Problems

It is believed that the population of Ruritania can be described as follows:

1. \(25\%\) are fair-haired and the rest are dark-haired;
2. \(20\%\) are green-eyed and the rest hazel-eyed;
3. the population can also be divided into narrow-headed and broad-headed;
4. no narrow-headed person has green eyes and fair hair;
5. those who are green-eyed are as likely to be narrow-headed as broad-headed;
6. those who are green-eyed and broad-headed are as likely to be fair-headed as dark-haired;
7. half of the population is broad-headed and dark-haired;
8. a hazel-eyed person is as likely to be fair-haired and broad-headed as dark-haired and narrow-headed.

Show Solution

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Solution:

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Conditions tell us: \begin{align*} && a+b+d+e &= 0.25 \\ && b+c+e+f &= 0.2 \\ && e &= 0 \\ && b+c &= e + f \\ && b &= c \\ && c+h &= 0.5 \\ && a &= g \\ \end{align*}

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So \(4b = 0.2 \Rightarrow b = 0.05\)

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And \begin{align*} && 0.25 &= a + d + 0.05 \\ && 1 &= 2a + d + 0.65 \\ \Rightarrow && a &= 0.15 \\ && d &= 0.05 \end{align*}

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So the proportion who are narrow-headed is \(30\%\). It's obviously relatively unlikely for your six Ruritanian friends to all be broad-headed if it's a random sample, but friendship groups are are likely to be biased so it's not too surprising. Assuming there is a sufficiently large number of Ruritanians, we might model the number of narrow-headed Ruritanians from a sample of \(200\) as \(X \sim B(200, 0.3)\). Computing \(\mathbb{P}(X \leq 50)\) by hand is tricky, so let's use a binomial approximation to obtain: \(X \approx N(60, 42)\) and \begin{align*} \mathbb{P}(X \leq 50) &\approx \mathbb{P} \left (Z \leq \frac{50 - 60+0.5}{\sqrt{42}} \right) \\ &\approx \mathbb{P} \left (Z \leq -\frac{9.5}{6.5} \right) \\ &\approx \mathbb{P} \left (Z \leq -\frac{3}{2} \right) \\ &\approx 5\% \end{align*} (actually this approximation gives \(7.1\%\) and the binomial value gives \(7.0\%\)). This also seems somewhat surprising

The parliament of Laputa consists of 60 Preservatives and 40 Progressives. Preservatives never change their mind, always voting the same way on any given issue. Progressives vote at random on any given issue.

1. A randomly selected member is known to have voted the same way twice on a given issue. Find the probability that the member will vote the same way a third time on that issue.
2. Following a policy change, a proportion \(\alpha\) of Preservatives now consistently votes against Preservative policy. The Preservative leader decides that an election must be called if the value of \(\alpha\) is such that, at any vote on an item of Preservative policy, the chance of a simple majority would be less than 80\%. By making a suitable normal approximation, estimate the least value of \(\alpha\) which will result in an election being called.