4MA0 Higher June 2013 Paper 4R Q18
18 Here are ten counters.
Each counter has a number on it.
Fern puts the ten counters in a bag.
She takes at random a counter from the bag.
Fern puts the counter back into the bag.
Then Rajan takes at random one of the ten counters from the bag.
He does not put the counter back into the bag.
He then takes at random a second counter from the bag.
| Scheme | Marks |
|---|---|
| \(\dfrac{6}{10} + \dfrac{3}{10}\) | M1 |
| \(\dfrac{9}{10}\) oe | A1 |
| (2) |
| Scheme | Marks |
|---|---|
| \(\dfrac{6}{10} \times \dfrac{5}{9}\) oe and no other terms | M1 |
| \(\dfrac{30}{90}\) oe inc \(\dfrac{1}{3}\) | A1 |
| (2) |
Notes
A1: Sample space method – award 2 marks for a correct answer; otherwise no marks
| Scheme | Marks |
|---|---|
| \(\dfrac{6}{10} \times \dfrac{3}{9}\) oe or for \(\dfrac{3}{10} \times \dfrac{1}{9}\) oe or for \(\dfrac{7}{10} \times \dfrac{3}{9}\) | M1 |
| \(\dfrac{6}{10} \times \dfrac{3}{9} + \dfrac{3}{10} \times \dfrac{6}{9} + \dfrac{3}{10} \times \dfrac{1}{9} + \dfrac{1}{10} \times \dfrac{3}{9}\) oe or for \(\dfrac{7}{10} \times \dfrac{3}{9} + \dfrac{3}{10} \times \dfrac{7}{9}\) oe | M1 |
| \(\dfrac{42}{90}\) oe inc \(\dfrac{7}{15}\) | A1 |
| (3) |
Notes
A1: Sample space method – award 3 marks for a correct answer; otherwise no marks
SC
M1 for \(\dfrac{6}{10} \times \dfrac{3}{10}\) oe or \(\dfrac{3}{10} \times \dfrac{1}{10}\) oe
M1 for \(\dfrac{6}{10} \times \dfrac{3}{10} + \dfrac{3}{10} \times \dfrac{6}{10} + \dfrac{3}{10} \times \dfrac{1}{10} + \dfrac{1}{10} \times \dfrac{3}{10}\) oe