Computing FOLLOW sets for a LL(1) grammar
The follow set of something translates to asking: what symbol can come next in the string we are trying to read with the grammar rules the machine has?

Let's say we take a string "abba". The input tape symbol will end the string with $ as the delimiter. So we can say that 'a', 'b', 'b', 'a' and '$' are the symbols that the read/write head of the parser will see while evaluating this string. I am taking the below production rules for this grammar.

S-&gt;AB
A-&gt;ab|a
B-&gt;baB|E


The string "abba" can be evaluated in the following manner:

S-&gt;AB
 -&gt;abB
 -&gt;abbaB
 -&gt;abbaE


In the above case, Follow(A) = {b,$}.

'b' is part of the set because B's production follows A's production, S-&gt;AB, and whatever is the first letter that B produces will follow the last letter A produces.

Similarly, $ is part of the set because of B's epsilon production. 
When B is evaluated using rule B-&gt;E, then for Follow(A) there is no first letter produced by B.

In such a case where the Follow(A) is an epsilon production,

S-&gt;AE | using B-&gt;E


Then Follow(A) becomes Follow(S) which is $. 

If you look at the string "abba", once the last a is read the only symbol that can come after the string is $. 

As a generalization, if X-&gt;ABC and C-&gt;E, then Follow(B) = Follow(X).

In your question's example, evaluating Follow(A') results in only two possibilities where A' is on the RHS and can have a following symbol. Below are the two productions that can result:

A-&gt;bA'
A'-&gt;aA'


Unfortunately, out of the two one is Follow(A') = Follow(A') as per the rule I have mentioned in last line of the answer.

Hence we can only consider the other follow which is

Follow(A') = Follow(A)


i.e.

Follow(A) = First(B)


In a practical sense, if we have 

S-&gt;ABC


and 

A-&gt; bA'
A'-&gt;aA'|E


Then whenever we have an A production, it will further lead to either A -&gt; baA' or A-&gt;b (if we take A'-&gt;E).

In both cases the only thing that can follow A' is whatever is the first of B. 

In a similar manner we can compute the same for Follow(B') and Follow(C').