To solve this problem, we need to find the probability that all identical alphabets are found together when the alphabets of the word "ALLAHABAD" are arranged at random.

Let's start by understanding the given word:

Word: ALLAHABAD

Number of alphabets in the word: 8
Number of distinct alphabets: 5 (A, L, H, B, D)
Number of identical alphabets: 3 (A, L, A)

Now, let's consider the identical alphabets as one group. So, we have 4 groups in total: {A, L, A}, {L}, {H}, {B, A, D}

Finding the total number of arrangements:
To find the total number of arrangements, we treat each group as a distinct entity. So, we have 4 entities to arrange. The number of arrangements of these entities can be found using the concept of permutation.

Total arrangements = 4!

Finding the number of favorable arrangements:
In the favorable arrangements, all identical alphabets are together. So, we treat the group of identical alphabets as one entity. We now have 3 entities to arrange: {A, L, A}, {H}, {B, D}

Number of arrangements of the entities = 3!

However, within the group of identical alphabets, the two 'A's can be arranged in 2! ways.

Number of arrangements within the group = 2!

So, the number of favorable arrangements = (3!)(2!)

Finding the probability:
Probability = Number of favorable arrangements / Total arrangements

Probability = (3!)(2!) / 4!
= (3 * 2)(2) / (4 * 3 * 2 * 1)
= 12 / 24
= 1 / 2

However, we need to consider that the two 'A's can be arranged in two different ways: AA or AA. So, we multiply the probability by 2.

Final probability = 2 * (1 / 2)
= 1

Therefore, the probability that all identical alphabets are found together in the randomly arranged word "ALLAHABAD" is 1/63, which is option A.

(AAAA), (LL), HBD