We focus our attention on counting length 6 sequences with available entries $\{0,1,2,\dots,9,X\}$.
Simply fix the far left entry of the sequence to be $X$. As there is only ever the one entry with an $X$, this will allow us to treat all rotationally equivalent arrangements the same.
$X~\underline{~}~\underline{~}~\underline{~}~\underline{~}~\underline{~}$
Now... let us ignore the $X$ again and look solely at the five remaining positions. If we were not interested in worrying about reflective symmetry we would find that there are simply $10^5$ possible sequences that can be filled in these remaining positions. As we are concerning ourselves with reflective symmetry however, more care needs to be applied.
We ask ourselves, how many sequences of 5 digits exist unique up to reflection? To count this, we might say that every sequence was counted differently twice: once as $abcde$ and again as $edcba$, so we would have $\frac{10^5}{2}$ such unique sequences however this ignores palindromes.
So, to correct the count, let us first take our $10^5$ sequences, remove the $10^3$ palindromes, divide the result in half to remove duplicates, and then add back in the palindromes.
This gives us a total of $(10^5-10^3)\cdot\frac{1}{2}+10^3=50500$ unique up to reflection sequences of length five. Appending an $X$ at the front of each of these sequences gives us a unique up to reflection and unique up to rotation bracelet.
Side note: Since we were given that every bracelet has exactly one bead $X$, this made things considerably easier for us. We were always able to rotate any bracelet until our $X$ bead was at the far left. Further, when considering reflections, we only ever needed to consider reflecting across the axis containing $X$ since reflection across any other axis would move the $X$ bead from the "left half" to the "right half" of the bracelet or vice versa.
As for what went wrong with your attempt, there were many leaps in logic that I did not understand so I cannot pinpoint everything that you did wrong. One thing that stood out to me was your section:
Then to generate unique bracelets from these necklaces, I have three options to place the X bead for each. Ex. Original Necklace = 00001
Unique Bracelets Generated = X00001, 0X0001, 00X001
If we were to generate a necklace ahead of time and try to place our X bead within it, some will have 3 options for where to place X as in your example giving 3 bracelets formed from this necklace, but many others will not.
For example if your original necklace was 00000 then every place where you would insert X would result in the same bracelet giving only one bracelet formed from this necklace.
For another example if your original necklace was 12345 then every place where you would insert X would result in a different bracelet giving six different bracelets formed using this necklace.
Because we are asked about probability, and not about the number of combinations, and because we are choosing with repetitions - we can ignore the identity of the individual beads and instead treat each choice as a $\frac{4}{20} = 20\%$ probability to choose yellow and $80\%$ probability to choose non-yellow.
For any given choice of the 5 places where yellow beads are chosen, the probability to choose it is $$
0.2^5\cdot 0.8^{25}
= \frac{2^5\cdot 8^{25}}{10^5\cdot 10^{25}}
= \frac{2^5\cdot 2^{75}}{10^5\cdot 10^{25}}
= \frac{2^{80}}{10^{30}}
$$
Since each such case is disjoint to the others, we can calculate how many ways there are to choose these 5 places and multiply: $$\boxed{
\frac{2^{80}}{10^{30}} \cdot \binom{30}{5}
= \frac{2^{80}}{10^{30}} \cdot \frac{30!}{5!\cdot 25!}
\approx 17.2279\%
}$$
For the second, the first and last places always have yellow beads so we only choose 3 places out of 28:
$$\boxed{
\frac{2^{80}}{10^{30}} \cdot \binom{28}{3}
= \frac{2^{80}}{10^{30}} \cdot \frac{28!}{3!\cdot 25!}
\approx 0.396\%
}$$
Best Answer
Paul Raff gave a formula for both bracelets and necklaces so in my answer, I will provide a general method that you can use for this kind of problem. It works also if you want to colour a cube for example.
As Paul Raff pointed out, you did get mix up between bracelet and necklace so in my answer I will include the answer for both of them.
Where did you get it wrong?
It seems to me that you are counting number of ways to colour a bracelet rather than a necklace so I checked your calculation with respect to colouring a bracelet. I would say that the main problem in your counting is that even for each case of the base, you cannot always guarantee to get the final result for that case by multiplying as you did. For example, in the case of $A^2B^2C^2$, we consider two following iterations:
The one on the left gives $\frac{6\cdot5\cdot 4}{3!}=\frac{120}{6}$ ways to choose three colours $A,B,C$, which is what you gave in your table. However, the one on the right gives $\frac{6 \cdot 5 \cdot 4}{2}=\frac{120}{2}$ ways to choose three colours $A,B,C$.
Necklace colouring
You can use Burnside lemma where you can count the number of ways to colour the object by looking at its group of symmetry $G$. For the necklace, the group $G$ can be:
Let $X$ be the set of all possible colouring for the necklace at a fixed orientation. This follows $|X|=6^6$ as there are $6$ possible colours for each bead.
Now, in Burnside lemma, we essentially want to count number of colourings from $X$ that remains unchanged under actions from $G$. In particular:
The same question can be asked for $120^{\circ}$ rotation: This happens when three of the beads (each is one bead away from the other) have the same colour and the remaining three beads have the same colour. There are $6$ possible colours for the first three beads and there are $6$ other possible colours for remaining $6$ beads. This gives us $6^2$ possible colourings.
Similarly, with $180^{\circ}$ rotation, a colouring is fixed under this rotation when any pair of opposite beads have the same colour. There are $6$ ways to colour each pair of opposite beads so there are $6^3$ possible colourings.
With $240^{\circ}$ rotation, it's the same as $120^{\circ}$ so we have $6^2$ possible colourings. With $300^{\circ}$, it's the same as $60^{\circ}$ so we have $6$ possible colourings.
With the "do nothing" action then every colouring remains unchanged after this action so there are $6^6$ possible colourings.
The Burnside lemma says that you can add all these numbers up and divide by number of elements of $G$ (which is $6$) to obtain all possible colourings. Hence, the answer for colouring a necklace is $$\frac{6^6 \cdot 1 +6 \cdot 2+ 6^2 \cdot 2+6^3 \cdot 1}{6}=7826.$$
Bracelet colouring
The difference between bracelets and necklaces is in the group of symmetry $G$. In particular, for bracelets, $G$ has some extra elements: Two colourings of the bracelet are considered same if from one colouring, we can reflect the bracelet through a line to obtain the other colouring. There are two types of lines:
This time $G$ has $12$ elements.
Next, we do the same thing with necklace, i.e. we count number of colourings that remains fixed under these reflections:
Now, applying Burnside's lemma, we sum up all these numbers counted for each element in $G$ then we divide by number of elements in $G$ (which is $12$). The final answer is $$\frac{1}{12}\left( 6^6 \cdot 1+6 \cdot 2+ 6^2 \cdot 2+6^3 \cdot 1+6^3 \cdot 3+ 6^4 \cdot 3 \right)=4291.$$