Problem source

The probability of throwing a 6 with a biased die is $p\,$. It is known that  $p$ is equal to one or other of the numbers $A$ and $B$ where  $0 < A < B < 1 \,$.  Accordingly the following  statistical 
test of the hypothesis $H_0: \,p=B$ against the alternative hypothesis $H_1: \,p=A$ is performed.
The die is thrown  repeatedly until a 6 is obtained. Then if $X$ is the total number of throws,  $H_0$ is accepted if $X \le M\,$, where $M$ is a given positive integer;  otherwise $H_1$ is accepted.
Let ${\alpha}$ be the probability that $H_1$ is accepted if $H_0$ is true, and let ${\beta}$ be  the probability that $H_0$ is accepted if $H_1$ is true.  
Show that ${\beta} = 1- {\alpha}^K,$ where $K$ is independent of $M$ and is to be determined in terms of $A$ and $B\,$.  Sketch the graph of ${\beta}$ against ${\alpha}\,$.

Solution source

$X \sim Geo(p)$. 

$\alpha = \mathbb{P}(X > M | p = B) = (1-B)^{M}$

$\beta = \mathbb{P}(X \leq M | p = A) = 1 - \mathbb{P}(X > M | p = A) = 1 - (1-A)^{M}$

\begin{align*}
\ln \alpha &= M \ln(1-B) \\
\ln (1-\beta) &= M \ln(1-A) \\
\frac{\ln \alpha}{\ln (1-\beta)} &= \frac{\ln(1-B)}{\ln(1-A)} \\
\ln(1-\beta) &= \ln \alpha \frac{\ln (1-A)}{\ln(1-B)} \\
\beta &= 1- \alpha^{ \frac{\ln (1-A)}{\ln(1-B)} }
\end{align*}

and $K = \frac{\ln (1-A)}{\ln(1-B)} $

Since $0 < A < B < 1$ we must have that $0 < 1 - B < 1-A < 1$ and $\ln(1-B) < \ln(1-A) < 0$ so $0 < K < 1$

\begin{center}
    
\begin{tikzpicture}[scale=3]
    \draw[->] (0, -.1) -- (0, 1.2);
    \draw[->] (-.1, 0) -- (1.2, 0);

    \draw[domain = 0:1, samples=400, variable = \x]  plot ({\x},{1-(\x)^0.4}); 

    \node at (1.3, 0) {$\alpha$};
    \node at (0, 1.3) {$\beta$};
\end{tikzpicture}
\end{center}