We have,
EXAMINATION
AAEIIMNNOTX
Given word has A′s→2,I′s→2,N′s→2 and rest other are distinct alphabets.
Total number of ways to form words using the alphabets of EXAMINATION is
=2!⋅2!⋅2!11!
Now, putting M at 4th position
.−−−M−−−−
Total words with M at fourth Place =2!2!2!10!
Required probability =11!10!=111