Assume there is an Ideal Random Number Generator which generates any real number between 0 and given integer. Two numbers are generated from the above generator using integer A and B, let's assume the numbers generated are X1 and X2. There is another integer C. What is the probability that summation of X1 and X2 is less than C.
[Math] Probability that sum of 2 random numbers are less than an integer
probabilityrandom variables
Related Solutions
A picture to go with Did's excellent hint.
Yes, your answer is fundamentally wrong. Let me point at that it is not even right in the finite case. In particular, you are using the following false axiom:
If two sets of outcomes are equally large, they are equally probable.
However, this is wrong even if we have just two events. For a somewhat real life example, consider some random variable $X$ which is $1$ if I will get married exactly a year from today and which is $0$ otherwise. Now, clearly the sets $\{1\}$ and $\{0\}$ are equally large, each having one element. However, $0$ is far more likely than $1$, although they are both possible outcomes.
The point here is probability is not defined from cardinality. It is, in fact, a separate definition. The mathematical definition for probability goes something like this:
To discuss probability, we start with a set of possible outcomes. Then, we give a function $\mu$ which takes in a subset of the outcomes and tells us how likely they are.
One puts various conditions on $\mu$ to make sure it makes sense, but nowhere do we link it to cardinality. As an example, in the previous example with outcomes $0$ and $1$ which are not equally likely, one might have $\mu$ defined something like: $$\mu(\{\})=0$$ $$\mu(\{0\})=\frac{9999}{10000}$$ $$\mu(\{1\})=\frac{1}{10000}$$ $$\mu(\{0,1\})=1$$ which has nothing to do with the portion of the set of outcomes, which would be represented by the function $\mu'(S)=\frac{|S|}2$.
In general, your discussion of cardinality is correct, but it is irrelevant. Moreover, the conclusions you draw are inconsistent. The sets $(0,1)$ and $(0,\frac{1}2]$ and $(\frac{1}2,1)$ are pairwise equally large, so your reasoning says they are equally probable. However, the number was defined to be in $(0,1)$ so we're saying all the probabilities are $1$ - so we're saying that we're certain that the result will be in two disjoint intervals. This never happens, yet your method predicts that it always happens.
On a somewhat different note, but related in the big picture, you talk about "uncountably infinite sets" having the property that any non-trivial interval is also uncountable. This is true of $\mathbb R$, but not all uncountable subsets - like $(-\infty,-1]\cup \{0\} \cup [1,\infty)$ has that the interval $(-1,1)=\{0\}$ which is not uncountably infinite. Worse, not all uncountable sets have an intrinsic notion of ordering - how, for instance, do you order the set of subsets of natural numbers? The problem is not that there's no answer, but that there are many conflicting answers to that.
I think, maybe, the big thing to think about here is that sets really don't have a lot of structure. Mathematicians add more structure to sets, like probability measures $\mu$ or orders, and these fundamentally change their nature. Though bare sets have counterintuitive results with sets containing equally large copies of themselves, these don't necessarily translate when more structure is added.
Best Answer
Hint:
Draw the sample space on axes $X_1, X_2$. You should get a rectangle of area $AB$. Now shade the portion $X_1 + X_2 < C$ in this rectangle.
Now consider cases where $C \in \,$ one of the following intervals $(0 < \min(A, B)), [\min(A, B), \max(A, B)], (\max(A, B), A+B), (A+B, \infty)$...