In the word "MISSISSIPPI",

 Number of M's = 1 

Number of I's = 4 

Number of S's = 4 

Number of P's = 2 

Total number of letters = 11 

So total number of permutations = 11! / (1! 4! 4! 2!) = 34650

 Now, the total number of permutations taking all the letters except the S's = 7! / (1! 4! 2!) = 105

 After these 7 letters are permuted in a row, a total of 8 places (6 in between positions 2 end positions) are obtained, in which remaining 4 S's can be placed so that no two S's occur together. 

The number of ways in which 4 S's can be placed in 8 positions is 8C4 = 70.

 So the number of permutations taking all the letters so that no two S's occur together is = 105 x 70 = 7350 Therefore, the required probability = 7350 / 34650 = 7 / 33 

Note: Calculations are done instantly to minimize typing. We should do the calculations at the last to avoid large multiplications and divisions.