June 2025 Paper 2 Q12
12 Sam has 9 cards, each with a different non-zero digit printed on it.
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
Sam chooses 5 cards at random and places them in a random order in a straight line, to form a 5-digit number.
| Scheme | Marks | AO |
|---|---|---|
| (i) \(\dfrac{5}{9}\) | B1 | 3.1a |
| [1] | ||
| (ii) (6or7or8or9) nnnn: \(\frac{4}{9}\) | M1 | 3.1a |
| 59 nnn: \(\frac{1}{9} \times \frac{1}{8} \left(= \frac{1}{72}\right)\) | M1 | 1.1 |
| 58 (6or7or9) nn: \(\frac{1}{9} \times \frac{1}{8} \times \frac{1}{7} \times 3 \left(= \frac{3}{504} = \frac{1}{168}\right)\) | M1 | 2.1 |
| \(\mathrm{P}(\gt 58600) = \frac{4}{9} + \frac{1}{72} + \frac{1}{168} = \frac{13}{28}\) AG | A1 | 2.1 |
| [4] |
Notes
(a)(i)
B1: awrt 0.556 (3sf)
(a)(ii)
M1: Allow \(1 - \frac{4}{9}\) or \(\frac{5}{9} - \cdots\) (where \(\cdots\) is any other correct work)
M1: May be included in an expression of the form \(1 - \cdots\).
M1: May be included in an expression of the form \(1 - \cdots\).
A1: Some indication of summation must be seen
Or \(1 - \left(\frac{4}{9} + 6 \times \frac{1}{9} \times \frac{1}{8} + 4 \times \frac{1}{9} \times \frac{1}{8} \times \frac{1}{7}\right) = \frac{5}{9} - \frac{6}{72}\left(\frac{1}{12}\right) - \frac{4}{504}\left(\frac{1}{126}\right)\)
Alternative method 1 (counting possibilities for first three digits)
| Scheme | Marks | AO |
|---|---|---|
| (6or7or8or9) (one from eight) (one from 7): \(4 \times 8 \times 7\) | M1 | |
| 59 (1or2or3or4or6or7or8): 7 | M1 | |
| 58 (6or7or9): 3 | M1 | |
| Total \(= 234\) \(\div\) total all possible numbers: \(9 \times 8 \times 7 = 504\) \(\mathrm{P}(\gt 58600) = \frac{234}{504} = \frac{13}{28}\) AG | A1 |
A1: AG
Alternative method (using permutations)
| Scheme | Marks | AO |
|---|---|---|
| (6or7or8or9) nnnn: \(4 \times {}^8\mathrm{P}_4 = 6720\) | M1 | |
| 59 nnn: \({}^7\mathrm{P}_3 = 210\) | M1 | |
| 58 (6or7or9) nn: \(3 \times {}^6\mathrm{P}_2 = 90\) | M1 | |
| Total \(= 7020\) \(\mathrm{P}(\gt 58600) = \frac{7020}{{}^9\mathrm{P}_5} = \frac{7020}{15120} = \frac{13}{28}\) AG | A1 |
A1: AG
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{13}{28} \div \dfrac{5}{9}\) | M1 | 3.1a |
| \(= \dfrac{117}{140}\) or 0.836 (3 sf) | A1FT | 1.1 |
| [2] |
Notes
M1: M1 for \(\frac{13}{28} \div\) their answer to (i) soi. Allow 0.464 for \(\frac{13}{28}\)
A1FT: FT (i) (but A0 for answers outside \([0,1]\))