There are ten balls of each colour. Under uniform sampling without replacement, all $\left(\genfrac{}{}{0ex}{}{20}{12}\right)$ subsets of twelve labelled balls are equally likely. A favourable subset chooses four of the ten white balls and eight of the ten black balls, giving $\left(\genfrac{}{}{0ex}{}{10}{4}\right)\left(\genfrac{}{}{0ex}{}{10}{8}\right)$ choices. The hypergeometric distribution therefore givesThus the requested integers are $\boxed{N_1=N_2=10,m_1=4,m_2=8}$, with $N=20$ and $k=12$.

With equally many balls of each colour, uniform sampling without replacement givesHere an admissible population producing positive probability for three white balls has even $N\ge 6$, so the cancellation is legitimate. Equating this to $2/19$ gives $19(N-4)=16(N-1)$, hence

The probability of two black balls is the complement of the other two possibilities:The probability of equal colours is $P(WW)+P(BB)=1-2x+2p_w$. Thus the required conditional probability isThis is defined when the conditioning event has positive probability. If $N=2$ with one ball of each colour, that event is impossible and the conditional probability is undefined.

The probability that the first ball is white is $x$. Partitioning that event according to the second colour shows that $P(WB)=x-p_w$. Uniform sampling without replacement makes the two draw positions exchangeable, so $P(BW)=P(WB)$. ThereforeFor a direct count, the same value is $2(xN)((1-x)N)/[N(N-1)]$.

Interchanging the two colours preserves the probability distribution when their population counts are equal. Thus the probability of two black balls is also $p_w$. The three mutually exclusive possibilities are two white, two black, or different colours, so

With $N/2$ white balls, uniform sampling without replacement gives the probabilityRearranging, $4p_w(N-1)=N-2$, soA finite admissible population has even $N\ge 2$ and $0\le p_w<1/4$, so the denominator is nonzero.

Put $h=x-\pi /2$. The sine addition identity gives $\sin x=\cos h=1-h^2/2+h^4/24+O(h^6)$. Applying the exponential series givesThus the first three nonzero terms in the Taylor series about $\pi /2$ are

The Maclaurin series of the hyperbolic cosine is $\cosh x=1+x^2/2+x^4/24+O(x^6)$. The binomial series with exponent $-1/2$ gives $(1+u{)}^{-1/2}=1-u/2+3u^2/8+O(u^3)$. Substituting $u=x^2/2+x^4/24+O(x^6)$ yieldsThese are the first three nonzero terms; the remainder is written in Big O notation.

Multiplying the ordinary differential equation by the nonzero integrating factor from part (i) makes its left-hand side $dF(x,y(x))/dx$. Hence the potential is constant:Since $16+x^2>0$ for every real $x$, this is equivalent to the explicit general solutionIt is defined for all real $x$. There are no lost solutions, since the integrating factor and the coefficient of $y^{\prime }$ are everywhere nonzero.

Let $P=\mu (x)(xy-16x-x^3)$ and $Q=\mu (x)(16+x^2)$. An exact differential form must satisfy $P_y=Q_x$. These partial derivatives giveTaking antiderivatives produces $\ln |\mu |=-\frac{1}{2}\ln (16+x^2)+C$. A convenient nonzero integrating factor for a differential one-form is thereforeAn explicit potential, which also proves global exactness, isIndeed $F_x=P$ and $F_y=Q$ for the chosen integrating factor.

The zero solution satisfies both the ordinary differential equation and the initial condition. The right-hand side is continuously differentiable in $y$ near $(x,y)=(2,0)$, so the Picard-Lindelöf theorem gives local uniqueness; equivalently the nonzero family in part (a) can never take the value zero at $x=2$. ThusThe restriction $x>0$ is the maximal interval containing $2$ on which the original ordinary differential equation is defined.

Using the general solution in part (a), the initial condition givesTherefore the solution of this initial value problem isThe formula is used on the interval containing $x=1$ where the original ordinary differential equation and the displayed expression are defined.

From the same multiplication, the imaginary part isHere the planar vectors are embedded in three dimensions with zero third coordinate. Its magnitude is the area of their parallelogram; its sign is positive when the oriented angle from ${\mathbf{v}}_2$ to ${\mathbf{v}}_1$ has positive sine. The reversed order matters: $\Im (z_1{z}_2^{\ast })$ is the negative of the usual signed area for the order $({\mathbf{v}}_1,{\mathbf{v}}_2)$.

Write $z_j=x_j+i{y}_j$. Multiplication givesTherefore the real part is the dot product of the corresponding vectors:For nonzero vectors, this is $|z_1||z_2|\cos (\arg z_1-\arg z_2)$, measuring their alignment.

With $z_1=a+ib$ and $z_2=c+id$, direct multiplication gives $z_1{z}_2=(ac-bd)+i(ad+bc)$. Taking the complex conjugate yields

Write $z_1=a+ib$ and $z_2=c+id$ with real $a,b,c,d$. By the definition of complex conjugation,

Equality in the Cauchy-Schwarz inequality requires $\sqrt{r}v(r)=\Omega r^{3/2}$ for one constant $\Omega$. Thus $v(r)=\Omega r$ andThis is solid-body rotation. At fixed $L$, the constant is $\Omega =2L/(V{a}^2)$. The value at the single axis point $r=0$ does not affect the integrals; if $\omega$ is continuous, it has the same constant value there as well.

Using the same cylindrical-shell volume,For unit density, this is the axial angular momentum. To bound it, apply the integral Cauchy-Schwarz inequality to $f(r)=\sqrt{r}v(r)$ and $g(r)=r^{3/2}$:Multiplication by $(2\pi h{)}^2$ gives $L^2\le \pi h{a}^{4}K$. Since the chamber's volume is $V=\pi a^{2}h$, this is precisely

The volume form in cylindrical coordinates is $dV=rdrd\varphi dz$. Integrating around the axis and over the height gives cylindrical-shell volume $2\pi hrdr$. Hence the volume integral isFor unit density, this is the kinetic energy of the rotational motion.

Since $D^6=\mathrm{diag}(1,4^6)=\mathrm{diag}(1,4096)$,