Multiple Events: WJEC GCSE Higher June 2023 Unit 2 Q13

Higher June 2023 Unit 2 Q13

WJECHigherCurrent spec5 marksMultiple Events

13. Catherine has three spinners, as shown below.

She spins each spinner once.

(a) Calculate the probability that all three spinners land on prime numbers. [2]

(b) The numbers that the three spinners land on are added together.
Calculate the probability that the total is greater than 4. [3]

Mark scheme (a)

Scheme	Marks	Comments
\(\frac{2}{4} \times \frac{3}{5} \times \frac{3}{6}\)	M1
\(= \frac{18}{120}\) ISW (\(= \frac{3}{20}\) or 0·15) oe	A1	If no marks, award SC1 for: \(\frac{a}{4} \times \frac{b}{5} \times \frac{c}{6}\) correctly evaluated with at least two of a, b and c correct OR an answer of \(\frac{48}{120}\) (\(= \frac{2}{5}\) or 0·4 or equivalent) from assuming that 1 is prime, \(\frac{3}{4} \times \frac{4}{5} \times \frac{4}{6}\) OR an answer of \(\frac{4}{120}\) (\(= \frac{1}{30}\) or 0·03(3…) or equivalent) from excluding 2 as prime, \(\frac{1}{4} \times \frac{2}{5} \times \frac{2}{6}\)

Mark scheme (b)

Scheme	Marks	Comments
[1 – P(1, 1, 1) – 3×P(1, 1, 2) = ] \(1 - 4 \times \frac{1}{4} \times \frac{1}{5} \times \frac{1}{6}\) OR \(\frac{120-4}{120}\)	M2	Award M1 for sight of: \(4 \times \frac{1}{4} \times \frac{1}{5} \times \frac{1}{6}\) or equivalent (from P(1, 1, 1) + 3×P(1, 1, 2)) OR \(1 - 2 \times \frac{1}{4} \times \frac{1}{5} \times \frac{1}{6}\) or equivalent (from 1 – P(1, 1, 1) – P(1, 1, 2)) OR \(1 - 3 \times \frac{1}{4} \times \frac{1}{5} \times \frac{1}{6}\) or equivalent (from 1 – 3×P(1,1,2) ) OR \(1 - \frac{1}{4} \times \frac{1}{5} \times \frac{1}{6}\) or equivalent (from 1 – P(1,1,1) ), note: this may include further incorrect work, e.g. \(\frac{117}{120}\left(= \frac{39}{40}\right)\) from \(1 - \frac{1}{4} \times \frac{1}{5} \times \frac{1}{6} - \frac{1}{4} \times \frac{1}{5} \times \frac{2}{6}\) or equivalent OR 120–4 (120 permutations – the 4 permutations not allowed)
\(= \frac{116}{120}\) ISW \(\left(= \frac{29}{30}\right)\) oe	A1	CAO Allow \(0{\cdot}9\dot{6}\) OR 0·966(6…) OR 0·967 oe. Allow 0·97 from correct working. If no marks award SC1 for an answer of \(\frac{114}{120}\left(= \frac{19}{20}\right)\) from 1–3×P(1,1,1)–3×P(1,1,2) *Note: award M2 for a fully correct method of listing all the ways leading to totals greater than 4, but M0 for a partial attempt.