Let $C={\overline{B}}_1\cap B_2$. Under replacement its joint probability iswhereas without replacement it isIn the latter case the first removed ball is non-blue, so all $N_B$ blue balls remain. Applying Bayes' theorem and cancelling the positive common factors yields the posterior replacement probability after different-colour draws:It does not depend on $N_B$. To prove the requested bound directly,Equality holds at $r=0$ and $r=1$, and the inequality is strict between them. Its sketch joins $(0,0)$ to $(1,1)$, stays below the diagonal, and is increasing and convex becauseThe interpretation is that removing a non-blue ball increases the chance of blue on the second draw; seeing the two different colours therefore favours the no-replacement explanation.

Use the intended model of uniform draws and a replacement decision independent of the first ball's colour, with probability $r$. The marginal specification $P(R)=r$ alone would not determine the experiment if replacement depended on that colour. This independence assumption is used throughout the question.

Write ${\overline{B}}_j$ for a non-blue draw and $\overline{R}$ for no replacement. The sample space consists of the eight ordered outcomes obtained by choosing one of $B_1,{\overline{B}}_1$, then one of $R,\overline{R}$, then one of $B_2,{\overline{B}}_2$. The chain rule for probabilities gives their probabilities:For example, the first entry is $(2/5)r(2/5)$. The fifth is $(2/5)(1-r)(1/4)$ because one blue ball has been removed. With no replacement after a non-blue first draw, the two remaining colours each have two balls, explaining the last two entries. The first four entries sum to $r$ and the last four to $1-r$, so the full sample space has total probability one. At $r=0$ or $1$, the corresponding impossible outcomes simply have zero probability.

The finite form of Taylor's theorem iswhere $\xi$ lies between $a$ and $x$ when $x\ne a$. Sufficient hypotheses are continuous derivatives up to order $n-1$ on the interval joining the endpoints and an $n$th derivative in its interior. The first $n$ terms end at degree $n-1$; the Taylor remainder uses the $n$th derivative. At $x=a$ the remainder is zero. This finite Taylor polynomial identity does not require convergence of an infinite Taylor series.

Work on an open domain with $xy>0$ and $x\ne 0$, where the displayed square root is real and differentiable. Split $u=v+h$ with $v=\varphi (xy)$ and $h=\sqrt{xy}\psi (y/x)$. For the first term, the chain rule givesso $x^2{v}_{xx}-y^2{v}_{yy}=0$.

For the second term, set $t=y/x$. Its first partial derivatives areThey show $x{h}_x+y{h}_y=h$, the degree-one case of Euler theorem for homogeneous functions. Differentiating that identity with respect to each variable yieldsThe mixed partial derivatives agree here by the chain rule. Multiplying these equations by $x$ and $y$, respectively, proves the second derivative identity for a degree-one homogeneous function:Adding the results for $v$ and $h$ verifies the partial differential equationEquivalently, $h(\lambda x,\lambda y)=\lambda h(x,y)$ for positive $\lambda$, so the same proof follows directly from its homogeneous function property. The statement is local to a domain where the given expression is defined; it does not require assigning values on $x=0$.

Write the scalar field as $u=R(r)Y(\theta ,\varphi )$, where $R=r+r^{-2}$ and $Y=\sin \theta \cos \varphi$. Its radial contribution to the Laplacian in spherical coordinates isFor the angular contribution, $Y_{\theta }=\cos \theta \cos \varphi$ and $Y_{\varphi \varphi }=-\sin \theta \cos \varphi$, soCombining the radial and angular terms provesThe calculation uses $0<\theta <\pi$ to avoid coordinate singularities. In Cartesian coordinates, $u=x(1+r^{-3})$, which extends smoothly across the polar axis for $r>0$. It therefore satisfies Laplace's equation there as well. The origin is a genuine singularity and is excluded.

Let $M=2x+e^y$ and $N=x{e}^y-\cos y$. Their cross partial derivatives agree: $M_y=e^y=N_x$. On the simply connected plane, this makes the differential form exact. Integrating $F_x=M$ gives $F=x^2+x{e}^y+h(y)$; matching $F_y=N$ then requires $h^{\prime }(y)=-\cos y$, so one potential isThe ordinary differential equation states $dF/dx=0$ along a solution. Its initial condition fixes the constant to $F(1,\pi /2)=e^{\pi /2}$. Hence the solution is specified implicitly byAt the initial point $F_y=e^{\pi /2}\ne 0$, so the implicit function theorem guarantees a unique local branch $y(x)$ through that point.

In the $x,z$ plane, the embankment lies above the parabola $z=H{x}^2/L^2$ and below $z=H$. The bore has circular cross-sectionIt touches the valley floor at the origin. To check that this entire disk lies inside the projected embankment, its equation gives $x^2\le Dz-z^2\le Dz$. Since $HD/L^2<1$, it follows that $H{x}^2/L^2\le z$, with strict inequality away from the bottom contact. Also $z\le D<H$. Thus no circular segment needs to be subtracted from the tunnel section.

For each $(x,z)$ in the disk, the length of bore inside the material is the local width $2b(2-z/H)$. The required volume integral is thereforeThe disk has area $A=\pi D^2/4$ and centroid height $D/2$, so ${\iint }_{\mathcal{D}}zdxdz=AD/2$. Applying the volume of a solid with affine thickness formula givesThe width varies along the circular section; multiplying its area by the top width alone would miss this variation.

Integrate by horizontal slices. At a fixed height $0\le z\le H$, the parabolic boundary gives $|x|\le L\sqrt{z/H}$; the other inequality gives $|y|\le b(2-z/H)$. Thus the cross-section is a rectangle of areaThe volume integral, followed by integration by substitution $u=z/H$, isThis has units of volume. As a check on the geometry, the width decreases with height while the length of the horizontal slice increases from zero to $2L$.