Higher November 2021 Paper 1 Q15
15 Magnus and Garry play 2 games of chess against each other.
The probability that Magnus beats Garry in any game is \(\dfrac{2}{9}\)
The probability that any game between Magnus and Garry is drawn is \(\dfrac{4}{9}\)
The result of any game is independent of the result of any other game.
For each game of chess,
the winner gets 2 points and the loser gets 0 points,
when the game is drawn, each player gets 1 point.
Magnus and Garry now play a third game of chess.
| Scheme | Marks |
|---|---|
| \(\dfrac{3}{9}\) | B1 |
| \(\dfrac{2}{9},\ \dfrac{4}{9},\ \dfrac{3}{9}\) | B1ft |
| (2) |
Notes
B1: for lower 1st game branch probability
B1ft: for all values correct on 2nd game branches
| Scheme | Marks |
|---|---|
| \(\left(\dfrac{2}{9} \times \dfrac{3}{9}\right)\) or \(\left(\dfrac{3}{9} \times \dfrac{2}{9}\right)\) or \(\left(\dfrac{4}{9} \times \dfrac{4}{9}\right)\) oe or | M1 |
| \(\left(\dfrac{2}{9} \times \dfrac{3}{9}\right) + \left(\dfrac{3}{9} \times \dfrac{2}{9}\right) + \left(\dfrac{4}{9} \times \dfrac{4}{9}\right)\) oe | M1 |
Correct answer scores full marks (unless from obvious incorrect working) Answer: \(\dfrac{28}{81}\) | A1 |
| (3) |
Notes
M1: ft from their tree diagram for one correct product from WL or L W or DD
(allow probabilities to 2 dp truncated or rounded)
M1: ft for a fully correct method
A1: Allow 0.345 ... (2 dp truncated or rounded) or 34.5% (2 sf truncated or rounded)
| Scheme | Marks |
|---|---|
| \(\left(\dfrac{2}{9} \times \dfrac{4}{9} \times \dfrac{3}{9}\right)\) or \(\left(\dfrac{4}{9} \times \dfrac{4}{9} \times \dfrac{4}{9}\right)\) | M1ft |
| \(6 \times \left(\dfrac{2}{9} \times \dfrac{4}{9} \times \dfrac{3}{9}\right) + \left(\dfrac{4}{9} \times \dfrac{4}{9} \times \dfrac{4}{9}\right)\) | M1ft |
Correct answer scores full marks (unless from obvious incorrect working) Answer: \(\dfrac{208}{729}\) | A1 |
| (3) | |
| (8 marks) |
Notes
M1ft: from their tree diagram for any combination of WLD or DDD
(allow probabilities to 2 dp truncated or rounded)
M1ft: for a fully correct method
A1: Allow 0.285 ... (2 dp truncated or rounded) or 28.5% (2 sf truncated or rounded)
| Scheme | Marks |
|---|---|
(a) Answer: \(\dfrac{3}{9}\) | B1 |
| \(\dfrac{2}{9},\ \dfrac{4}{9},\ \dfrac{3}{9}\) | B1ft |
(b) 1 and \(\left(\dfrac{2}{9} \times \dfrac{2}{9}\right)\) or \(\left(\dfrac{4}{9} \times \dfrac{2}{9}\right)\) or \(\left(\dfrac{4}{9} \times \dfrac{3}{9}\right)\) or \(\left(\dfrac{3}{9} \times \dfrac{3}{9}\right)\) oe | M1ft |
| \(1 - \left[\left(\dfrac{2}{9} \times \dfrac{2}{9}\right) + 2\left(\dfrac{4}{9} \times \dfrac{2}{9}\right) + 2\left(\dfrac{4}{9} \times \dfrac{3}{9}\right) + \left(\dfrac{3}{9} \times \dfrac{3}{9}\right)\right]\) oe | M1ft |
Correct answer scores full marks (unless from obvious incorrect working) Answer: \(\dfrac{28}{81}\) | A1 |
(c) 1 and \(\left(\dfrac{2}{9} \times \dfrac{2}{9} \times \dfrac{2}{9}\right)\) or \(\left(\dfrac{2}{9} \times \dfrac{2}{9} \times \dfrac{4}{9}\right)\) or \(\left(\dfrac{2}{9} \times \dfrac{2}{9} \times \dfrac{3}{9}\right)\) or \(\left(\dfrac{2}{9} \times \dfrac{4}{9} \times \dfrac{4}{9}\right)\) or \(\left(\dfrac{2}{9} \times \dfrac{3}{9} \times \dfrac{3}{9}\right)\) or \(\left(\dfrac{4}{9} \times \dfrac{4}{9} \times \dfrac{3}{9}\right)\) or \(\left(\dfrac{4}{9} \times \dfrac{3}{9} \times \dfrac{3}{9}\right)\) or \(\left(\dfrac{3}{9} \times \dfrac{3}{9} \times \dfrac{3}{9}\right)\) oe | M1ft |
\(1 - \left[\left(\dfrac{2}{9} \times \dfrac{2}{9} \times \dfrac{2}{9}\right) + 3\left(\dfrac{2}{9} \times \dfrac{2}{9} \times \dfrac{4}{9}\right) + 3\left(\dfrac{2}{9} \times \dfrac{2}{9} \times \dfrac{3}{9}\right) + 3\left(\dfrac{2}{9} \times \dfrac{4}{9} \times \dfrac{4}{9}\right)\right.\) \(\left. + 3\left(\dfrac{2}{9} \times \dfrac{3}{9} \times \dfrac{3}{9}\right) + 3\left(\dfrac{4}{9} \times \dfrac{4}{9} \times \dfrac{3}{9}\right) + 3\left(\dfrac{4}{9} \times \dfrac{3}{9} \times \dfrac{3}{9}\right) + \left(\dfrac{3}{9} \times \dfrac{3}{9} \times \dfrac{3}{9}\right)\right]\) oe | M1ft |
Correct answer scores full marks (unless from obvious incorrect working) Answer: \(\dfrac{208}{729}\) | A1 |
Notes
B1: (a) for lower 1st game branch probability
B1ft: (a) for all values correct on 2nd game branches
M1ft: (b) from their tree diagram for 1 and one correct product from WW, DW, DL or LL
(allow probabilities to 2 dp truncated or rounded)
M1ft: (b) for a fully correct method
A1: (b) Allow 0.345 ... (2 dp truncated or rounded) or 34.5% (2 sf truncated or rounded)
M1ft: (c) from their tree diagram for 1 and one correct product from WWW or WWD or WWL or WDD or WLL or DDL or DLL or LLL
(allow probabilities to 2 dp truncated or rounded)
M1ft: (c) for a fully correct method
A1: (c) Allow 0.285 ... (2 dp truncated or rounded) or 28.5% (2 sf truncated or rounded)