Bayes theorem, confusion between $P(X|Z)$ and $P(Z|X)$ In a problem I'm trying to solve it states:

There is a machine displays the inserted credit. Unfortunately, the display is broken. At time $k = 3$, we can infer from the display that the inserted credit is either $2$ or $3$, but certainly not $0$ or $1$.
Without taking into account the additional information provided by the display, the state x(3) has the probability distribution

$f(x_3=0) = 0.1,$

$f(x_3=1) = 0.2,$

$f(x_3=2) = 0.2,$

$f(x_3=3) = 0.5.$

What is the probability distribution of $x(3)$ given the information from the display?

Now my reasoing was the following: the sentence we can infer from the display that the inserted credit is either $2$ or $3$, but certainly not $0$ or $1$ is equivalent to say: given the state of the display, we can say there is $50\%$ probability that the coin is $2$ and $50\%$ probability that the coin is $3$. 
So the probability distribution asked would be simply $50\%$ for $2$ and $50\%$ for $3$.

But from the solution of the exercise, it looks at the problem from another perspective. It states that if we can infer something from the display means that we can depict the conditional probability of $z$ given $x$ (and not the opposite as I thought) and then uses the prior information $f(x_3)$ and this likelihood ($f(z|x_3)$) with the bayes theorem to arrive to the solution. 

Why is my reasoning wrong?
You're applying the principle of indifference where it doesn't apply. There are two requirements for that principle to apply. First, there must be symmetry among the alternatives (or, as the Wikipedia article puts it, they must be &ldquo;indistinguishable except for their names&rdquo;). This is not the case here &ndash; the credit states $2$ and $3$ are distinguishable, as you can readily convince yourself by replacing $3$ with $1000$: You probably wouldn't believe that the credit states $2$ and $1000$ are equiprobable. The second requirement is the one you state, that we don't have further information. In the present case, we do have further information, namely, the given a priori distribution, which again distinguishes between the credit states $2$ and $3$. Thus the principle of indifference doesn't apply.
OK this is an interesting problem. So we know that inserted credit equals $2$ or $3$. Let us try to calculate the probability that the inserted credit is $3$ given that we know it is $2$ or $3$. I will just call the inserted credit $x$. We use the formula $P(A|B)=P(A \wedge B)/P(B)$. Where $A$ is $x=3$ and $B$ is $x=2 \vee x=3$. This becomes $P(x=3 | (x=2 \vee x=3))$. The fact that $A$ implies $B$ and so $P(A \wedge B)=P(A)$ is important to keep in mind. We can calculate this as $0.5/0.7=0.71$. By a similar method, we can work out that the probability that the display read $2$ is $0.2/0.7=0.29$. These add up to $1$ which is good news. It must be one of them! Another way to have come at this would have been to say the answer must be $2$ or $3$ so the probabilities of these scores must split in the ration $0.2:0.5$ or $2:5$ or $0.29:0.71$.