Real number probability

There are 9 possible ranges for the first number, with each having an equal probability:

0-0.5

0.5-1

1-1.5

1.5-2

2-2.5

2.5-3

3-3.5

3.5-4

4-4.5

4 of these ranges will make the sum of the two rounded numbers equal 5:

0.5-1

1.5-2

2.5-3

3.5-4

Therefore, the answer is $\boxed{\frac{4}{9}}$

Let $S(x):=\lfloor x\rceil+\lfloor 4.5-x\rceil$ where $\lfloor x\rceil$ is the result of rounding $x.$ We're effectively trying to find the probability that a randomly selected value $x$ on the range $0\leq x\leq4.5$ has $S(x)=5.$

One can easily see that this occurs $\tfrac49$ of the time by looking at the graph of $S(x)$ on the interval (as shown below); however, another way to do it is by looking at the conditions on which $S(x)$ is 5. We want both $x$ and $4.5-x$ to round up, which means that for each rounded value of $4.5-x,\ x$ must be in the upper-half of its permissible range. For example, if we consider the case where $S(x)=1+4=5,$ we want $4.5-x$ to round to $4,$ implying $0\leq x\leq 1$ , but then $x$ only rounds to $1$ if it's in the upper-half of that interval. This occurs four times (once for each value from $0$ to $4$ ), but because of the last value being cut in half, we remove one of the $5$ possible instances where the sum would be $5.$ Thus, $\tfrac49.$