Probability of word LOLLIES Consider this probability question.
"A four letter 'word' is chosen randomly from the letters of the word LOLLIES. What is the probability of this word containing exactly one L?"
Now computing the total number of 4 letters words is straightforward by considering cases.
Case #1: The word contains 3 L's
Case #2: The word contains 2 L's
Case #3: The word contains no repetition. 
However, I feel that as tempting as it may be to do the following, it is not correct.
P(Exactly one L) = (Number of words with exactly one L)/(Total number of words).
Because the events are no longer equally likely depending on the case.
So how would you go about doing this problem? My gut feeling is telling me that it is far simpler than initially perceived using combinations. But could this please be confirmed?
 A: Calculate the total number of different $4$-letter words:


*

*Number of words with L-count equal $0$ is $\dfrac{4!}{0!}\cdot{\dbinom{4}{4}}=24$

*Number of words with L-count equal $1$ is $\dfrac{4!}{1!}\cdot{\dbinom{4}{3}}=96$

*Number of words with L-count equal $2$ is $\dfrac{4!}{2!}\cdot{\dbinom{4}{2}}=72$

*Number of words with L-count equal $3$ is $\dfrac{4!}{3!}\cdot{\dbinom{4}{1}}=16$


So the probability of choosing a word containing exactly one L is $\dfrac{96}{24+96+72+16}=\dfrac{6}{13}$

The solution above is under the assumption that the order of the letters in the word matters.
A: Let $X$ denote the number of L's in the chosen word. Then $$P\left\{ X=k\right\} =\binom{3}{k}\binom{4}{4-k}\binom{7}{4}^{-1}$$
This under the assumption that a specific letter cannot be chosen twice and that any letter has the same probability to be chosen. Here the $3$ L's are looked at as distinct letters.  
The $3$ corresponds with the number of L's in LOLLIES and the upper $4$ with the number of non-L's. The lower $4$ corresponds with the number of letters that are chosen, and $7$ with the total number of letters in LOLLIES.
For special case $k=1$ this leads to $P(X=1)=\frac{12}{35}$
A: I think you can find the probability of exactly one L like this:
Let's take the case that the first letter was an L. There is a $\frac{3}{7}$ probability of this. (Of $7$ possible letters, $3$ are Ls.) Then there is a $\frac{4}{6}$ probability that the second letter is not L, $\frac{3}{5}$ for the third, and $\frac{2}{4}$ for the fourth. The probability that only the first letter is an L is therefore $\frac{3}{7}\cdot\frac{4}{6}\cdot\frac{3}{5}\cdot\frac{2}{4} = \frac{3}{35}$.
It shouldn't be difficult to show that there is a $\frac{3}{35}$ probability for the L to be the second, third, or fourth letter instead.
So the total probability that there is exactly one L is $\frac{3}{35} + \frac{3}{35} + \frac{3}{35} + \frac{3}{35} = \frac{12}{35}$.
