 — B1
- (b) \(5 + 4 - 2\) or a list of the union — M1
- (b) 7 — A1
Quick check
-
1
What does \(A \cap B\) mean?
Show answerHide answer
D: The items in both \(A\) and \(B\)
\(\cap\) is the intersection, the overlap.
-
2
What does \(A \cup B\) mean?
Show answerHide answer
C: The items in \(A\) or \(B\) or both
\(\cup\) is the union, everything in either circle.
-
3
What does \(A'\) mean?
Show answerHide answer
B: The items not in \(A\)
\(A'\) is the complement of \(A\).
-
4
\(A = \{2, 4, 6, 8, 10\}\) and \(B = \{3, 6, 9\}\). What is \(A \cap B\)?
Show answerHide answer
A: \(\{6\}\)
Only 6 is in both sets.
-
5
18 students play football and 6 of them also play tennis. How many play football only?
Show answerHide answer
D: \(12\)
\(18 - 6 = 12\).
-
6
30 students: 18 play football, 14 play tennis and 6 play both. How many play neither?
Show answerHide answer
C: \(4\)
\(12 + 6 + 8 = 26\) play at least one, so \(30 - 26 = 4\).
-
7
In the same survey, what is the probability that a student chosen at random plays football only?
Show answerHide answer
B: \(\dfrac{2}{5}\)
\(\dfrac{12}{30} = \dfrac{2}{5}\). The fraction \(\dfrac{18}{30}\) would include those who play both.
-
8
In a class of 40, 22 play football, 19 play tennis and 5 play neither. How many play both?
Show answerHide answer
A: \(6\)
\(40 - 5 = 35\) play at least one, and \(22 + 19 - 35 = 6\).
-
9
\(n(A) = 15\), \(n(B) = 12\) and \(n(A \cap B) = 5\). What is \(n(A \cup B)\)?
Show answerHide answer
D: \(22\)
\(15 + 12 - 5 = 22\).