Year: 2012
Paper: 2
Question Number: 12
Course: LFM Stats And Pure
Section: Tree Diagrams
There were just over 1000 entries for paper II this year, almost exactly the same number as last year. Overall, the paper was found marginally easier than its predecessor, which means that it was pitched at exactly the level intended and produced the hoped-for outcomes. Almost 50 candidates scored 100 marks or more, with more than 400 gaining at least half marks on the paper. At the lower end of the scale, around a quarter of the entry failed to score more than 40 marks. It was pleasing to note that the advice of recent years, encouraging students not to make attempts at lots of early parts to questions but rather to spend their time getting to grips with the six that can count towards their paper total, was more obviously being heeded in 2012 than I can recall being the case previously. As in previous years, the pure maths questions provided the bulk of candidates' work, with relatively few efforts to be found at the applied ones. Questions 1 and 2 were the most popular questions, although each drew only around 800 "hits" – fewer than usual. Questions 3 – 5 & 8 were almost as popular (around 700), with Q6 attracting the interest of under 450 candidates and Q7 under 200. Q9 was the most popular applied question – and, as it turned out, the most successfully attempted question on the paper – with very little interest shown in the rest of Sections B or C.
Difficulty Rating: 1600.0
Difficulty Comparisons: 0
Banger Rating: 1500.7
Banger Comparisons: 2
A modern villa has complicated lighting controls. In order for the light in the swimming pool to be on, a particular switch in the hallway must be on and a particular switch in the kitchen must be on. There are four identical switches in the hallway and four identical switches in the kitchen. Guests cannot tell whether the switches are on or off, or what they control.
Each Monday morning a guest arrives, and the switches in the hallway are either all on or all off. The probability that they are all on is $p$ and the probability that they are all off is $1-p$. The switches in the kitchen are each on or off, independently, with probability
$\frac12$.
\begin{questionparts}
\item On the first Monday, a guest presses one switch in the hallway at random and one switch in the kitchen at random. Find the probability that the swimming pool light is on at the end of this process. Show that the probability that the guest has pressed the swimming pool light switch in the hallway, given that the light is on at the end of the process, is $\displaystyle \frac{1-p}{1+2p}$.
\item On each of seven Mondays, guests go through the above process independently of each other, and each time the swimming pool light is found to be on at the end of the process. Given that the most likely number of days on which the swimming pool light switch in the hallway was pressed is 3, show that $\frac14 < p < \frac{5}{14}$.
\end{questionparts}\begin{questionparts}
\item $\,$
\begin{align*}
&& \mathbb{P}(\text{hall switch on}) &= \underbrace{p \cdot \frac34 }_{\text{already on and not flipped}}+ \underbrace{(1-p) \cdot \frac14}_{\text{not on and flipped}} \\
&&&= \frac14 +\frac12 p\\
&& \mathbb{P}(\text{kitchen on}) &= \frac12 \\
\Rightarrow && \mathbb{P}(\text{pool is on}) &= \frac18 + \frac14p
\end{align*}
\begin{align*}
&& \mathbb{P}(\text{flipped hall switch} | \text{pool on}) &= \frac{\mathbb{P}(\text{flipped hall and pool on})}{\mathbb{P}(\text{pool on})} \\
&&&= \frac{(1-p)\frac14 \cdot \frac 12}{\frac18 + \frac14 p} \\
&&&= \frac{1-p}{1+2p}
\end{align*}
\item The number of days the swimming pool light was pressed is $X = B\left (7, \frac{1-p}{1+2p} \right)$, and we have that $\mathbb{P}(X = 2) < \mathbb{P}(X = 3) > \mathbb{P}(X=4)$ (since the binomial is unimodal). Let $q = \frac{1-p}{1+2p} $
\begin{align*}
&& \mathbb{P}(X = 2) &< \mathbb{P}(X = 3) \\
\Rightarrow && \binom{7}{2} q^2(1-q)^5 &< \binom{7}{3}q^3(1-q)^4 \\
\Rightarrow && 21(1-q) &< 35q \\
\Rightarrow && 21 &< 56q \\
\Rightarrow && \frac{3}{8} &< \frac{1-p}{1+2p} \\
\Rightarrow && 3+6p &< 8-8p \\
\Rightarrow && 14p &< 5\\
\Rightarrow && p &< \frac5{14} \\
\\
&& \mathbb{P}(X = 3) &> \mathbb{P}(X = 4) \\
\Rightarrow && \binom{7}{3} q^3(1-q)^4 &> \binom{7}{4}q^4(1-q)^3 \\
\Rightarrow &&(1-q)&> q \\
\Rightarrow && \frac12 &> q \\
\Rightarrow && \frac12 &> \frac{1-p}{1+2p} \\
\Rightarrow && 1+2p &> 2-2p \\
\Rightarrow && 4p &> 1\\
\Rightarrow && p &> \frac1{4}
\end{align*}
Therefore $\frac14 < p < \frac{5}{14}$ as required.
\end{questionparts}
This was not a popular question, and most attempts petered out after part (i), which was usually handled very well, even when the situation was split into more cases than was strictly necessary. Indeed, few made much of a serious attempt at (ii), mainly because they were either finding the range of p such that (i)'s given answer was equal to 7/3, or solving 2.5 < E(X) < 3.5, where X was the number of days that the light was on.