STEP I, II, III 2000 solutions

III/13:

If the first die to show a six does so on the rth roll then there'll (r-1) rolls where all dice fail to show a six = $(q^{r-1})^n = q^{rn - n}$

On the rth roll: P(1 or more shows a six) = 1 - P(none shows a six) = 1 - q^n. Overall probability = $q^{rn-n}(1 - q^n) = q^{rn - n} - q^{rn} = q^{rn}(q^{-n}-1)$, as required.

Let G(t) denote the probability generating function. r can take values 1, 2, 3, ... so $\text{G}(t) = q^n(q^{-n}-1)t + q^{2n}(q^{-n} - 1)t^2 + ...$ Geometric series, a = $(1-q^n)t$, r = $q^nt$. Sum to infinity = $\dfrac{(1-q^n)t}{1-q^nt}$.


When n = 2, p = 1/6, q = 5/6 so E(R) = $\dfrac{25/36}{1 - 25/36} + 1 = \frac{25}{11} + 1 = \frac{36}{11}$.

$P(S = r) = P(S \le r) - P(S \le r - 1)$.
Less than or equal to r for one die is 1 - q^r; not n failures. So for n-dice: $(1-q^r)^n$, and similarly for r - 1: $(1-q^{r-1})^n$. Overall probability: $(1-q^r)^n - (1-q^{r-1})^n$

$(1-q^r)^2 - (1-q^{r-1})^2 = q^{2r} - q^{2r-2} + 2q^{r-1} - 2q^r$ r can take values 1, 2, 3, ...

=> $\displaystyle \text{G}(t) = \sum_{r=1}^{\infty}(q^{2r} - q^{2r-2} + 2q^{r-1} - 2q^r)t^r = \sum_{r=1}^{\infty} \left[ t^rq^{2r} - t^rq^{2r-2} + 2t^rq^{r-1} - 2q^rt^r \right]$

$= \dfrac{tq^2}{1-tq^2} - \dfrac{t}{1-tq^2} + \dfrac{2t}{1-tq} - \dfrac{2qt}{1-qt} = \dfrac{tq^2 - t}{1 - tq^2} + \dfrac{2t-2qt}{1-qt}$.

$\text{G'}(1) = \text{E}(S) = \dfrac{2}{1-q} = \dfrac{2}{p}$ = 12 when p = 1/6. (I haven't LaTeXed the derivative of G(t), it's too late and it's just tedious.)

II / 8

$\displaystyle - \int \frac{dy}{(y+3)^\frac{1}{2}} = -2 \int -2xe^{-x^2} dx \implies -2(y+3)^{\frac{1}{2}} = -2e^{-x^2} + C \implies y = e^{-2x^2} + 2ke^{-x^2} + (k^2 - 2)$

After subbing (0,6) ; $\displaystyle (k+4)(k-2) = 0$

$\displaystyle \therefore y = e^{-2x^2} - 8e^{-x^2} + 13$ and case 2 $\displaystyle y = e^{-2x^2} + 4e^{-x^2} + 1$

There is a vlaue of k such that as x tends to $\displaystyle \infty$ y tends to $\displaystyle 1$

(ii) wip. Got
$\displaystyle y^2e^{-3x^2}$ but the answer requires $\displaystyle ye^{-3x^2}$

2000 paper 3 nos. 4,6 & 7

In STEP III Q7, in the last part could anyone help me understand how its possible for L to equal e. We've just shown that P(n) < e for all n so i'm just having trouble understanding why the limit as n -> infinity can be equal to e.

II / 2

$\displaystyle p[x] = (x-a)^2q[x] \Rightarrow p'[x] = (x-a)(2q[x] + (x-a)q'[x])$

$\displaystyle g[x] = (x-a)^4q[x] \Rightarrow p'[x] = (x-a)^3(4q[x] + (x-a)q'[x]) *$

Given p(x) we differentiate to find;

The roots of the equation $(x-\alpha)^3(4q[x] + (x-\alpha)q'[x]) = 6x^5 + 20x^4 - 20x^3 - 120x^2 - 80x + 32$

Inspection reults in $\displaystyle \alpha = -2$ So if p[x] denotes the polynomial given then p[-2] = 0. Using the factor and remainder theorem; k = 48

Damn. :P Is this definitely all you need to do for this question? I reasoned that it wanted us to prove something different for the $(x - a)^{4}$ bit, so showed that p'''(a) = 0.

Then for the second part, I found the values of x for which p'''(x) = 0, and said that a must be one of these, and put them into p''(x) and p'(x) to eliminate two of them, giving me a = -2, which I then finished like you did. How did you know when you solved,

$(x-\alpha)^3(4q[x] + (x-\alpha)q'[x]) = 6x^5 + 20x^4 - 20x^3 - 120x^2 - 80x + 32$

that the root you'd found was definitely a?