Problem source

It is believed that the population of Ruritania can be described as follows: 
\begin{questionparts}
\item $25\%$ are fair-haired and the rest are dark-haired; 
\item $20\%$ are green-eyed and the rest hazel-eyed; 
\item the population can also be divided into narrow-headed and broad-headed; 
\item no narrow-headed person has green eyes and fair hair; 
\item those who are green-eyed are as likely to be narrow-headed as broad-headed; 
\item those who are green-eyed and broad-headed are as likely to be fair-headed
as dark-haired; 
\item half of the population is broad-headed and dark-haired; 
\item a hazel-eyed person is as likely to be fair-haired and broad-headed
as dark-haired and narrow-headed. 
\end{questionparts}
Find the proportion believed to be narrow-headed. 
I am acquainted with only six Ruritanians, all of whom are broad-headed.
Comment on this observation as evidence for or against the given model. 
A random sample of 200 Ruritanians is taken and is found to contain 50 narrow-heads. On the basis of the given model, calculate (to a reasonable approximation) the probability of getting 50 or fewer narrow-heads.
Comment on the result.

Solution source

\begin{center}
    \begin{tikzpicture}[scale=2.5]
        \draw (0,0) -- (0, 3) -- (6, 3) -- (6, 0) -- cycle;
        \draw (3, 1) circle [x radius=1.4, y radius =0.8];
        \draw (2, 2) circle [x radius=1.4, y radius =0.8];
        \draw (4, 2) circle [x radius=1.4, y radius =0.8];

        \node at (3, 0.35) {narrow-headed};
        \node at (2, {3-0.35}) {fair-haired};
        \node at (4, {3-0.35}) {green-eyed};

        \node at (2, 2) {$a$};
        \node at (3, 2) {$b$};
        \node at (4, 2) {$c$};
        \node at (2.5, 1.6) {$d$};
        \node at (3, 1.6) {$e$};
        \node at (3.5, 1.6) {$f$};
        
        \node at (3, 1) {$g$};
        \node at (1, 1) {$h$};
    \end{tikzpicture}
\end{center}

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*}

\begin{center}
    \begin{tikzpicture}[scale=2.5]
        \draw (0,0) -- (0, 3) -- (6, 3) -- (6, 0) -- cycle;
        \draw (3, 1) circle [x radius=1.4, y radius =0.8];
        \draw (2, 2) circle [x radius=1.4, y radius =0.8];
        \draw (4, 2) circle [x radius=1.4, y radius =0.8];

        \node at (3, 0.35) {narrow-headed};
        \node at (2, {3-0.35}) {fair-haired};
        \node at (4, {3-0.35}) {green-eyed};

        \node at (2, 2) {$a$};
        \node at (3, 2) {$b$};
        \node at (4, 2) {$b$};
        \node at (2.5, 1.6) {$d$};
        \node at (3, 1.6) {$0$};
        \node at (3.5, 1.6) {$2b$};
        
        \node at (3, 1) {$a$};
        \node at (1, 1) {$0.5-b$};
    \end{tikzpicture}
\end{center}

So $4b = 0.2 \Rightarrow b = 0.05$

\begin{center}
    \begin{tikzpicture}[scale=2.5]
        \draw (0,0) -- (0, 3) -- (6, 3) -- (6, 0) -- cycle;
        \draw (3, 1) circle [x radius=1.4, y radius =0.8];
        \draw (2, 2) circle [x radius=1.4, y radius =0.8];
        \draw (4, 2) circle [x radius=1.4, y radius =0.8];

        \node at (3, 0.35) {narrow-headed};
        \node at (2, {3-0.35}) {fair-haired};
        \node at (4, {3-0.35}) {green-eyed};

        \node at (2, 2) {$a$};
        \node at (3, 2) {$0.05$};
        \node at (4, 2) {$0.05$};
        \node at (2.5, 1.6) {$d$};
        \node at (3, 1.6) {$0$};
        \node at (3.5, 1.6) {$0.1$};
        
        \node at (3, 1) {$a$};
        \node at (1, 1) {$0.45$};
    \end{tikzpicture}
\end{center}

And

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

\begin{center}
    \begin{tikzpicture}[scale=2.5]
        \draw (0,0) -- (0, 3) -- (6, 3) -- (6, 0) -- cycle;
        \draw (3, 1) circle [x radius=1.4, y radius =0.8];
        \draw (2, 2) circle [x radius=1.4, y radius =0.8];
        \draw (4, 2) circle [x radius=1.4, y radius =0.8];

        \node at (3, 0.35) {narrow-headed};
        \node at (2, {3-0.35}) {fair-haired};
        \node at (4, {3-0.35}) {green-eyed};

        \node at (2, 2) {$0.15$};
        \node at (3, 2) {$0.05$};
        \node at (4, 2) {$0.05$};
        \node at (2.5, 1.6) {$0.05$};
        \node at (3, 1.6) {$0$};
        \node at (3.5, 1.6) {$0.1$};
        
        \node at (3, 1) {$0.15$};
        \node at (1, 1) {$0.45$};
    \end{tikzpicture}
\end{center}

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