Higher June 2023 Paper 2 Q23

23 Here are three sets of cards.

In a game, a player has two options.

Option 1
Pick two cards from Set A

Option 2
Pick one card from Set B
and
pick one card from Set C

The cards are picked at random.

The player wins if the total of their two cards is exactly 10

Which option gives a better chance of winning?

Show working to support your answer. [4 marks]

Mark scheme

Answer	Mark	Comments
\(\dfrac{3}{8}\ (\times)\ \dfrac{2}{7}\) or \(\dfrac{6}{56}\) or \(\dfrac{3}{28}\)	M1	oe fraction, decimal or percentage allow \(\dfrac{2}{7}\) to be [0.285, 0.286] or [28.5, 28.6]% allow \(\dfrac{6}{56}\) to be [0.107, 0.107143] or [10.7, 10.7143]% may be seen on a tree diagram allow 6 out of 56
\(\dfrac{1}{7}\ (\times)\ \dfrac{1}{4}\ (\times 2)\) or \(\dfrac{1}{28}\ (\times 2)\) or \(\dfrac{2}{28}\) or \(\dfrac{1}{14}\)	M1	oe fraction, decimal or percentage allow \(\dfrac{1}{7}\) to be [0.142, 0.143] or [14.2, 14.3]% allow \(\dfrac{1}{28}\) to be [0.035, 0.036] or [3.5, 3.6]% allow \(\dfrac{2}{28}\) to be [0.071, 0.07143] or [7.1, 7.143]% may be seen on a tree diagram allow 1 out of 28 or 2 out of 28
\(\dfrac{6}{56}\) and \(\dfrac{2}{28}\)	A1	oe fractions, decimals or percentages allow 6 out of 56 and 2 out of 28
Probabilities in comparable form and Option 1	A1ft	ft their \(\dfrac{6}{56}\) and their \(\dfrac{2}{28}\) with M2A0 correct comparisons include \(\dfrac{3}{28}\) and \(\dfrac{2}{28}\) \(\dfrac{6}{56}\) and \(\dfrac{4}{56}\) 0.107 and 0.071 10.7% and 7.1% 6 out of 56 and 4 out of 56

Up to M2 may be awarded for correct work with no answer or incorrect answer, even if this is seen amongst multiple attempts
3 ways to win in Option 1 and 2 ways to win in Option 2 so Option 1	M0M0A0A0
\(\dfrac{3}{8} \times \dfrac{2}{7} = \dfrac{6}{56} \qquad \dfrac{1}{7} \times \dfrac{1}{4} = \dfrac{1}{28}\) \(\dfrac{6}{56}\) and \(\dfrac{2}{56}\) and Option 1	M1M1 A0A1ft
Assuming replacement can score a maximum of M0M1A0A0
Choosing Option 1 cannot be implied by inequalities