Question

Die $A$ has $4$ red and $2$ white faces whereas die $B$ has $2$ red and $4$ white faces. A coin is flipped once. If it shows a head, the game continues by throwing die $A$, if it shows tail, then die $B$ is to be used. If the probability that die $A$ is used is $\frac{64}{66}$ where it is given that red turns up every time in first n throws, then $n$ is

Answer 1

Let $R$ be the event that a red face appears in each of the first $n$ throws
$E_1:$ Die $A$ is used when head has already fallen.
$E_2:$ Die $B$ is used when tail has already fallen.
$P\left(R|E_1\right)={\left(\frac{2}{3}\right)}^n$ and $P\left(R|E_2\right)={\left(\frac{1}{3}\right)}^n$
Given, $P\left(E_1|R\right)=\frac{64}{66}$
$\Rightarrow \frac{P\left(E_1\right)P\left(R|E_1\right)}{P\left(E_1\right).P\left(R|E_1\right)+P\left(E_2\right).P\left(R|E_2\right)}=\frac{64}{66}$
$\frac{\frac{1}{2}{\left(\frac{2}{3}\right)}^n}{\frac{1}{2}.{\left(\frac{2}{3}\right)}^n+\frac{1}{2}.{\left(\frac{1}{3}\right)}^n}=\frac{32}{33}$ (Bayes' theorem)
$\Rightarrow \frac{2^n}{2^n+1}=\frac{32}{33}$

This can be written as $\frac{2^n}{2^n+1}=\frac{2^5}{2^5+1}$
$\Rightarrow n=5$