A2 June 2022 Q6
6. The discrete random variable \(V\) has probability distribution
| \(v\) | 2 | 3 | 4 |
|---|---|---|---|
| \(\mathrm{P}(V = v)\) | \(\dfrac{9}{25}\) | \(\dfrac{12}{25}\) | \(\dfrac{4}{25}\) |
The discrete random variable \(W\) has probability generating function
\[\mathrm{G}_W(t) = t\left(\frac{2}{5}t + \frac{3}{5}\right)^5\]Given that \(V\) and \(W\) are independent,
The discrete random variable \(Y = 2X + 3\)
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{G}_V(t) = \dfrac{9}{25}t^2 + \dfrac{12}{25}t^3 + \dfrac{4}{25}t^4\) or \(t^2\left(\dfrac{9}{25} + \dfrac{12}{25}t + \dfrac{4}{25}t^2\right)\) | M1 | 1.1b |
| \(= t^2\left(\dfrac{2}{5}t + \dfrac{3}{5}\right)^2\)* | A1* cso | 2.1 |
| (2) |
Notes
M1 A correct un-simplified pgf based on \(\sum t^v\mathrm{P}(V = v)\)
A1* cso must see an un-simplified version i.e. M1 scored and no incorrect working seen
| Scheme | Marks | AO |
|---|---|---|
| (i) \(\mathrm{G}_W{}^{\prime}(t) = 2t\left(\dfrac{2}{5}t + \dfrac{3}{5}\right)^4 + \left(\dfrac{2}{5}t + \dfrac{3}{5}\right)^5\) | M1 | 2.1 |
| [\(\mathrm{G}_W{}^{\prime}(1) =\)] \(\underline{\mathbf{3}}\) | A1 | 1.1b |
| (ii) \(\mathrm{G}_W{}^{\prime\prime}(t) = 2\left(\dfrac{2}{5}t + \dfrac{3}{5}\right)^4 + \dfrac{16}{5}t\left(\dfrac{2}{5}t + \dfrac{3}{5}\right)^3 + 2\left(\dfrac{2}{5}t + \dfrac{3}{5}\right)^4\) oe | M1 | 2.1 |
| \(\mathrm{G}_W{}^{\prime\prime}(1) = \dfrac{36}{5}\) | A1 | 1.1b |
| \(\mathrm{Var}(W) = \text{“}\dfrac{36}{5}\text{”} + \text{“}3\text{”} - (\text{“}3\text{”})^2\) | M1 | 2.1 |
| \(= \dfrac{6}{5}\) | A1 | 1.1b |
| (6) |
Notes
(i) M1 Differentiating using the product rule to find \(\mathrm{G}_W{}^{\prime}(t)\) Allow un-simplified e.g. \(5 \times \dfrac{2}{5}t\)
Need two terms added and at least one correct. If they expand we need 3 correct.
A1 3 from a correct derivative
(ii) 1st M1 Attempt \(\mathrm{G}_W{}^{\prime\prime}(t)\) ft their \(\mathrm{G}_W{}^{\prime}(t)\) [must be at least 2 terms or a product], one correct ft term, same rule for differentiating a product
1st A1 \(\dfrac{36}{5}\) or 7.2 from a correct derivative
2nd M1 \(\mathrm{G}_W{}^{\prime\prime}(1) + \mathrm{G}_W{}^{\prime}(1) - \left(\mathrm{G}_W{}^{\prime}(1)\right)^2\) ft their \(\mathrm{G}_W{}^{\prime\prime}(t)\) if different from \(\mathrm{G}_W{}^{\prime}(t)\) and \(\mathrm{G}_W(t)\)
2nd A1 Dep on M3A2 \(\dfrac{6}{5}\) or 1.2
Alternative for (b)
| Scheme | Marks | AO |
|---|---|---|
| \(W = P + 1\) where \(P \sim \mathrm{B}(5, 0.4)\) so \(\mathrm{Var}(W) = \mathrm{Var}(P)\) | ||
| (i) \(\mathrm{G}_P{}^{\prime}(t) = 2\left(\dfrac{2}{5}t + \dfrac{3}{5}\right)^4\) | M1 | 2.1 |
| \(\mathrm{G}_W{}^{\prime}(1) = 2 + 1 = 3\) | A1 | 1.1b |
| (ii) \(\mathrm{G}_P{}^{\prime\prime}(t) = \dfrac{16}{5}\left(\dfrac{2}{5}t + \dfrac{3}{5}\right)^3\ ;\ \ \mathrm{G}_P{}^{\prime\prime}(1) = \dfrac{16}{5}\) | M1; A1 | 2.1 1.1b |
| \(\mathrm{Var}(W) = \text{“}\dfrac{16}{5}\text{”} + \text{“}2\text{”} - (\text{“}2\text{”})^2\ ;\ \ = \dfrac{6}{5}\) | M1; A1 | 2.1 1.1b |
SC MR They use \(\mathrm{G}_V(t)\) instead of \(\mathrm{G}_W(t)\) Provided some correct differentiation seen:
Award B1 for \(\mathrm{E}(V) = \dfrac{14}{5}\) and B1 for \(\mathrm{Var}(V) = \dfrac{12}{25}\) score as M0A1M0A0M0A1
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{G}_X(t) = t^2\left(\dfrac{2}{5}t + \dfrac{3}{5}\right)^2 \times t\left(\dfrac{2}{5}t + \dfrac{3}{5}\right)^5\) | M1 | 3.1a |
| \(= t^3\left(\dfrac{2}{5}t + \dfrac{3}{5}\right)^7\) | A1 | 1.1b |
| (2) |
Notes
M1 Realising the need to use \(\mathrm{G}_X(t) = \mathrm{G}_V(t) \times \mathrm{G}_W(t)\)
A1 \(t^3\left(\dfrac{2}{5}t + \dfrac{3}{5}\right)^7\)
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{G}_Y(t) = t^3 \times (t^2)^3 \times \left(\dfrac{2}{5}t^2 + \dfrac{3}{5}\right)^7\) | M1 | 3.1a |
| \(= t^9\left(\dfrac{2}{5}t^2 + \dfrac{3}{5}\right)^7\) | A1 | 1.1b |
| (2) |
Notes
M1 Realising the need to multiply through by \(t^3\) or substitute \(t^2\) for \(t\) or sight of \(t^3\mathrm{G}_X(t^2)\)
A1 \(t^9\left(\dfrac{2}{5}t^2 + \dfrac{3}{5}\right)^7\) oe Need not be in its simplest form
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{P}(Y = 15)\) is coefficient of \(t^{15}\) ie \(\ldots + t^9 \times {}^7C_3\left(\dfrac{2}{5}t^2\right)^3\left(\dfrac{3}{5}\right)^4 + \ldots\) or \(\mathrm{P}(X = 6)\) need coefficient of \(t^6\) i.e. \(\ldots + t^3 \times {}^7C_3\left(\dfrac{2}{5}t\right)^3\left(\dfrac{3}{5}\right)^4 + \ldots\) | M1 | 1.1b |
| [\(\mathrm{P}(Y = 15) =\)] \(\dfrac{22680}{78125} = \dfrac{4536}{15625} = 0.290304\) | A1 | 1.1b |
| (2) | ||
| (14 marks) |
Notes
M1 Attempting to find correct coefficient of \(t^n\) or identify \(Y = 2J + 9\) where \(J \sim \mathrm{B}(7, 0.4)\)
Need an expression can ft their \(\mathrm{G}_Y(t)\) or \(\mathrm{G}_X(t)\) of the form \(t^n(at^m + b)^k\)
Allow a statement that \(\mathrm{P}(Y = 15) = 0\) if it follows from their pgf
A1 For a correct exact answer or allow awrt 0.2903 Allow 0.29 from correct expression