"What is the probability the word contains "H" once?"
This is slightly ambiguous. Do you mean the word i) contains at least one H or ii) contains exactly one H?
i) has a trick. There are are $29^5$ total words. There are $28^5$ words that contain no H's whatsoever (-- each letter can be any one of the remaining 28 letters). So there are $29^5 - 28^5$ words that have one or more Hs.
ii) there is a naive way and a sophisticated way. First the naive way: There are $1*28*28*28*28$ ways to type a five letter word a five letter word starting with H and the rest of the letters are not H. There are $28*1*28*28*28$ ways to type a word where the 2nd letter is H but the rest are not H. There are $28*28*1*28*28$ where the third letter is H and so on.
So in total there are $1*28*28*28*28 + 28*1*28*28*28 + ..... + 28*28*28*28*1 = 5*28^4$.
... or the sophisticated way: The number of ways to type a 5-letter word where a specific letter slot must be H and the remaining 4 letter slots must not be H is $1$ for the dedicated letter slot and $28^4$ for the remaining 4 letters slot (each of which can be any of the 28 remaining letters remaining). That is $28^4$ possible ways. Now there are 5 possible choices for which dedicated letter slot is the H. That means there are $5*28^4$ possible words. (5 choices for which letter is the H-- and 28 choices for each of the remaining four letters. Multiply the choices... $5*28^4$.
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" 2*2/3*3. I don't see the connection to the 5 in the answer"
If you need to figure out $n$ letter word with $m$ choices of letters the number of words with exactly one ~ is $n*(m-1)^{n-1}$. So the answer with 3 letters and a 3 letter word is $3*(2)^{2} = 12$ [ABB,ABC, ACB,ACC, BAB,BAC, CAB,CAC, BBA,BCA,CBA,CCA].
One 2 corresponds the 28. The other to the 4. The 3 corresponds to the 5.
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Two post-scripts.
I just realized I gave the answers to "What is the NUMBER OF WAYS" rather than "What is the probability".
Well, Probability of Event = (Number of Ways Event can Happen)/(Number of total things that can happen).
So for i) Number of ways to get at least one H = $29^5 - 28^5$. Number of total things that can happen = $29^5$.
So probability is $(29^5 - 28^5)/29^5 = 1 - (28/29)^5$.
for ii) Number of ways to get exactly one H = $5*28^4$. Number of total things that can happen = $29^5$ so probability is $5*28^4/29^5$.
For your 3 letters and a three letter word: Number of ways to get exactly one A = $3*2^2 = 12$. Total ways to type three letters = $3^3 = 27$.
So probability = $3*2^2/3^3 = 2^2/3^2$. Note: because you had 3 letters total and 3 letter length words one of the 3s cancelled. Which is probably why you didn't see the corespondence.
If you had say a four letter word of ABC with exactly one A then the total ways would be $4*2^3/3^4$. (Do you see why) and no canceling. That you chose the same number of letters and length of words was a misleading coincidence.
2nd postcript as per aarons comment.... well, I have to go cook dinner now. I'll get to it later but... It's to introduce the idea of choicing m out of n options. i.e. how to choice which position the one H is. Or if you had to calculate the probability of a five letter word with exactly TWO Hs--- how to choice which of the 5 positions we can place the 2 Hs....
Think about it. I'll get back to it.
For a concrete example, let's use your example, the string MAMA, typed on a $26$=key keyboard.
As you say, the chance that the first four letters typed by one monkey will spell MAMA is $\left(\frac1{26}\right)^4 \approx 2.2\times 10^{-6}.$
The chance that the monkey does not type MAMA is
$1 - \left(\frac1{26}\right)^4 \approx 0.9999978.$
But now let's put a million monkeys in front of typewriters and examine the first four letters that each one of them types.
When we look at the first four letters typed by any single monkey, there is still only a $2.2\times 10^{-6}$ chance that those four letters spell MAMA.
But the chance that at least one of the million monkeys typed MAMA is one minus the probability that ever single monkey typed something else.
The probability that any single monkey typed something else is still
$1 - \left(\frac1{26}\right)^4 \approx 0.9999978,$
but for all one million of them to type things other than MAMA, the probability is
$$ \left(1 - \left(\frac1{26}\right)^4\right)^{1000000} \approx 0.112$$
and therefore the probability that at least one will type MAMA is
approximately $0.888.$
If you increase the number of monkeys to ten million, the probability they all type something other than MAMA is
$$ \left(1 - \left(\frac1{26}\right)^4\right)^{10000000} \approx 3\times10^{-10}$$
and therefore the probability of at least one MAMA is
approximately $0.9999999997.$
Often the problem is formulated so that you not only have an extraordinary number of monkeys (if not infinite) but they keep on typing forever and you only need to find MAMA somewhere in one of the monkey's output, not necessarily in the first four letters.
This probability converges to $1$ very quickly.
If you choose to look for a much longer string, such as the entire works of Shakespeare, the probability of observing it from a finite number of monkeys will be less for the same number of monkeys and the same amount of time,
but it too eventually converges to $1.$
In any event, the typical formulation has nothing to do with assembling the string from the output of multiple monkeys. You have to find the string in the output of just one monkey. The trick is that you get to throw away and ignore the output of all the other monkeys who typed something else.
Best Answer
The 2 examples you made (the bible and the sequence of the number $\frac{1}{3}$) are similar, but deeply different. In fact, the bible is a finite sequence of characters, whereas the decimal digits of $\frac{1}{3}$ are infinitely many.
So, the probability that the monkey writes exactly the bible during its typing is 1, while the probability that the monkey will write exactly all the digits of $\frac{1}{3}$ is 0.
Digits of $\frac{1}{3}$
In order to write exactly the digits of $\frac{1}{3}$, from a certain digits the monkey must push the key 0 and then the key 3 infinitely many times... up to infinity! This has probability (not rigorous) $\frac{1}{10} * \frac{1}{10} * \dots * \frac{1}{10} = (\frac{1}{10})^\infty = 0$.
Sequence of letters of the bible
In order to write exactly the sequence of letters of the bible, instead, the monkey should guess a finite number of characters.
The first thing we can notice is that the probability that this happens is larger than 0. In fact, assuming that the bible is 1.000.000 letters long, the probability that the bible starts from the first letter of the first page that the monkey writes is something like $p = \frac{1}{26} * \frac{1}{26} * \dots * \frac{1}{26} = (\frac{1}{26})^{1.000.000}$, which is a very small number, but it is larger than 0.
Then, think that the bible could start from the second letter of the first page, so we have another $p$ probability that the monkey will write exactly the bible. And consider that the monkeys has infinitely many letters from which it can start to write the bible, so it has infinitely many chances to succeed in the challenge, each of them having $p$ probability to work.
Note that these infinitely many events (each corresponding to which letter the monkey is starting to write the bible from) are not independent, so the total probability will not be $\infty * p = \infty$, but they are enough to guarantee that the monkey will achieve it anyway, with probability 1.
The rigorous proof that this probability is 1 is not even difficult, but I will not write it here. You can find it on Wikipedia.
The idea of the proof is to estimate the probability that the monkey will not write the bible and eventually you can proof that that probability is 0, meaning that it is almost impossible (but still not impossible) that the monkey doesn't write the bible.