Here's the issue: when we speak of a "random number from $0$ to $1$," for instance, we are usually speaking of a uniform probability distribution, which means that "every number has an equal probability of getting picked." That's actually a very weird thing to say, because any particular number has a $0$ probability of being picked, so we have to go back and explore what it means to have a random number from $0$ to $1$.
If I have a die, it has six sides, each with $\frac{1}{6}$ probability of getting picked. A probability density function $p(x)$ tells us that the probability of the result of the die roll being $n$ is $p(n)$. Obviously the die has to give some result with a $100\%$ ($p=1$) probability. So,
$$\sum_{i=1}^{6}p(x_i)=1$$
if there are $6$ possible results and $x_1=1$, $x_2=2$, $x_3=3$ etc. If $n$ is not a possible result, $p(n)= 0$. The natural generalization of this is to continuous random variables, and you probably know that the generalization of discrete Riemann sums is integrals, so we are forced to take
$$\int_{-\infty}^{\infty}p(x)\ dx = 1$$
A uniform probability distribution is one where the probability density function $p(x)$ is just a constant value (the same way the probability density function for a die is always $\frac{1}{6}$ for any number from $1$ through $6$). We can always find a uniform probability distribution over finite intervals. For instance, for $[1,3]$, we must have $p(x)=C$ such that
$$\int_1^3 C\ dx = 1 \Rightarrow 3C-C=1 \Rightarrow C=\frac{1}{2}$$
But, for infinite intervals, this will give us an indeterminate value. If we want a uniform probability distribution over the reals, we need
$$\int_{-\infty}^{\infty}C \ dx = 1$$
for some real $C$. But, you see,
$$\int_{-\infty}^{\infty}C \ dx = \lim_{b \to \infty}\int_{-b}^{b}C\ dx = \lim_{b\to \infty} 2Cb$$
which is an indeterminate limit! There's no way $2Cb=1$ for any $C$! The same problem presents if we attempt to find a uniform probability distribution over $\mathbb{R}^2$, which is what's being assumed in this problem. We need
$$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}C\ dx=1$$
but writing out the limits gives us
$$\lim_{b\to \infty}\int_{-b}^{b}\int_{-b}^{b}C\ dx = \lim_{b\to \infty} 4Cb^2$$
which is also indeterminate, so it can never be equal to $1$ for any real $C$!! That's why this problem is malformed, as the distribution you would default to doesn't exist, and can't be answered unless you ask for the real probability distribution over the plane.
