A2 October 2020 Q6
6. A discrete random variable \(X\) has probability generating function given by
\[\mathrm{G}_X(t) = \frac{1}{64}\,(a + bt^2)^2\]where \(a\) and \(b\) are positive constants.
Given that \(\mathrm{P}(X = 4) = \dfrac{25}{64}\)
The random variable \(Y = 3X + 2\)
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{P}(X = 3) = \underline{\mathbf{0}}\) | B1 | 1.1b |
| (1) |
Notes
B1: 0 (Since there is no term in \(t^3\))
| Scheme | Marks | AO |
|---|---|---|
| (i) Coefficient of \(t^4 = \tfrac{1}{64}b^2\) | M1 | 2.1 |
| \(\tfrac{1}{64}b^2 = \tfrac{25}{64}\) | M1 | 1.1b |
| \(b = 5\) (reject \(b = -5\) since \(b \gt 0\)) | A1 | 2.3 |
| \(\mathrm{G}_X(1) = 1\) \(\tfrac{1}{64}(a + \text{“}5\text{”})^2 = 1\) | M1 | 2.1 |
| \(a = 3\) (reject \(a = -13\) since \(a \gt 0\)) | A1 | 1.1b |
| \(\mathrm{P}(X = 2) =\) coefficient of \(t^2 = \tfrac{1}{64}(2ab)\) | M1 | 3.4 |
| \(= \underline{\tfrac{15}{32}}\) | A1 | 1.1b |
| (7) | ||
| (ii) \(\mathrm{E}(X) = \mathrm{G}'_X(1)\) | M1 | 2.1 |
| \(\mathrm{G}'_X(t) = \tfrac{2}{64}(\text{“}3\text{”} + \text{“}5\text{”}t^2) \times \text{“}10\text{”}t\) or \(\mathrm{G}'_X(t) = \tfrac{1}{64}(\text{“}60\text{”}t + \text{“}100\text{”}t^3)\) | M1 | 1.1b |
| \(\mathrm{G}'_X(1) = 2.5\) | A1ft | 1.1b |
| (3) |
Notes
(i) M1: Realising that \(\tfrac{1}{64}b^2\), the coefficient of \(t^4\), is needed
M1: Equating their coefficient of \(t^4\) to \(\tfrac{25}{64}\) with an attempt to find \(b\)
A1: \(b = 5\) only
M1: Realising that \(\mathrm{G}_X(1) = 1\) is required
A1: \(a = 3\) only
M1: Finding coefficient of \(t^2\) with their \(a \gt 0\) and \(b \gt 0\)
A1: \(\tfrac{15}{32}\) (condone awrt 0.469)
(ii) M1: Realising \(\mathrm{G}'_X(1)\) is needed
M1: Attempt to differentiate \(\mathrm{G}_X(t)\) with their values of \(a\) and \(b\)
A1ft: 2.5 (ft (3sf) their values of \(a\) and \(b\), \(a \gt 0\) and \(b \gt 0\)) \(\mathrm{E}(X) = \tfrac{ab + b^2}{16}\)
Alternative:
M1: Realising \(X = 0\), 2 and 4 only
M1: \([0 \times \mathrm{P}(X = 0)] + 2 \times \mathrm{P}(X = 2) + 4 \times \mathrm{P}(X = 4)\)
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{G}_Y(t) = t^2\mathrm{G}_X(t^3)\ \left[= \tfrac{t^2}{64}(a + b(t^3)^2)^2\right]\) | M1 | 3.1a |
| \(\mathrm{G}_Y(t) = \tfrac{t^2}{64}(\text{“}3\text{”} + \text{“}5\text{”}t^6)^2\) | A1ft | 1.1b |
| (2) | ||
| (13 marks) |
Notes
M1: either \(\mathrm{G}_X(t^3)\) or \(\times t^2\) or using \(Y = 2, 8, 14\)
A1ft: ft their values of \(a\) and \(b\), \(a \gt 0\) and \(b \gt 0\)
\(\mathrm{G}_Y(t) = \tfrac{t^2}{64}(\text{“}3\text{”} + \text{“}5\text{”}t^6)^2\) or \(\mathrm{G}_Y(t) = \tfrac{t^2}{64}(\text{“}9\text{”} + \text{“}30\text{”}t^6 + \text{“}25\text{”}t^{12})\) or
\(\mathrm{G}_Y(t) = \tfrac{1}{64}(\text{“}9t^2\text{”} + \text{“}30\text{”}t^8 + \text{“}25\text{”}t^{14})\)