There are two shooters with different guns and bullets. Each shooter shoots a bullet to a different target hanging on a wall. The hit of each bullet follows a normal distribution centered on its target. Each bullet removes a piece of the wall whose volume follows a normal distribution centered on ten times the volume of the bullet.
What is the probabilty that a volume $V_1$ was extracted in a point $X_1$ on the wall by the shooters (i.e., $P(V_1,X_1)$)?
What happens if we study two different points (i.e., $P(V_1, X_1 \ and \ V_2, X_2)$)?
Since your distributions are continuous densities, gaussian in this case, the probability of any point is 0. That is the answer to your questions as stated.
To get probabilities, you need to specify a region on $P(V_1,X_1)$ whose area is greater than zero. For example, lets say you want to know $P(V_1\pm\Delta V,X_1\pm\Delta X)$ Where $\Delta X, \Delta V >0$. What you want is the probability that either the bullet from shooter 1 or shooter 2 or both hit a location within $X\pm\Delta X$ and remove $V\pm\Delta V$ volume.
Let X be the modpoint of the interval where the wall will be hit by a bullet, and V be the midpoint of the volume interval for the piece removed. $B_i$ be the X-coordinate of bullet $i = 1,2$ $\mu_i$ be the X-coordinate of target $i$, $\sigma_{bi}$ be the standard deviation of the location of $B_i$,and $\sigma_{vi}$ the standard deviation of the volume removed. Then you need to get the probability for the folowing three situations:
(1) $P(|B_1-X|\leq\Delta X\cap|B_2-X|>\Delta X\cap |V_1-V|\leq \Delta V)$
(2) $P(|B_2-X|\leq\Delta X\cap|B_1-X|>\Delta X\cap |V_2-V|\leq \Delta V)$
(3) $P(|B_1-X|\leq\Delta X\cap|B_2-X|\leq\Delta X\cap |V_1+V_2-V|\leq \Delta V)$
You can now calculate each using the normal CDF function.