Understanding the Problem
To find the probability that no two vowels in the word "AMBIGUITY" are together, we first identify the components of the word. The word "AMBIGUITY" consists of 9 letters, including 4 vowels (A, I, U, I) and 5 consonants (M, B, G, T, Y).

Step 1: Arranging the Consonants
- We begin by arranging the 5 consonants (M, B, G, T, Y).
- The number of arrangements for the consonants is:
\(5! = 120\)

Step 2: Placing the Vowels
- Next, we need to place the 4 vowels in such a way that no two vowels are adjacent.
- After arranging the consonants, we create gaps for the vowels. There are 6 possible gaps (before the first consonant, between consonants, and after the last consonant):
- _ C _ C _ C _ C _ C _
- We need to choose 4 out of these 6 gaps to place the vowels. The number of ways to choose 4 gaps from 6 is given by:
\( \binom{6}{4} = 15\)
- Since there are 4 vowels, which include a repetition of 'I', the arrangements of the vowels will be calculated as:
\( \frac{4!}{2!} = 12\)

Step 3: Total Arrangements
- Thus, the total arrangements where no two vowels are together is:
\(120 \times 15 \times 12 = 21600\)

Step 4: Total Arrangements of the Word
- The total arrangements of the word "AMBIGUITY" (considering the repeated 'I') is:
\( \frac{9!}{2!} = 181440\)

Step 5: Probability Calculation
- The probability that no two vowels are together is given by:
\[ P = \frac{21600}{181440} = \frac{5}{42} \]
The final probability that no two vowels are together in the word "AMBIGUITY" is:
\[ \frac{5}{42} \]