Higher January 2021 Paper 2R Q21
21 A bag contains \(n\) beads.
6 of the beads are red and the rest are blue.
Ravi is going to take at random 2 beads from the bag.
The probability that the 2 beads will be of the same colour is \(\dfrac{9}{17}\)
Using algebra, and showing each stage of your working, calculate the value of \(n\).
(6)
| Scheme | Marks |
|---|---|
\(\dfrac{6}{n} \times \dfrac{5}{n - 1}\) or \(\dfrac{n - 6}{n} \times \dfrac{n - 7}{n - 1}\) oe or \(\dfrac{6}{n} \times \dfrac{n - 6}{n - 1}\) | M1 |
\(\dfrac{6}{n} \times \dfrac{5}{n - 1}\) and \(\dfrac{n - 6}{n} \times \dfrac{n - 7}{n - 1}\) oe or \(2 \times \dfrac{6}{n} \times \dfrac{n - 6}{n - 1}\) oe | M1 |
\(\dfrac{6}{n} \times \dfrac{5}{n - 1} + \dfrac{n - 6}{n} \times \dfrac{n - 7}{n - 1} = \dfrac{9}{17}\) oe or \(2 \times \dfrac{6}{n} \times \dfrac{n - 6}{n - 1} = 1 - \dfrac{9}{17}\) oe | M1 |
| \(2n^2 - 53n + 306\;(= 0)\) oe | A1 |
\((2n - 17)(n - 18)\;(= 0)\) or \(\dfrac{-{-53} \pm \sqrt{(-53)^2 - 4 \times 2 \times 306}}{2 \times 2}\) or \(\left(n - \dfrac{53}{4}\right)^2 - \left(\dfrac{53}{4}\right)^2 + 153 = 0\) oe | M1 |
| Working required Answer: 18 | A1 |
| (6) | |
| (6 marks) |
Notes
M1: for red, red or blue, blue
This may be seen as part of an equation
allow eg \(n - 6 - 1\) in place of \(n - 7\)
or for red, blue
M1: for both products, with no other products
This may be seen as part of an equation
or for red, blue + blue, red
M1: Correct equation
or correct equation using the complementary event.
A1: Correct simplification of equation to a 3 term quadratic.
eg \(8n^2 - 212n + 1224\;(= 0)\)
A1: cao dep M3
do not award if non-integer solution also given.