A2 October 2021 Q6
6. The probability generating function of the random variable \(X\) is
\[\mathrm{G}_X(t) = k(1 + 2t)^5\]where \(k\) is a constant.
The probability generating function of the random variable \(Y\) is
\[\mathrm{G}_Y(t) = \frac{t(1 + 2t)^2}{9}\]Given that \(X\) and \(Y\) are independent,
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{G}_X(1) = 1\) | M1 | 2.1 |
| \(k \times 3^5 = 1 \quad \therefore k = \dfrac{1}{243}\)* | A1*cso | 1.1b |
| (2) |
Notes
M1: Stating \(\mathrm{G}_X(1) = 1\) eg \(\mathrm{G}_X(1) = k(1 + 2)^5 = 1 \quad k(1 + 2)^5 = 1\)
Allow Verification \(\dfrac{1}{243} \times 3^5 = 1\)
A1*: Fully correct proof with no errors Substituting \(t = 1\)
Verification need therefore \(\mathrm{G}_X(1) = 1\)
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{P}(X = 2)\) is coefficient of \(t^2\) so \(\mathrm{G}_X(t) = k\left(\ldots + {}^5C_2(2t)^2 + \ldots\right)\) | M1 | 1.1b |
| \(\mathrm{P}(X = 2) = \dfrac{40}{243}\) | A1 | 1.1b |
| (2) |
Notes
M1: Attempting to find the coefficient of \(t^2\)
A1: \(\dfrac{40}{243}\) or awrt 0.165
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{G}_W(t) = \dfrac{t^3}{243}\left(1 + 2(t^2)\right)^5\) | M1 | 3.1a |
| \(\mathrm{G}_W(t) = \dfrac{t^3}{243}\left(1 + 2t^2\right)^5\) | A1 | 1.1b |
| (2) |
Notes
M1: Realising the need to multiply through by \(t^3\) or subst \(t^2\) for \(t\)
A1: \(\dfrac{t^3}{243}\left(1 + 2t^2\right)^5\) oe eg \(\dfrac{t^3}{243}\left(1 + 10t^2 + 40t^4 + 80t^6 + 80t^8 + 32t^{10}\right)\)
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{G}_U(t) = \dfrac{1}{243}(1 + 2t)^5 \times \dfrac{t(1 + 2t)^2}{9}\) | M1 | 3.1a |
| \(= \dfrac{t(1 + 2t)^7}{2187}\) | A1 | 1.1b |
| (2) |
Notes
M1: Realising the need to use \(\mathrm{G}_U(t) = \mathrm{G}_X(t) \times \mathrm{G}_Y(t)\)
A1: \(\dfrac{t(1 + 2t)^7}{2187}\) oe
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{G}_U{}^{\prime}(t) = \dfrac{14t(1 + 2t)^6}{2187} + \dfrac{(1 + 2t)^7}{2187}\) | M1 | 2.1 |
| \(\mathrm{G}_U{}^{\prime}(1) = \dfrac{17}{3}\) | A1ft | 1.1b |
| \(\mathrm{G}_U{}^{\prime\prime}(t) = \dfrac{168t(1 + 2t)^5}{2187} + \dfrac{14(1 + 2t)^6}{2187} + \dfrac{14(1 + 2t)^6}{2187}\) | M1 | 2.1 |
| \(\mathrm{G}_U{}^{\prime\prime}(1) = 28\) | A1 | 1.1b |
| \(\mathrm{Var}(U) = \text{“}28\text{”} + \text{“}\dfrac{17}{3}\text{”} - \left(\text{“}\dfrac{17}{3}\text{”}\right)^2\) | M1 | 2.1 |
| \(= \dfrac{14}{9}\) | A1 | 1.1b |
| (6) | ||
| (14 marks) |
Notes
M1: For an attempt to differentiate G (\(u\)) e.g \(\mathrm{G}_U{}^{\prime}(t) = At(1 + 2t)^6 + B(1 + 2t)^7\) ft their part(d) if in the form \(kt(1 + 2t)^n\) where \(n \geqslant 5\)
A1ft: \(\dfrac{17}{3}\) or awrt 5.67
M1: For attempting second derivative eg \(\mathrm{G}_U{}^{\prime\prime}(t) = Ct(1 + 2t)^5 + D(1 + 2t)^6\) ft their part(d) if in the form \(kt(1 + 2t)^n\) where \(n \geqslant 5\)
A1 28
M1: Using \(\mathrm{G}_U{}^{\prime\prime}(1) + \mathrm{G}_U{}^{\prime}(1) - \left(\mathrm{G}_U{}^{\prime}(1)\right)^2\) ft their values
A1: \(\dfrac{14}{9}\) or awrt 1.56
Alternative (e)
| Scheme | Marks |
|---|---|
| \(\mathrm{G}_X{}^{\prime\prime}(t) = A(1 + 2t)^3\) | M1 |
| \(\mathrm{G}_X{}^{\prime}(1) = \dfrac{10}{3}\) and \(\mathrm{G}_X{}^{\prime\prime}(1) = \dfrac{80}{9}\) | A1ft |
| \(\mathrm{G}_Y{}^{\prime\prime}(t) = H(8 + 24t)\) | M1 |
| \(\mathrm{G}_Y{}^{\prime}(1) = \dfrac{7}{3}\) and \(\mathrm{G}_Y{}^{\prime\prime}(1) = \dfrac{32}{9}\) | A1 |
| Using \(\mathrm{G}_U{}^{\prime\prime}(1) + \mathrm{G}_U{}^{\prime}(1) - \left(\mathrm{G}_U{}^{\prime}(1)\right)^2\) to find \(\mathrm{Var}(X)\), Var \(Y\) and Var \(U\) | M1 |
| \(\dfrac{14}{9}\) or awrt 1.56 | A1 |