I am studying for tomorrows exam and I have checked many tutorials telling how to convert NFA to Regex but I can't seem to confirm my answers. Following the tutorials, I solved that NFA

My solution was:
aba
Am I correct?
I am studying for tomorrows exam and I have checked many tutorials telling how to convert NFA to Regex but I can't seem to confirm my answers. Following the tutorials, I solved that NFA

My solution was:
aba
Am I correct?
Your answer a*ba* is Correct. I can drive your answer from NFA in given image as follows:
There is a self loop on start state q0 with label a. So there can be any number of as are possible at initial (prefix) including null ^ in RE. So Regular Expression(RE) start with a*.
You need only one b to reach to final state. Actually for an accepting string; there must be at-least one b in string of a and b. So RE a*b to reach to either q1 or q2. Both are final states.
Once you reach to a final state (q1 or q2). No other b is possible in string (there is no outgoing edge for b from q1 and q2).
Only symbol is a can be possible at q1 and q2. Also for, a at q1 or at q2 move switch between q1 , q2 and both are final. So after symbol b any number of as can be in suffix. (So string ends with a* ).
And RE is a*ba*.
Also, its DFA is as follows:
DFA:
======
a- a-
|| ||
▼| ▼|
--►(q0)---b---►((q1))
a* b a* :RE
====
Any number of as at q0 that is: a*
once you get b you can switch to final state q1: b
at final state any number of a is possible: a*
And its a Minimized DFA!
Here is some more interesting answer by me on FAs and REs, I believe will be useful to you:
That answer is correct, in that both of the following are true:
However, I can't check your work because you haven't posted any.