Higher June 2018 Paper 2R Q16
16 There are 8 counters in a bag.
There is a number on each counter.
Fiona takes at random three of the counters.
She adds the numbers on the three counters to get her total.
Work out the probability that her total is an odd number.
(4)
| Scheme | Marks |
|---|---|
| M1 | |
eg P(o, o, o) = \(\dfrac{5}{8} \times \dfrac{4}{7} \times \dfrac{3}{6} \left(= \dfrac{60}{336} = \dfrac{5}{28} = 0.178(571\ldots)\right)\) or P(e, e, o) = \(\dfrac{3}{8} \times \dfrac{2}{7} \times \dfrac{5}{6} \left(= \dfrac{30}{336} = \dfrac{5}{56} = 0.0892(857\ldots)\right)\) | M1 |
| M1 | |
| \(\dfrac{25}{56}\) | A1 |
| (4) | |
| (4 marks) |
Notes
M1: for \(\dfrac{a}{8} \times \dfrac{b}{7} \times \dfrac{c}{6}\) where \(a \lt 8\), \(b \lt 7\), \(c \lt 6\)
M1: for a complete method to find P(o, o, o) or P(o, e, e) or P(e, o, e) or P(e, e, o)
M1: for a complete method to find P(o, o, o) and at least one of P(o, e, e), P(e, o, e), P(e, e, o)
A1: oe \(\dfrac{150}{336}\), 0.446(428571..)
SC B2 for \(\dfrac{260}{512} \left(= \dfrac{65}{128} = 0.507(8125)\right)\), B1 for \(\dfrac{170}{512} \left(= \dfrac{85}{256} = 0.332(03125)\right)\)