Year: 2015
Paper: 2
Question Number: 12
Course: LFM Stats And Pure
Section: Tree Diagrams
As in previous years the Pure questions were the most popular of the paper with questions 1, 2 and 6 the most popular. The least popular questions on the paper were questions 8, 11 and 13 with fewer than 250 attempts for each of them. There were many examples of solutions in this paper that were insufficiently well explained given that the answer to be reached had been provided in the question.
Difficulty Rating: 1600.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
Four players $A$, $B$, $C$ and $D$ play a coin-tossing game with a fair coin. Each player chooses a sequence of heads and tails, as follows:
Player A: HHT; Player B: THH; Player C: TTH; Player D: HTT.
The coin is then tossed until one of these sequences occurs, in which case the corresponding player is the winner.
\begin{questionparts}
\item Show that, if only $A$ and $B$ play, then $A$ has a probability of $\frac14$ of winning.
\item If all four players play together, find the probabilities of each one winning.
\item Only $B$ and $C$ play. What is the probability of $C$ winning if the first two tosses are TT?
Let the probabilities of $C$ winning if the first two tosses are HT, TH and HH be $p$, $q$ and $r$, respectively. Show that $p=\frac12 +\frac12q$.
Find the probability that $C$ wins.
\end{questionparts}\begin{questionparts}
\item The only way $A$ can win is if the sequence starts HH, if it does not start like this, then the only way HHT can appear is after a sequence of THH...H, but then THH has already appeared and $B$ has won. Therefore the probability is $\frac14$
\item If HH appears before TT then either $A$ or $B$ will win. If HH appears first, then $A$ has a $\frac14$ probability of winning. So $A$: $\frac18$, $B:$, $\frac38$, $C:$, $\frac18$, $D: \frac38$
\item If the first two tosses are TT then $C$ will win.
If the first two tosses are HT, then either the next toss is T and $C$ wins, or the next toss is H, and it's as if we started TH. ie $p = \frac12 + \frac12 q$.
If the first two tosses are TH, then either the next toss is H and $C$ losses or the next toss is T and it's like starting HT. So $q = \frac12 p$.
Therefore $p = \frac12 + \frac14p \Rightarrow p = \frac13$
If the first two tosses are HH, then eventually a T appears, and it's the same as starting HT. Therefore the probability $C$ wins is:
$\frac14 + \frac14 \cdot \frac13 + \frac14 \cdot \frac16 + \frac14 \cdot \frac13 = \frac{11}{24}$
\end{questionparts}
Many solutions to this question did not include sufficient explanation to gain full credit. In the first part, marks were not awarded simply for stating that the value of ¼ could be achieved by multiplying ½ by ½ (often with an additional multiplication by 1) – an explanation of where this calculation comes from was also required. In the second part a number of candidates stated that it was symmetric and so the answer must be ¼ but with insufficient explanation why. In part (iii), some candidates obtained a geometric series which was then summed to get the probability of C winning if the first two tosses are TT. In the final part some correct answers were offered, but without explanation of the method. A number of candidates made incorrect assumptions such as that p+q=1, or p+q+r=1. When finding the probability that C wins a lot of candidates were able to achieve some of the marks by working out the probability in terms of q.