4400 Higher November 2004 Paper 4 Q23
23 There are 10 beads in a box.
\(n\) of the beads are red.
Meg takes one bead at random from the box and does not replace it.
She takes a second bead at random from the box.
The probability that she takes 2 red beads is \(\frac{1}{3}\).
Show that \(n^2 - n - 30 = 0\)
(4)
| Scheme | Marks |
|---|---|
| \(\dfrac{n}{10} \times \dfrac{n - 1}{9} = \dfrac{1}{3}\) | B1 |
| M1 | |
| \(3n(n - 1) = 90\) or \(n(n - 1) = 30\) | M1 |
| \(3n^2 - 3n = 90\) or \(n^2 - n = 30\) | A1 |
| (4) | |
| (4 marks) |
Notes
B1: for \(\dfrac{n}{10}\) and \(\dfrac{n - 1}{9}\) seen
M1: for \(\dfrac{n}{10} \times \dfrac{n - 1}{9} = \dfrac{1}{3}\)