hey guys some help here plz
Note that questions before 2001 do not reflect well the difficulty of the current HSC exams. Here are sketch solutions.
i) This is a classic case of recursion. We can split the cases depending on if the first throw was a head or a tail. If the first was a head, we'd still need \(r-1\) heads and \(s\) tails in their appropriate positions, giving \(P(r-1,s)\). And vice versa if a tail was first. Since the probability that the first throw is \( \frac12\) for a head or a tail, we get \( \frac12 P(r-1,s) +\frac12 P(r,s-1) \).
ii) We have \(P(2,3) = \frac12P(1,3) + \frac12P(2,2) \)
For \(P(1,3)\), we need one head
before we reach three tails, so we'd have either \(H\), \(TH\) or \(TTH\).
For \(P(2,2)\), we need two heads
before we reach two tails, so we'd have either \(HH\), \(THH\) or \(HTH\)
iii) For the induction, the base case at \(n=0\) would consequently imply that \(r+s-1 = 0\). Since \(s \geq 1\), the only case is \( \boxed{r=0, s=1} \). This should thus be an easy check.
Then assume that the statement holds for \(n=k\). This means that it holds
for all choices of \(r\) and \(s\) that satisfy the property \(r+s-1=k\). We need to prove it for \(n=k+1\), i.e. all choices of \(r\) and \(s\) that satisfy the property \( r+s-1=k+1\).
If \(r=0\), we have \(s=k+2\). This case should be handled separately, as it should fall out immediately.
Otherwise, \(r \geq 1\) and we can take care of all of those in the one go with the inductive assumption.
\begin{align*}P(r,s) &= \frac12 P(r-1, s) + \frac12 P(r, s-1)\\ &=\frac12 \times \frac1{2^k} \left[ \binom{k}{0} + \binom{k}{1} +\dots +\binom{k}{s-2}+ \binom{k}{s-1} \right]\\ &\quad + \frac12 \times\frac1{2^k} \left[ \binom{k}{0} + \binom{k}{1} + \dots + \binom{k}{s-2} \right]\tag{assumption} \end{align*}
At this point, combine the brackets together, and make use of the property \( \binom{N+1}{K+1} = \binom{N}{K} + \binom{N+1}{K+1} \) and \( \binom{N}{0} = \binom{N+1}{0} = 1 \) to complete the proof of this case. And after this case is proven, the induction is complete.
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