Multiple Events: WJEC GCSE Higher November 2021 Unit 1 Q18

Higher November 2021 Unit 1 Q18

WJECHigherCurrent spec5 marksMultiple Events

18.

Three of the seven cards shown above are selected at random, without being replaced.

(a) Find the probability that the first card selected is a 3, the second is a 4 and the third is a 5. [2]

(b) Find the probability that the sum of the numbers selected is greater than 9. [3]

Mark scheme (a)

Scheme	Marks	Comments
\(\tfrac{4}{7} \times \tfrac{1}{6} \times \tfrac{2}{5}\) or equivalent	M1	Penalise once only throughout for a repeated error in calculating the denominator (of 210)
8/210 (= 4/105)	A1	ISW

Mark scheme (b)

Scheme	Marks	Comments
1 – P (3, 3, 3)	M1
\(1 - \tfrac{4}{7} \times \tfrac{3}{6} \times \tfrac{2}{5}\)	M1
\(\tfrac{186}{210}\ \left(= \tfrac{93}{105} = \tfrac{62}{70} = \tfrac{31}{35}\right)\)	A1	ISW If no other marks, award SC1 for an answer of \(\tfrac{279}{343}\) (from working ‘with replacement’) OR SC1 for sight of \(\tfrac{4}{7} \times \tfrac{3}{6} \times \tfrac{2}{5}\)

Scheme	Marks	Comments
(P(total10)+P(total11)+P(total12)+P(total13)+P(total14)=) P(3,3,4) × 3 + P(3,3,5) × 3 + P(3,4,5) × 6 + P(3,5,5) × 3 + P(4,5,5) × 3	M1	M0 if orderings are not considered
\(= \tfrac{4}{7} \times \tfrac{3}{6} \times \tfrac{1}{5} \times 3 + \tfrac{4}{7} \times \tfrac{3}{6} \times \tfrac{2}{5} \times 3 + \tfrac{4}{7} \times \tfrac{1}{6} \times \tfrac{2}{5} \times 6 + \tfrac{4}{7} \times \tfrac{2}{6} \times \tfrac{1}{5} \times 3 + \tfrac{1}{7} \times \tfrac{2}{6} \times \tfrac{1}{5} \times 3\)	M1
\(\tfrac{186}{210}\ \left(= \tfrac{93}{105} = \tfrac{62}{70} = \tfrac{31}{35}\right)\)	A1	ISW If no marks awarded, award SC1 for the correct method for calculating any individual total, e.g. P(total 10) = \(\tfrac{4}{7} \times \tfrac{3}{6} \times \tfrac{1}{5} \times 3\) or equivalent For information only: P(10)= \(\tfrac{36}{210}\ \left(= \tfrac{6}{35}\right)\) P(11)= \(\tfrac{72}{210}\ \left(= \tfrac{12}{35}\right)\) P(12)= \(\tfrac{48}{210}\ \left(= \tfrac{8}{35}\right)\) P(13)= \(\tfrac{24}{210}\ \left(= \tfrac{4}{35}\right)\) P(14)= \(\tfrac{6}{210}\ \left(= \tfrac{1}{35}\right)\) OR award SC1 for a calculation leading to an answer of \(\tfrac{54}{210}\) (from adding probabilities without accounting for different ordering) OR award SC1 for an answer of \(\tfrac{279}{343}\) (from working ‘with replacement’).