4MA0 Higher June 2014 Paper 4R Q15
15 Maria has two bags.
In bag A, there are 5 white counters and 2 red counters.
In bag B, there are 3 white counters and 2 red counters.
Maria is going to take at random one counter from bag A and one counter from bag B.
(a) Complete the probability tree diagram.
(b) Work out the probability that both counters will be white. (2)
(c) Work out the probability that exactly one of the counters will be white. (3)
| Scheme | Marks |
|---|---|
| \(\dfrac{2}{7}\) then two pairs of branches with \(\dfrac{3}{5}\), \(\dfrac{2}{5}\) on the W and R branches respectively | M1 |
| fully correct tree diagram | A1 |
| (2) |
Notes
M1: \(\dfrac{2}{7}\) on lower LH branch OR two additional pairs branches labelled with white and red
A1: for fully correct tree diagram with all probabilities and labels
NB: Accept 0.28571… rounded or truncated to 3 or more sig figs for \(\dfrac{2}{7}\)
| Scheme | Marks |
|---|---|
| \(\dfrac{5}{7} \times \text{“}\dfrac{3}{5}\text{”}\) | M1 |
| \(\dfrac{15}{35}\) | A1 |
| (2) |
Notes
M1: ft from their tree diagram
A1: oe eg. \(\dfrac{3}{7}\) or 0.428571… rounded or truncated to 3 or more sig figs
Accept 0.43 if working shown
| Scheme | Marks |
|---|---|
| \(\dfrac{5}{7} \times \text{“}\dfrac{2}{5}\text{”}\) or \(\dfrac{2}{7} \times \text{“}\dfrac{3}{5}\text{”}\) | M1 |
| \(\dfrac{5}{7} \times \text{“}\dfrac{2}{5}\text{”} + \dfrac{2}{7} \times \text{“}\dfrac{3}{5}\text{”}\) | M1 |
| \(\dfrac{16}{35}\) | A1 |
| (3) | |
| (7 marks) |
Notes
M1: for a correct product only ft probabilities < 1
M1: for full method
A1: oe accept decimal answer 0.457142… rounded or truncated to 3 or more sig figs
Accept 0.46 if working shown
M2 for \(1 - \text{“}\dfrac{15}{35}\text{”} - \text{“}\dfrac{2}{7}\text{”} \times \text{“}\dfrac{2}{5}\text{”}\)