Unit 7

5 marks · Normal

Let .

The universal set .

Let .

A number is chosen at random from .

(a)

Show that can be written as a fraction in its simplest form, and state the recurring digit.

[2 marks]

(b)

List the elements of set and set .

[1 mark]

(c)

Calculate the probability that a randomly chosen number from is in . Give your answer as a fraction in its lowest terms.

[2 marks]

Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.

Show answer

(a) and the recurring digit is 7. · must show Equate x to decimal, Multiply by 10, Subtract equations, Simplify fraction, Identify recurring digit

(b) A = {7, 17 (not in \xi), so A = {7}; B = {3, 6, 9, 12, 15} · must show List A correctly: units digit is 7, within \xi; List B correctly: multiples of 3 in \xi

Common mistakes: Incorrect conversion of recurring decimal; Omitting 0 in set A; Not listing all multiples of 3; Incorrect denominator for universal set calculation

Show mark scheme and worked solution

Marking points

M1 (a) · 1 mark · Forms and solves correct equations for x = 0.2\dot{7}, multiplies by 10, subtracts, and expresses as a fraction.

A1 (a) · 1 mark · Simplifies fraction to lowest terms and correctly states the recurring digit.

B1 (b) · 1 mark · Lists all members of set A (digits in {1,...,15} with same units as recurring digit) and set B (multiples of 3).

M2 (c) · 1 mark · Finds intersection A∩B and calculates probability as n(A∩B)/n(\xi). (follow-through)

A2 (c) · 1 mark · Gives correct probability as a fraction in lowest terms (0 or another valid answer based on earlier work). (follow-through)

Worked solution

a) Let .

Subtract from : .

The recurring digit is 7.

b) The recurring digit from part (a) is 7, so A = {7} (with ). B = {3, 6, 9, 12, 15}.

c) A \cap B = { } (no number from 1–15 is in both A and B).