| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Year | 2016 |
| Session | June |
| Marks | 5 |
| Topic | Probability Generating Functions |
| Type | Derive standard distribution PGF |
| Difficulty | Standard +0.8 This is a Further Maths probability generating functions question requiring derivation of the binomial PGF from first principles, manipulation of PGFs for independent random variables, and algebraic simplification to find a specific probability. While PGFs are standard Further Maths content, the algebraic manipulation in part (ii) and extracting the coefficient requires careful work across multiple steps, placing it moderately above average difficulty. |
| Spec | 5.02b Expectation and variance: discrete random variables5.02d Binomial: mean np and variance np(1-p) |
(i) $\sum t^r \text{P}(R=r) = \sum_{r=0}^{n} t^r\, {}^nC_r\, p^r(1-p)^{n-r}$
$= \sum_{r=0}^{n}(pt)^r\, {}^nC_r\,(1-p)^{n-r}$
$= (1-p+pt)^n$ **AG**
- M1: Use $\Sigma t^r \text{P}(R=r)$ and binomial probabilities
- A1: Indicate correct final term
- A1: Collect $p^r$ and $t^r$ and correctly obtain given expression
OR $(1-p+pt)^1$, M1A1; answer, A1
**[3]**
(ii) $\left(\tfrac{3}{4}+\tfrac{1}{4}t\right)^8\!\left(\tfrac{1}{4}+\tfrac{3}{4}t\right)^8$
$=(3+t)^8(1+3t)^8/4^{16}$
$= \left(\tfrac{3}{16}+\tfrac{1}{16}t(10+3t)\right)^8$ **AG**
$= \left(\tfrac{3}{16}\right)^8\!\left(1+\left[\tfrac{10}{3}+t\right]t\right)^8$
$t$ term: $\left(\tfrac{3}{16}\right)^8\!\left[8\times\tfrac{10}{3}\right] = \mathbf{4.07\times10^{-5}}$
- M1: Substitute and multiply
- A1: Correctly obtain given answer
- M1M1: Select $t$ term; method for expansion formula
- A1: Answer
OR: attempt to find $G'(0)$: M1; $8\!\left(\tfrac{3}{16}\right)^8\!\left(\tfrac{10}{3}+2t\right)\!\left(1+\left[\tfrac{10}{3}+t\right]t\right)^7$: A1; $= 4.07\times10^{-5}$: A1
**[5]**3 (i) Show that the probability generating function of a random variable with the distribution $\mathrm { B } ( n , p )$ is $( 1 - p + p t ) ^ { n }$.\\
(ii) $R$ and $S$ are independent random variables with the distributions $\mathrm { B } \left( 8 , \frac { 1 } { 4 } \right)$ and $\mathrm { B } \left( 8 , \frac { 3 } { 4 } \right)$ respectively. Show that the probability generating function of $R + S$ can be expressed as
$$\left( \frac { 3 } { 16 } + \frac { 1 } { 16 } t ( 10 + 3 t ) \right) ^ { 8 }$$
and use this result to find $\mathrm { P } ( R + S = 1 )$.
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2016 Q3 [5]}}