4400 Higher June 2009 Paper 3 Q17
17 Here are five counters.
Each counter has a number on it.
Layla puts the five counters in a bag.
She takes two counters at random from the bag without replacement.
Calculate the probability that
(i) both counters will have the number 3 on them,
(ii) the sum of the numbers on the two counters will be 6
(5)
| Scheme | Marks |
|---|---|
| (i) \(\frac{3}{5} \times \frac{2}{4}\) | M1 |
| \(\frac{6}{20}\) or \(\frac{3}{10}\) | A1 |
| (ii) \(\frac{1}{5} \times \frac{1}{4} \times 2 +\) “\(\frac{6}{20}\)” or \(\frac{2}{5} \times \frac{1}{4} +\) “\(\frac{6}{20}\)” | M1 |
| M1 | |
| \(\frac{8}{20}\) or \(\frac{2}{5}\) oe | A1 |
| (5) | |
| (5 marks) |
Notes
M1: Sample space method - award 2 marks for a correct answer, otherwise no marks
M1: for \(\frac{1}{5} \times \frac{1}{4}\) or \(\frac{2}{5} \times \frac{1}{4}\)
Award M0 M0 A0 for \(\frac{1}{5} + \frac{1}{5} = \frac{2}{5}\)
Sample space method - award 3 marks for a correct answer, otherwise no marks
M1: for complete sum
SC
M1 for \(\frac{1}{5} \times \frac{1}{5}\) or \(\frac{1}{25}\)
M1 for \(\frac{1}{5} \times \frac{1}{5} \times 2 +\) their (i)
Sample space method - award 2 marks for \(\frac{11}{25}\), otherwise no marks