Probability that $a+b+c+d = 1.5$ if $a+2b+3c+4d = 3.5$ Let $(x_1, x_2, x_3, x_4) \in [0,1]^4$ be a vector chosen uniforly at random.  What is the probability that we observe?
\begin{eqnarray*} x_1 + 2 x_2  + 3 x_3  + 4 x_4 &=& \;\;\,3.5 \pm 0.01\\
x_1 + \;\,x_2 +\;\, x_3 +\;\, x_4 &=& 1.5 \pm 0.01 \end{eqnarray*}
The $\pm 1$ is to ensure a 4 dimensional region.  Then the probability is also a volume.

This problem could be stated over integers: Let $(x_1, x_2, x_3, x_4) \in \{ 1, 2, \dots, 100\}^4$ be chosen randomly.  What is the probability that: 
\begin{eqnarray*} x_1 + 2 x_2  + 3 x_3  + 4 x_4 &=& 350 \pm 1\\
x_1 + \;\,x_2 +\;\, x_3 +\;\, x_4 &=& 150 \pm 1\end{eqnarray*}
The $\pm 1$ is not necessary since there are exactly $10^6$ possiblities (1 million) 

A simpler problem of this kind:

AIME 1998/10 Two mathematicians take a morning coffee break each day. They arrive at the cafeteria independently, at random times between 9 a.m. and 10 a.m., and stay for exactly $m$ mintues. The probability that either one arrives while the other is in the cafeteria is $40 \%,$ and $m=a-b\sqrt{c},$ where $a, b,$ and $c$ are positive integers, and $c$ is not divisible by the square of any prime. Find $a+b+c.$

A close reading of AIME problems (and beyond) could invite more solutions of this kind.
 A: Let $u_1, u_2, u_3, u_4$ be four distinct positive numbers.
For any $p, q \in \mathbb{R}$ and $\epsilon > 0$, let $V(p,q), V_\epsilon(p,q)$ be the region
$$\begin{align}
\Omega(p,q) &= \left\{ (x_1, x_2, x_3, x_4 ) \in [0,\infty)^4 : \sum_{k=1}^4 x_k \le p, \sum_{k=1}^4 u_k x_k \le q \right\}\\
\Omega_\epsilon(p,q) &= 
\left\{ (x_1, x_2, x_3, x_4 ) \in [0,\infty)^4 : \left| \sum_{k=1}^4 x_k - p \right| < \epsilon, \left|\sum_{k=1}^4 u_k x_k - q \right| < \epsilon\right\}
\end{align}
$$
Let $\Delta(p,q)$ and $\Delta_\epsilon(p,q)$ be their volume and $\rho(p,q) = \frac{\partial^2}{\partial p\partial q}\Delta(p,q)$.  
It is clear $\Delta(p,q)$ and $\rho(p,q)$ vanish unless $p, q > 0$.  
For any $\alpha, \beta \in \mathbb{C}$ with $\Re(\alpha + \beta u_k) > 0$ for all $k$, it is easy to see
$$\int e^{-(\alpha p + \beta q)}\rho(p,q) dp dq
= \int e^{-\sum_{k=1}^4 (\alpha + \beta u_k)x_k}\, d^4 x
= \frac{1}{\prod\limits_{k=1}^4(\alpha + \beta u_k)}
$$
This is essentially the Laplace transform of $\rho(p,q)$. Apply inverse transform, one can show that for $p, q > 0$,
$$\rho(p,q) = \sum_{k=1}^4 \frac{(q - pu_k)_{+}^2}{2!}\prod_{j=1,\ne k}^4 \frac{1}{u_j - u_k}
\quad\text{ where }\quad
(u)_{+}^2 = \begin{cases}
u^2, & u > 0\\
0,   & \text{otherwise}
\end{cases}
$$
When $\epsilon > 0$ is small enough such that the square
$[p-\epsilon,p+\epsilon] \times [q-\epsilon,q+\epsilon]$ 
lies completely in first quadrant and disjoint from those lines $q' - p'u_k = 0$ which doesn't pass through $(p,q)$. We can integrate the expression above to
get 
$$\begin{align}
\Delta_\epsilon(p,q) 
&= \int_{-\epsilon}^{\epsilon}\int_{-\epsilon}^{\epsilon} \rho(p+x,q+y) dxdy\\
&= 2\epsilon^2\sum_{k=1}^4 
\left(\prod_{j=1,\ne k}^4 \frac{1}{u_j - u_k}\right)\times
\begin{cases}
(q - p u_k)^2 + \frac{\epsilon^2}{3}(1 + u_k^2), & q - p u_k > 0\\
\frac{\epsilon^2}{6}(1 + u_k^2), & q - p u_k = 0\\
0, & \text{ otherwise }.
\end{cases}
\end{align}
$$
Using inclusion-exclusion principle, we can express the volume
$\Omega_\epsilon(p,q) \cap [0,1]^4$ in terms of $\Delta_\epsilon(\cdots)$.
More precisely, we have:
$$\verb/Vol/(\Omega_\epsilon(p,q) \cap [0,1]^4)
= \sum_{(\eta_1,\eta_2,\eta_3,\eta_4) \in \{0,1\}^4}
(-1)^{\sum_{k=1}^4\eta_k}
\Delta_\epsilon\left(p - \sum_{k=1}^4 \eta_k, q - \sum_{k=1}^4 u_k\eta_k \right)
$$
For the problem at hand, we choose $(u_1,u_2,u_3,u_4) = (1,2,3,4)$ and $(p,q) = (\frac32,\frac72)$.  
After throwing away terms that won't contributed, the desired volume for small enough $\epsilon$ is
$$\Delta_\epsilon\left(\frac32,\frac72\right)-\sum_{k=1}^3\Delta_\epsilon\left(\frac32-1,\frac72-k\right)$$
With help of a CAS, this can be simplified to $\epsilon^2 - \frac{16}{9}\epsilon^4$.
When $\epsilon = \frac{1}{100}$, this evaluates to $\frac{703}{7031250} \approx 0.00009982222$.
As a double check, I have counted the number of integer solutions 
of following problem:
$$
(x_1,x_2,x_3,x_4) \in \{ 1,\ldots,100 \}^4
\quad\text{ and }\quad
\begin{cases} 
x_1 + 2 x_2  + 3 x_3  + 4 x_4 &= 350 \pm 1\\
x_1 + \;\,x_2 +\;\, x_3 +\;\, x_4 &= 150 \pm 1
\end{cases}
$$
If we approximate each integer choice on RHS by a box of side $0.01$ in $(p,q)$ space, above integer problem corresponds to $\epsilon = 0.015$ in our analysis. 
This means the number of solutions should be around $1.5^2 \times 0.0001 \times 10^8 \approx 22500$. This is reasonably close to the actual number of solutions, $23096$, obtained by brute force.
