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STEP II 2007 question 12 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP II 2007 question 12 solution

(i) First observe that p(first six after r throws of a single die) = pq^{r-1}.

P(stop after r throws) = p(no sixes in first r-1 throws and 2 sixes on rth throw) + \displaystyle\sum_{k=1}^{r-1} p(no sixes in k-1 throws, single six after k throws and 2nd six on rth throw (i.e. r-k throws of single die))

\displaystyle=(q^2)^{r-1}p^2 + \sum_1^{r-1} (q^2)^{k-1} 2pq pq^{r-k-1}

\displaystyle=q^{2r-2}p^2 + 2p^2\sum_1^{r-1} q^{2k-2+1+r-k-1}

\displaystyle=q^{2r-2}p^2 + 2p^2\sum_1^{r-1} q^{k+r-2}

\displaystyle=q^{2r-2}p^2 + 2p^2q^{r-1}\frac{1-q^{r-1}}{1-q}

\displaystyle=q^{2r-2}p^2 + 2pq^{r-1}(1-q^{r-1})

\displaystyle=pq^{r-1}[pq^{r-1} + 2(1-q^{r-1})

\displaystyle=pq^{r-1}[2+(1-q)q^{r-1} -2q^{r-1}

\displaystyle=pq^{r-1}[2 - q^{r-1} -q^r]

Then E(T) = \sum_0^\infty r pq^{r-1}[2 - q^{r-1} -q^r]

\displaystyle=\frac{2p}{q}\sum_0^\infty rq^r -p\frac{1+q}{q^2} \sum_0^\infty rq^{2r}

\displaystyle=\frac{2pq}{q(1-q)^2} -p\frac{1+q}{q^2} \frac{q^2}{(1-q^2)^2}

\displaystyle=\frac{2p}{(1-q)^2} - \frac{p(1+q)}{(1-q^2)^2}

\displaystyle=\frac{2p}{p^2} - \frac{p(1+q)}{(1-q)^2(1+q)^2}

\displaystyle=\frac{2p}{p^2} - \frac{p}{p^2(1+q)} = \frac{2}{p} - \frac{1}{p(2-p)}

\displaystyle=\frac{3-2p}{p(2-p}.

So if E(T)=m, we have p(2-p)m = 3-2p

mp^2-2p(m+1)+3 = 0

\displaystyle p = \frac{m+1 \pm \sqrt{(m+1)^2-3m}}{m}.

Since p \leq 1 we need the -ve root, and so we end up with \displaystyle p = \frac{m+1 - \sqrt{m^2-m+1}}{m}.

Solution by DFranklin.

Retrieved from "http://thestudentroom.co.uk/wiki/STEP_II_2007_question_12_solution"
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