= Solution
Take the spatial <Fourier transform> of the <wave equation>. The transformed initial-value problem is
$$
\widetilde u_{tt}+k^2\widetilde u=0,
\qquad
\widetilde u(k,0)=\widetilde f(k),
\qquad
\widetilde u_t(k,0)=\widetilde g(k),
$$
and hence
$$
\widetilde u(k,t)
=\widetilde f(k)\cos(kt)
+\widetilde g(k)\frac{\sin(kt)}k.
$$
Part (c) and the <convolution theorem> turn the first term into
$$
\frac12\bigl[f(x+t)+f(x-t)\bigr].
$$
For the second term, choose $p$ with $p'=g$. Part (b) says $\widetilde g=ik\widetilde p$, so
$$
\widetilde g(k)\frac{\sin(kt)}k
=i\widetilde p(k)\sin(kt).
$$
Part (c) now makes its inverse transform
$$
\frac12\bigl[p(x+t)-p(x-t)\bigr]
=\frac12\int_{x-t}^{x+t}g(\xi)\,d\xi.
$$
Combining the two terms gives the <D'Alembert formula>
$$
\boxed{
u(x,t)=\frac12\bigl[f(x+t)+f(x-t)\bigr]
+\frac12\int_{x-t}^{x+t}g(\xi)\,d\xi.
}
$$
Let $A=2V$ and $B=W-V$. Part (c) makes $A$ and $B$ independent identically distributed exponential variables of rate $\lambda$. The change of variables $(A,B)=(rs,(1-r)s)$ has Jacobian $s$, and integrating $\lambda^2se^{-\lambda s}$ over $s>0$ gives density one for $0<r<1$. Therefore the <uniform ratio of independent exponential variables> gives
$$
\boxed{\frac{2V}{W+V}=\frac A{A+B}\sim\operatorname{Uniform}(0,1).}
$$
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