A vowel is followed by a consonant 80% of the time and a consonant is followed by a vowel 50% of the time. Proportion between cs and vs?

Question

A text is such that the probability that a vowel is followed by a consonant is 80% and a consonant is followed by a vowel 50% of the time. What is the proportion of vowels in the text?

I don't know how to answer this without knowing how many letters there are in the text.

I can have vowels in two ways: either a vowel is followed by a vowel, or a consonant is followed by a vowel. The probability of the former is .2 and the probability of the latter is .5. If I do .5 + .2 I get .7, which isn't isn't any of the given options.

The answer is 5/13. Help?

Answer 1

To reiterate what I said in the comments, this problem is about Markov Chains, and more specifically about "Regular" Markov Chains. Full details should be included in the textbook you are using or in your class notes and it is beyond the scope of this site to provide a full and complete overview on a topic like this. Including all necessary definitions and proofs would generally take up a whole chapter in a textbook.

The final results to mention without proof are that given a Regular Markov Chain like this, regardless the initial state we will after enough time approach the "steady state," a scenario where further transitions do not continue to change the state. That is, the proportions of consonants and vowels will have reached a sort of equilibrium point.

That is, given a stochastic matrix $A$ and any initial state vector $v$ we have some steady state vector $\bar{v}$ such that $A\bar{v}=\bar{v} = \lim\limits_{n\to\infty}A^n v$ and further $\lim\limits_{n\to\infty} A^n$ has every column equal to $\bar{v}$.

Finding $\bar{v}$ is a matter of solving the matrix equation $A\bar{v}=\bar{v}$. Subtracting the one side to the other and including the identity matrix so we can factor yields $(A-I)\bar{v}=0$. In other words, $\bar{v}$ is in the kernel of $A-I$, so we can use standard Linear Algebra techniques to continue.

Alternately worded, we can avoid phrasing things in terms of matrices and solve the corresponding system of equations. Here, letting $v$ be the proportion of letters which are vowels and $c$ the proportion of letters which are consonants, we have the following system of equations:

$$\begin{cases} v = .2v + .5c\\c = .8v+.5c\\v+c=1\end{cases}$$

Note how the first two lines were crafted from the given information in the problem statement. Note also how the first two lines both convey the same information... if you were to move things to their own side, both lines are equivalent to $.8v = .5c$. Finally, note the final line that $v+c=1$. This last line comes about since we are discussing proportions or probabilities, or however you wish to describe things.

So, solving this by whatever method you choose, you will arrive at $v=\frac{5}{13}$

By substitution, $.8v=.5c$ so $1.6v = c$ and $v+c=1$ we have $v+1.6v=2.6v=1$ and so $v=\frac{1}{2.6}=\frac{10}{26}=\frac{5}{13}$

Answer 2

This is a Regular Markov Chain problem. We will first write the problem in a stochastic matrix $A$ that describes the transitions in the Markov chain. This may be done as follows: $$\begin{equation*} A = \begin{pmatrix} .5 & .5 \\ .8 & .2 \\ \end{pmatrix} \end{equation*}$$ where $A_{00}$ represents the case where a consonant is followed by a consonant, $A_{01}$ the case where a consonant is followed by a vowel, $A_{10}$ the case where a vowel is followed by a consonant, and $A_{11}$ the case where a vowel is followed by a vowel. Obviously, the rows should always sum to one.

The question is asking you to find a steady state distribution, where $v = vA$. We solve this as follows: $$ \begin{equation*} \begin{pmatrix} c & v \\ \end{pmatrix}= \begin{pmatrix} c & v \\ \end{pmatrix} \begin{pmatrix} .5 & .5 \\ .8 & .2 \\ \end{pmatrix} \end{equation*} $$ and from this we can get $c = 0.5c + 0.8v$ and $v = 0.5c + 0.2v$. Obviously, $c + v = 1$, so we can solve the linear system of equations to obtain $$ \begin{equation*} \begin{pmatrix} c & v \\ \end{pmatrix} = \begin{pmatrix} \frac{8}{13} & \frac{5}{13} \\ \end{pmatrix} \end{equation*} $$

Hence, the proportion of vowels in the text is $\frac{5}{13}$.