4MA0 Higher June 2016 Paper 3 Q21
21 There are 9 counters in a bag.
There is a number on each counter.
Kal takes at random 3 counters from the bag.
He adds together the numbers on the 3 counters to get his Total.
Work out the probability that his Total is 6
(5)
| Scheme | Marks |
|---|---|
| eg. \(\dfrac{3}{9} \times \dfrac{2}{8} \times \dfrac{1}{7} \left(= \dfrac{6}{504} = \dfrac{1}{84}\right)\) | M1 |
| eg. \(\dfrac{2}{9} \times \dfrac{3}{8} \times \dfrac{4}{7} \left(= \dfrac{24}{504} = \dfrac{1}{21}\right)\) | M1 |
| \(6 \times \text{"}\dfrac{24}{504}\text{"} \left(= \dfrac{144}{504} = \dfrac{6}{21} = \dfrac{2}{7}\right)\) | M1 |
| \(6 \times \dfrac{2}{9} \times \dfrac{3}{8} \times \dfrac{4}{7} + \dfrac{3}{9} \times \dfrac{2}{8} \times \dfrac{1}{7} \left(= \dfrac{6}{21} + \dfrac{1}{84}\right)\) | M1 |
| \(\dfrac{150}{504}\) | A1 |
| (5) |
Notes
M1: (probabilities from selecting 2, 2, 2)
allow \(\dfrac{3}{9} \times \dfrac{2}{9} \times \dfrac{1}{9} \left(= \dfrac{6}{729}\right)\) or \(\dfrac{3}{9} \times \dfrac{3}{9} \times \dfrac{3}{9} \left(= \dfrac{27}{729}\right)\)
M1: (probabilities from selecting 1, 2, 3)
allow \(\dfrac{2}{9} \times \dfrac{3}{9} \times \dfrac{4}{9} \left(= \dfrac{24}{729}\right)\)
M1: (probabilities for all combinations of 1, 2, 3)
allow \(6 \times \text{"}\dfrac{24}{729}\text{"} \left(= \dfrac{144}{729}\right)\)
M1: complete correct method
A1: oe eg. \(\dfrac{25}{84}\), 0.298, 0.297619…
(NB. An answer of \(\dfrac{150}{729} \left(= \dfrac{50}{243}\right)\) or \(\dfrac{171}{729} \left(= \dfrac{19}{81}\right)\) scores M1M1M1M0A0)