Problem: In a test paper, the questions are arranged so that 3/4's of the time a True is followed by a True and 2/3's of the time a False is followed by a False. You are confronted with a 100 questions text. Approximately what fraction of the answers will be true? [assume markov property]
Attempt: $p_T=p_T\times p_{T,T}+P_F\times p_{F,T}=p_T\times \frac{3}{4}+p_F\times \frac{1}{3}$, thus we have : $p_T=(1-p_T)\frac{1}{3}$, which implies $p_T=\frac{4}{7}$. Thus the required fraction is $\frac{4}{7}$. Is my solution right? If it is, then what is the significance of the phrase "100 questions text" in the problem?
Remark: In my solution, "T" stands for True and "F" stands for false.
Thank you in anticipation.
Let $p_n$ be the probability of the $n$-th answer to be true, $n\geq 1$. Using the Markov property we have for all $n\geq 2$: $$p_n = p_{n-1}\cdot (1-1/4) + (1-p_{n-1})\cdot 1/3. $$ Let $s = \frac{1/3}{1/3+1/4}$ and note that $s=s\cdot (1-1/4)+(1-s)\cdot 1/3$. Then we get: $$p_n-s = (p_{n-1}-s)(1-1/3-1/4).$$ So, after taking absolute value on both sides and iterating: $$|p_n - s| = |p_1-s||1-1/3-1/4|^{n-1}.$$ For large $n$, noting that $|1-1/3-1/4|<1$, $p_n$ approaches fast $s$.