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(D) The word $EXAMINATION$ contains $11$ letters: $A, A, E, I, I, M, M, N, N, O, T$.The total number of arrangements of these $11$ letters is given by

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$\frac{1}{11}$

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(D) The word $EXAMINATION$ contains $11$ letters: $A, A, E, I, I, M, M, N, N, O, T$.The total number of arrangements of these $11$ letters is given by $n(S) = \frac{11!}{2! 2! 2!}$,where $2!$ accounts for the repetitions of $A, I, M,$ and $N$.To find the number of arrangements where $M$ is fixed at the fourth position,we fix one $M$ at the fourth spot and arrange the remaining $10$ letters $(A, A, E, I, I, M, N, N, O, T)$.The number of such arrangements is $n(A) = \frac{10!}{2! 2! 2!}$.The probability is given by $P(A) = \frac{n(A)}{n(S)} = \frac{\frac{10!}{2! 2! 2!}}{\frac{11!}{2! 2! 2!}} = \frac{10!}{11!} = \frac{1}{11}$.

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:

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