Say you have a word with 4 available slots, take the word "bake" as an example. You want to compute the probability of a letter (from the English alphabet) being in a slot of the word. So what's the probability that "b" will be in the first slot, "a" in the second slot, "k" in the third and "e" in the fourth slot. I need the individual probabilities of for a letter being in the correct slot of the word.
Every letter in the English alphabet is available, the repetition of a letter is limited to the desired word so, for the word bake since all letters are different, there is no repetition.
For some reason the first thing that came to mind was the binomial distribution, but I don't think that answers my question. Although I still do think I could use this probably as "the probability of getting all 4 letters of the word in the correct slots"
Is the probability of getting a letter in a particular slot just: $({1\over4}) $ $({1 \over 26}) = 0.009615$ ?
As for the binomial distribution, would this be correct:
$$P(X=4)= {4 \choose 4}{0.03846}^{4}(1-0.03846)^{0}$$
Where $p= {1\over26} = 0.03846$ and $n = 4, x=4$
Best Answer
There is a total of $4! = 24$ different permutations of
bakewhich you can generate, each occurs with equal frequency since all letters are different.You are asking what is a chance to actually end up with the word
bakeitself -- and it should be $1/24$ in total...