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Theoretical probability · 5 marks

A game of chance consists of spinning an arrow (see the figure.) which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8, and these are equally likely outcomes. What is the probability that it will point at

  1. (i) 8?
  2. (ii) An odd number?
  3. (iii) A number greater than 2?
  4. (iv) A number less than 9?
  5. (v) A multiple of 3?
12345678
Answer: (i) 18 (ii) 12 (iii) 34 (iv) 1 (v) 14

Step-by-step solution

Given: S = {1, 2, 3, 4, 5, 6, 7, 8}, equally likely; n(S) = 8

Idea: With 8 equally likely outcomes, P(event) = number of favourable numbers8. List the favourable numbers for each part.

(i) 8?

  1. Favourable: {8}, 1 outcome.½ mark
  2. P(8) = 18.½ mark
18

(ii) An odd number?

  1. Odd numbers: {1, 3, 5, 7}, 4 outcomes.½ mark
  2. P(odd) = 48 = 12.½ mark
12

(iii) A number greater than 2?

  1. Greater than 2: {3, 4, 5, 6, 7, 8}, 6 outcomes.½ mark
  2. P = 68 = 34.½ mark
34

(iv) A number less than 9?

  1. Every number on the spinner (1 to 8) is less than 9: 8 outcomes.½ mark
  2. P = 88 = 1. This is a certain event.½ mark
1

(v) A multiple of 3?

  1. Multiples of 3 up to 8: {3, 6}, 2 outcomes.½ mark
  2. P = 28 = 14.½ mark
14
(i) 1/8 (ii) 1/2 (iii) 3/4 (iv) 1 (v) 1/4

Check: (iii) P(not greater than 2) = P({1, 2}) = 28 = 14, and 34 + 14 = 1 ✓.

Answer to write in the exam

(i)

n(S) = 8; E = {8}; n(E) = 1

∴ P(8) = 18

(ii)

E = {1, 3, 5, 7}; n(E) = 4

∴ P(odd number) = 48 = 12

(iii)

E = {3, 4, 5, 6, 7, 8}; n(E) = 6

∴ P(number greater than 2) = 68 = 34

(iv)

E = {1, 2, 3, 4, 5, 6, 7, 8}; n(E) = 8

∴ P(number less than 9) = 88 = 1

(v)

E = {3, 6}; n(E) = 2

∴ P(multiple of 3) = 28 = 14

Common mistakes that cost marks

  • In (iii), including 2 (“greater than 2” starts at 3) and getting 78.
  • In (iv), answering 89. The numbers on the spinner are only 1 to 8; 9 is not an outcome, and every outcome is less than 9.
  • In (v), including 9 or forgetting 6. Only 3 and 6 are multiples of 3 between 1 and 8.

How this can come in the exam

MCQ (1 mark)

A spinner with 10 equal sectors numbered 1 to 10 is spun. The probability that it stops on a prime number is

  1. 310
  2. 25
  3. 12
  4. 35
Show answer

(B) 25
Primes: 2, 3, 5, 7: 410 = 25.

Short answer (2 marks)

An arrow on a spinner with 12 equal sectors numbered 1 to 12 is spun. Find the probability that it points at (a) a factor of 12, (b) a number greater than 12.

Show answer(a) Factors of 12: {1, 2, 3, 4, 6, 12}: 612 = 12. (b) No number is greater than 12: probability 0 (impossible event).

Try one yourself

The spinner with numbers 1 to 8 (equally likely) is spun. Find the probability that it points at a number that is a multiple of 4.

Show answer

{4, 8}: 28 = 14.

More questions like this

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