Difference between Conditional Probability and Bayes Theorem To ask my question I will present an example. Suppose we have $2$ boxes with yellow and blue balls. Box 1 has $5$ yellow and $4$ blue balls and Box 2 has $7$ yellow and $5$ blue balls.
We draw a ball at random from one of the boxes ($50-50$ chance) and the ball is yellow. What is the probability of the ball originating from Box 1?
To solve this, I used conditional probability which I understand is wrong and the correct answer would be to use Bayes theorem, which I didn't even know about. However because the chance is $50-50$ for either box the results were very similar.
My question is, what is the difference between the two formulas?
 A: First I explain why there is no real difference between both, then I tackle your example.
There is no fundamental difference between conditional probability and Bayes' theorem. Actually, Bayes' theorem is a simple rewrite of the definition of conditional probability.
Let $A$ and $B$ be two events. 
The conditional probability of event $A$ knowing $B$ occurred is:
$$P(A \mid B) = \frac{P(A \cap B)}{P(B)}$$
Basically this formula restricts the sample space to what happens within event B, and normalize the new probability measure by dividing by $P(B)$ to ensure the sum is $1$ as it should be.
If you manipulate this expression, you find:
$$P(A \cap B) = P(A\mid B)\,P(B)$$
Of course, by symmetry, replace $A$ by $B$ and you get:
$$P(B \cap A) = P(B\mid A)\,P(A)$$
Now, Bayes' theorem simply merges these two expressions in a different form. In a way, it's not really a "theorem" because its proof is trivial. But its implication for statistics are so wide that it's called "theorem"
Bayes theorem:
$$P(A \mid B) = \frac{P(B \cap A)}{P(B)} = \frac{P(B \mid A)\,P(A)}{P(B)}$$
Asking the difference between Bayes' theorem and conditional probability is like asking the difference between these two equations:
$$ x = \frac{a}{b} \quad \text{ and } \quad b\times x = a $$
Hope this helps

Edit:
to tackle your example:
Let $B_1$ the event "the ball is drawn from box 1" and $B_2$ the event "the ball is drawn from box 2". We know that $P(B_1) = P(B_2) = 0.5$.
In box 1, there are 5 yellow and 4 blue balls so if $Y$ is the event "we draw a yellow ball", then $P(Y \mid B_1) = 5 / (4+5) = 5/9$.
Likewise for box 2, $P(Y \mid B_2) = 7 / (7 + 5) = 7 / 12$
With this setup, we have:
$$P(Y) = P(Y \mid B_1)\,P(B_1) + P(Y \mid B_2)\,P(B_2)$$
$$P(Y) = 5/9 * 0.5 + 7/12 * 0.5 \approx 0.56944444$$
This formula I used is called "law of total probability". It can be used when two (or more) events partition the sample space $\Omega$. That is: $B_1 \cap B_2 = \emptyset$ and $B_1 \cup B_2 = \Omega$. 
Now, you can compute $P(B_1 \mid Y)$ in a similar fashion:
$$P(B_1 \mid Y) = \frac{P(B_1 \cap Y)}{P(Y)} = \frac{P(Y \mid B_1)\,P(B_1)}{P(Y)} = \frac{5/9 * 0.5}{5/9 * 0.5 + 7/12 * 0.5} = 0.4878...$$
