Higher June 2018 Paper 1 Q24
24. Steve and Bob are playing a card game with these 6 cards.
SSBBBB
In each round of the game:
- the 6 cards are shuffled and turned face down on a table
- Steve picks one card at random and keeps it
- Bob picks a card at random from the 5 remaining.
At the end of each round, Steve and Bob compare cards.
When
- both cards are S then Steve wins the game,
- both cards are B then Bob wins the game,
- the letters are different then neither of them wins the game and they play another round.
(a) Show that the probability that neither of them wins the first round is \(\dfrac{8}{15}\). [3]
(b) The game stops when someone wins.
Calculate the probability that Steve wins when exactly 2 rounds have been played. [3]
Calculate the probability that Steve wins when exactly 2 rounds have been played. [3]
| Scheme | Marks | Comments |
|---|---|---|
| (Probability 1st round draw = ) \(\left(\dfrac46\times\dfrac25\right)+\left(\dfrac26\times\dfrac45\right)\) or \(1-\dfrac26\times\dfrac15-\dfrac46\times\dfrac35\) or equivalent, seen or implied | M2 | Allow for a complete and correct sample space M1 for \(\dfrac46\times\dfrac25\left(=\dfrac{8}{30}\right)\) or equivalent, seen or implied or M1 for sight of \(1-\dfrac26\times\dfrac15\) or for sight of \(1-\dfrac46\times\dfrac35\) or equivalent |
| \(\dfrac{16}{30}\left(=\dfrac{8}{15}\right)\) or \(\dfrac{4}{15}+\dfrac{4}{15}\left(=\dfrac{8}{15}\right)\) or \(1-\dfrac{2}{30}-\dfrac{12}{30}\) or equivalent | A1 | Evidence must be seen as \(\dfrac{8}{15}\) is given e.g. \(\dfrac26\times\dfrac15+\dfrac46\times\dfrac35=\dfrac{14}{30}=\dfrac{7}{15}\) so answer is \(\dfrac{8}{15}\) implies M2 but is A0. |
| Scheme | Marks | Comments |
|---|---|---|
| Combines the probabilities correctly. \(\dfrac{8}{15}\times\dfrac26\times\dfrac15\) | M2 | M1 for sight of (Probability Steve wins = ) \(\dfrac26\times\dfrac15\) oe or \(\dfrac{2}{30}\) oe |
| \(\dfrac{16}{450}\) or equivalent ISW | A1 | Equivalent e.g. \(\dfrac{8}{225}\) |
| (6) |