Words with or without meaning are to be formed using all the letters of the word EXAMINATION. The probability that the letter M appears at the fourth position in any such word is:

The word EXAMINATION has 11 letters: E, X, A, M, I, N, A, T, I, O, N — with A appearing twice, I appearing twice, and N appearing twice, and the remaining letters M, E, X, T, O each appearing once.

The total number of distinct arrangements of all 11 letters is $$\frac{11!}{2! \cdot 2! \cdot 2!}$$.

To count arrangements where M is fixed at the 4th position, we arrange the remaining 10 letters (E, X, A, I, N, A, T, I, O, N) in the remaining 10 positions. The number of such arrangements is $$\frac{10!}{2! \cdot 2! \cdot 2!}$$.

The required probability is:

$$P = \frac{\dfrac{10!}{2! \cdot 2! \cdot 2!}}{\dfrac{11!}{2! \cdot 2! \cdot 2!}} = \frac{10!}{11!} = \frac{1}{11}$$