(a) $ξ = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\}$
$E = \{x: x$ is an even number$\}$
$M = \{x: x$ is a multiple of 3$\}$
(i) Complete the Venn diagram.
(ii) Write down $n(E \cup M)$.
(iii) A number is chosen at random from the universal set $\%$. Write down the probability that the number is in the set $E \cap M$.
(b) Meg says that an even number cannot be a prime number.
Is she correct?
Give a reason for your answer.
▶️ Answer/Explanation
(a)(i)
E circle contains: 2, 4, 6, 8, 10, 12
M circle contains: 3, 6, 9, 12
Intersection (E ∩ M): 6, 12
Outside both circles: 1, 5, 7, 11
(a)(ii) 8
Count all elements in E or M: 2,3,4,6,8,9,10,12
(a)(iii) $\frac{1}{6}$
There are 2 numbers in E ∩ M (6,12) out of 12 total numbers.
(b) No because 2 is even and a prime number.
Meg’s statement is incorrect as 2 is the only even prime number.
(a) E = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14}
F = {x: x is a factor of 14}
P = {x: x is a prime number less than 14}
(i) Write down the elements in set F.
(ii) Write down the elements in set P.

(iii)(a) Complete the Venn diagram.
(b) Write down \(n(F\cap P ) .\)
(c) A number is chosen at random from the universal set E.
Write down the probability that the number is in the set \(F\cup P, \)
(b) Write 195 as a product of its prime factors.
▶️ Answer/Explanation
(i) F = {1, 2, 7, 14}
Solution: Factors of 14 are numbers that divide it exactly (1×14, 2×7).
(ii) P = {2, 3, 5, 7, 11, 13}
Solution: Prime numbers less than 14 (only divisible by 1 and themselves).
(iii)(a)
Solution: F circle = {1,14}; intersection = {2,7}; P circle = {3,5,11,13}; outside = {4,6,8,9,10,12}.
(b) 2
Solution: Only {2,7} appear in both F and P sets.
(c) 4/7
Solution: F∪P has 8 elements, total elements in E are 14 → 8/14 simplifies to 4/7.
(b) 3 × 5 × 13
Solution: 195 ÷ 5 = 39 → 39 ÷ 3 = 13 (all primes).