Use the intended model of independent Bernoulli trials, with the same scoring probability $p$ for every shot. Equal marginal scoring probabilities alone would not establish the stated formula without independence. Let $N_1,N_2$ count misses in the two groups of five. They are independent random variables with binomial distribution $\mathrm{Bin}(5,1-p)$. A tie occurs precisely when $N_1=N_2=n$ for one of the mutually exclusive possibilities $n=0,\dots ,5$. Therefore the tie probability for independent binomial counts isAt $p=1/2$, Vandermonde's identity gives ${\sum }_n{\left(\genfrac{}{}{0ex}{}{5}{n}\right)}^2=\left(\genfrac{}{}{0ex}{}{10}{5}\right)=252$, soExchanging the two teams preserves the distribution and interchanges the two winning events. Their total probability is $1-\alpha$, so the first team wins at the end of the initial period with probability $193/512$. In this model the shooting order gives no advantage.

Let $\Omega$ be the interior of the tetrahedron, and take any constant vector field $\mathbf{q}$. Its divergence is zero, so the divergence theorem givesChoosing $\mathbf{q}$ to be each Cartesian basis vector proves ${\sum }_i{\mathbf{V}}_i=0$. This is the general fact that vector areas of a closed polyhedron sum to zero, specialized to four faces. The outward orientations are essential: the scalar face areas are all positive and do not sum to zero.

The volume of a parallelepiped is the absolute scalar triple product of its edge vectors. HereThe absolute value removes the orientation of the chosen edge ordering.

Take $P=(2,-1,2)$, $Q=(1,2,3)$ and $R=(4,1,0)$. The two plane directions are $Q-P=(-1,3,1)$ and $R-P=(2,2,-2)$. Their cross product is $(-8,0,-8)$, so a normal vector is $(1,0,1)$. The Cartesian equation of a plane is thereforeEach given point satisfies it, and the nonzero cross product confirms that the points determine a unique plane.

Differentiate $H=U+pV$ at fixed pressure and subtract the heat capacity at constant volume:Now use $U=U(T,V)$ and the multivariable chain rule along a constant-pressure path:Substitution cancels the fixed-volume derivative and gives the volume derivative form of the heat-capacity difference,For the ideal gas, the first term in brackets vanishes and $V=NkT/p$, so $(\partial V/\partial T{)}_p=Nk/p$. Consequently $\boxed{C_p-C_V=Nk}$ for the entire gas, with fixed particle number $N$. The extra heat capacity accounts for expansion work at constant pressure.

At fixed particle number, the differential of the internal energy identifies $T=(\partial U/\partial S{)}_V$ and $p=-(\partial U/\partial V{)}_S$. Equality of continuous mixed partial derivatives gives the first Maxwell relation:For the enthalpy, $dH=dU+pdV+Vdp=TdS+Vdp$. Thus $T=(\partial H/\partial S{)}_p$, $V=(\partial H/\partial p{)}_S$, and the same argument gives the enthalpy Maxwell relationThe smoothness and independent thermodynamic coordinates are understood within a single phase.

Let $D$ denote having the disease and $+$ a positive result. Assume I am sampled uniformly from the stated population, so $\mathbb{P}(D)=n/N$. By the law of total probability, $\mathbb{P}(+)=an/N+b(N-n)/N$. Thus Bayes' theorem givesThe denominator must be positive; otherwise the observed event has zero probability in this model and the conditional probability is undefined. The answer depends on disease prevalence as well as the test's two error rates.