4MA0 Higher January 2018 Paper 4R Q12
12 In a bowl, there are 20 grapes.
13 of these grapes are green and 7 of these grapes are red.
Jayesh takes at random a grape from the bowl and eats the grape.
He then takes at random another grape from the bowl and eats the grape.
Jayesh takes at random a third grape from the bowl and eats this grape.
| Scheme | Marks |
|---|---|
| B1 | |
| B1 | |
| Fully correct tree diagram | B1 |
| (3) |
Notes
B1: For \(\dfrac{7}{20}\) on lower LH branch
B1: Correct binary structure with 4 branches needed on RHS
B1: For fully correct tree diagram with all probabilities (\(\dfrac{7}{20}\), \(\dfrac{12}{19}\), \(\dfrac{7}{19}\), \(\dfrac{13}{19}\) and \(\dfrac{6}{19}\)) and labels.
| Scheme | Marks |
|---|---|
| \(\dfrac{13}{20} \times \dfrac{12}{19}\) | M1 |
| \(\dfrac{156}{380}\) | A1 |
| (2) |
Notes
M1: ft from their tree diagram in (a)
| Scheme | Marks |
|---|---|
| \(\dfrac{13}{20} \times \dfrac{12}{19} \times \dfrac{7}{18}\) or \(\dfrac{91}{570}\) or 0.15(9649...) oe | M1 |
\(\dfrac{13}{20} \times \dfrac{12}{19} \times \dfrac{7}{18} + \dfrac{13}{20} \times \dfrac{7}{19} \times \dfrac{12}{18} + \dfrac{7}{20} \times \dfrac{13}{19} \times \dfrac{12}{18}\) or \(3 \times \dfrac{13}{20} \times \dfrac{12}{19} \times \dfrac{7}{18}\) | M1 |
| \(\dfrac{91}{190}\) | A1 |
| (3) | |
| (8 marks) |
Notes
M1: Ft from (a)
M1: Ft from (a)
Dep. For full correct method
A1: oe
Accept 0.47(894...) rounded or truncated to at least 2dp.
ft method marks if probabilities <1
With Replacement
| Scheme | Marks |
|---|---|
| \(\dfrac{13}{20} \times \dfrac{13}{20} \times \dfrac{7}{20}\) or \(\dfrac{1183}{8000}\) or 0.14(7875) | M1 |
| \(3 \times \dfrac{13}{20} \times \dfrac{13}{20} \times \dfrac{7}{20}\) or \(\dfrac{3549}{8000}\) or 0.44(3625) | M1 |