Higher November 2021 Paper 2 Q16
16 A box contains 15 counters.
There are 4 red counters, 5 green counters and the rest are yellow counters.
Niklas takes at random a counter from the box and writes down the colour of his counter.
He then puts the counter back into the box.
Sasha then takes at random a counter from the box and writes down the colour of her counter.
Work out the probability that the counters taken by Niklas and Sasha both have the same colour.
(3)
| Scheme | Marks |
|---|---|
\(\dfrac{4}{15} \times \dfrac{4}{15}\) or \(\dfrac{5}{15} \times \dfrac{5}{15}\) or \(\dfrac{6}{15} \times \dfrac{6}{15}\) oe (where 6 = 15 – 4 – 5) | M1 |
\(\dfrac{4}{15} \times \dfrac{4}{15} + \dfrac{5}{15} \times \dfrac{5}{15} + \dfrac{6}{15} \times \dfrac{6}{15}\) oe eg \(\dfrac{16}{225} + \dfrac{1}{9} + \dfrac{4}{25}\) (where 6 = 15 – 4 – 5) | M1 |
Correct answer scores full marks (unless from obvious incorrect working) Answer: \(\dfrac{77}{225}\) | A1 |
| (3) | |
| (3 marks) |
Notes
M1: oe
for one correct product (allow decimals to 2 dp rounded or truncated)
\(\left(\dfrac{4}{15}\right)^2 = (0.26(6\ldots))^2 = 0.07(11\ldots)\)
\(\left(\dfrac{5}{15}\right)^2 = (0.33(3\ldots))^2 = 0.11(1\ldots)\)
\(\left(\dfrac{6}{15}\right)^2 = (0.4)^2 = 0.16\)
M1: oe
for the sum of all three correct products
A1: oe 0.34(222….) or 34.(222…)%
(if no marks awarded, SCB2 for \(\dfrac{31}{105}\) oe from non-replacement,
SCB1 for a fully correct method for non-replacement)