Problem source

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. 
	\begin{questionparts}
	\item 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. 
	\item 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.
\end{questionparts}

Solution source

\begin{questionparts}
\begin{align*}
\P(\text{pre} | \text{same votes}) &= \frac{\P(\text{pre} \cap \text{same votes})}{\P(\text{same votes})} \\
&= \frac{\frac{60}{100}}{\frac{60}{100} + \frac{40}{100} \cdot \frac{1}{2}} \\
&= \frac{6}{8} = \frac{3}{4} \\
\P(\text{votes same way} | \text{same votes}) &= \frac{3}{4} + \frac{1}{4} \cdot \frac{1}{2} \\
&= \frac{7}{8}
\end{align*}

\item The vote is will now be $60(1-\alpha)$ for, $60\alpha$ against and $X \sim B(40, \frac12)$ at random between those. For a majority, they need $60(1-\alpha) + X > 50$, ie $\P(X > 60\alpha - 10) \geq 0.8$. Using a normal approximation to the binomial, we need $X \approx N(20, 10)$, so

\begin{align*}
\P(X > 60 \alpha - 10) &= 1- \P(X \leq 60 \alpha - 10) \\
&\approx 1 - \P(\sqrt{10}Z+20 \leq 60\alpha - 10.5) \\
&\approx 1 - \P(Z \leq \frac{60\alpha - 30.5}{\sqrt{10}})
\end{align*}

If we want this to be less than $0.2$ we need $ \frac{60\alpha - 30.5}{\sqrt{10}} < -0.8416 \Rightarrow \alpha < 0.4639$.

This would correspond to 27 or fewer exiles or 33 or more remaining preservatives. 

[Actual computations using Binomial distribution shows we should expect at least 17 to randomly join 20% of the time, so 34 preservatives are required]
\end{questionparts}