4400 Higher November 2007 Paper 3 Q21
21 A coin is biased so that the probability that it shows heads on any one throw is \(p\).
The coin is thrown twice.
The probability that the coin shows heads exactly once is \(\dfrac{8}{25}\)
Show that \(25p^2 - 25p + 4 = 0\)
(3)
| Scheme | Marks |
|---|---|
| \(2p(1 - p) = \frac{8}{25}\) | M1 |
| \(p(1 - p) = \frac{4}{25}\) or \(p - p^2 = \frac{4}{25}\) | M1 |
| \(25p(1 - p) = 4\) or \(25(p - p^2) = 4\) | A1 |
| (3) | |
| (3 marks) |
Notes
M1: allow \(p(1 - p) = \frac{8}{25}\) for M1 only
M1: or \(50p(1 - p) = 8\) or \(50(p - p^2) = 8\)
A1: or \(25p - 25p^2 = 4\) oe, no fracs & 2 canc’ld
Alt 1
| \(2p(1 - p) = \frac{8}{25}\) oe | M1 |
| \(p = \frac{1}{5}\) or \(\frac{4}{5}\) | M1 |
| \((p - \frac{1}{5})(p - \frac{4}{5}) = 0\) or \((5p - 1)(5p - 4) = 0\) | A1 |
Alt 2
| solve equn | M1 |
| \(2 \times \frac{1}{5} \times \frac{4}{5}\) | M1 |
| \(= \frac{8}{25}\) | A1 |
\(p = \frac{1}{5}\) or \(\frac{4}{5}\) seen without \(2p(1 - p) = \frac{8}{25}\) or \(2 \times \frac{1}{5} \times \frac{4}{5}\): M0M0A0