4MA0 Higher June 2011 Paper 3 Q17
17 Here are seven counters.
Each counter has a number on it.
Ali puts the seven counters in a bag.
He takes, at random, a counter from the bag and does not replace the counter.
He then takes, at random, a second counter from the bag.
Calculate the probability that
(i) the number on the second counter is 2 more than the number on the first counter,
(ii) the number on the second counter is 1 more than the number on the first counter.
(5)
| Scheme | Marks |
|---|---|
| (i) \(\frac{1}{7} \times \frac{2}{6}\) and no other terms | M1 |
| \(\frac{2}{42}\) or \(\frac{1}{21}\) oe | A1 |
| (ii) \(\frac{1}{7} \times \frac{1}{6}\) or \(\frac{2}{7} \times \frac{3}{6}\) | M1 |
| \(\frac{1}{7} \times \frac{1}{6} + \frac{2}{7} \times \frac{3}{6}\) | M1 |
| \(\frac{7}{42}\) or \(\frac{1}{6}\) oe | A1 |
| (Total 5 marks) |
Notes
A1: Also accept 0.05, 0.04, 0.047, 0.048 etc
Sample space method – award 2 marks for a correct answer; otherwise no marks
M1: SC M1 for \(\frac{1}{7} \times \frac{1}{7}\) or \(\frac{2}{7} \times \frac{3}{7}\)
M1: M1 for \(\frac{1}{7} \times \frac{1}{7} + \frac{2}{7} \times \frac{3}{7}\)
A1: Also accept \(0.1\dot{6}\), 0.16, 0.17, 0.166, 0.167 etc but not 0.2
Sample space method – award 3 marks for a correct answer; otherwise no marks