Higher June 2018 Paper 1R Q15
15 There are two bags of counters, bag X and bag Y.
There are 20 counters in bag X.
11 of the counters are blue and the rest are red.
There are 16 counters in bag Y.
9 of the counters are blue and the rest are red.
Arkady takes at random a counter from bag X and takes at random a counter from bag Y.
| Scheme | Marks |
|---|---|
| \(\dfrac{9}{20}\) on first red branch | B1 |
 Answer: Correct binary structure | B1 |
\(\dfrac{9}{16}, \dfrac{7}{16}, \dfrac{9}{16}, \dfrac{7}{16}\) Answer: Labels and correct probabilities on all second branches | B1 |
| (3) |
Notes
B1: Correction: the printed scheme lists these as \(\dfrac{9}{20}, \dfrac{7}{16}, \dfrac{9}{20}, \dfrac{7}{16}\); bag Y has 9 blue counters out of 16, so the blue branches are \(\dfrac{9}{16}\), as the scheme’s own tree diagram and part (c) show.
| Scheme | Marks |
|---|---|
| \(\dfrac{\text{``}{9}\text{''}}{20} \times \dfrac{\text{``}{7}\text{''}}{16}\) | M1 |
| \(\dfrac{63}{320}\) or 0.196(875) | A1 |
| (2) |
Notes
A1: oe ft diagram
Accept 0.20 or better
| Scheme | Marks |
|---|---|
| M1 | |
| \(\dfrac{\text{``}{9}\text{''}}{20} \times \dfrac{\text{``}{7}\text{''}}{16} + \dfrac{11}{20} \times \dfrac{9}{16}\) | M1 |
| \(\dfrac{162}{320}\) or 0.506(25) | A1 |
| (3) | |
| (8 marks) |
Notes
M1: for \(\dfrac{11}{20} \times \dfrac{9}{16}\)
M1: for \(\dfrac{\text{``}{9}\text{''}}{20} \times \dfrac{\text{``}{7}\text{''}}{16} + \dfrac{11}{20} \times \dfrac{9}{16}\)
A1: oe
Accept 0.51 or better