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Use the usual model of independent games, each with Sheila-win probability $\theta$. After $m+n$ games, an outcome with $m$ wins and $n$ losses has $\left(\genfrac{}{}{0ex}{}{m+n}{m}\right)$ possible orders. Since both totals are below eleven, none of those sequences has already ended the match. ThusFor a completed Sheila win with $n$ losses, the last game must be her eleventh win. The preceding $10+n$ games contain ten wins and $n$ losses, so their probability is $\left(\genfrac{}{}{0ex}{}{10+n}{10}\right){\theta }^{10}(1-\theta {)}^n$, followed by one further factor $\theta$. The allowed loss counts are $0\le n\le 10$, and these terminal events are disjoint. HenceThis is the match win probability with a fixed winning score, obtained by the negative binomial stopping argument. Constant marginal game-win probabilities alone would not justify the product without the usual independence assumption.

Normalize the Rayleigh distribution by substituting $s=x^2/(2{\sigma }^2)$:For its sketch, the density starts at zero, is positive for $x>0$, and tends to zero at infinity. Differentiation givesIt therefore has a single maximum at $x=\sigma$, with height $e^{-1/2}/\sigma$.

The expected value isIntegrating by parts with $xdx{e}^{-x^2/(2{\sigma }^2)}=-{\sigma }^{2}d(e^{-x^2/(2{\sigma }^2)})$ makes the boundary term vanish and leaves ${\int }_0^{\infty }{e}^{-x^2/(2{\sigma }^2)}dx$. The Gaussian integral gives

Let $C$ mean control-group membership and $E$ the adverse event. Equal allocation gives $P(C)=P(C^c)=1/2$, so the law of total probability gives $P(E)=(p_0+p_1)/2$. Bayes' theorem then givesThis is defined when $p_0+p_1>0$. If both are zero, the conditioning event has zero probability and the stated conditional probability is undefined.

Integrate the probability density function over the specified interval:The endpoint inclusions make no difference for this continuous distribution.

The support is $x\ge 1$, and its normalized Pareto distribution density integrates to

Real points require $x\ge 0$, and the two branches areThis is a semicubical parabola, symmetric about the horizontal axis and with a cusp at the origin. The parametrization $x=t^2$, $y=t^3$ traces both branches. Their limiting slopes at the origin are zero, so the common tangent direction is the positive horizontal axis, even though the curve has a cusp there.

On $0\le x\le 1/2$, the quadratic polynomial is positive. The required area is therefore its definite integral above the horizontal axis:The units are square coordinate units.

The second derivative is $y^{\prime \prime }=6x$, so $y^{\prime \prime }(1)=6>0$. By the second-derivative test, the stationary point is a strict local minimum, at $(1,5)$. It is not a global minimum, since this cubic is unbounded below toward negative infinity.

Differentiate the polynomial and factor:The stationary points therefore have horizontal coordinatesFor a check, $y^{\prime \prime }=6x-4$ is negative at the first and positive at the second, giving a local maximum and a local minimum respectively.

This exponential function is positive everywhere, with $y(0)=1$, $y\to \infty$ as $x\to -\infty$ and $y\to 0$ as $x\to \infty$. Its derivatives areIt is strictly decreasing overall, with a horizontal stationary inflection point at $(0,1)$. A second, nonstationary inflection point occurs at $x=(2/3{)}^{1/3}$, where $y=e^{-2/3}$. The only horizontal asymptote is $y=0$ toward positive infinity; the graph is not symmetric about the vertical axis.

Where the original expression is defined, multiplication by $\cos x$ givesThus the graph repeats with period $\pi$ and has vertical asymptotesBetween consecutive asymptotes the extended cotangent branch decreases from $+\infty$ to $-\infty$, since $y^{\prime }=-1/(\cos x+\sin x{)}^2<0$. It passes through $(k\pi ,1)$ and $(\pi /4+k\pi ,1/2)$.

It is essential to preserve the domain when simplifying a function. At $x=\pi /2+k\pi$, the original tangent is undefined. The finite limiting value is zero, so the graph has holes at $(\pi /2+k\pi ,0)$, rather than actual zeros. The simplified expression extends across those points, but the original one does not.

Interpret the stated side lengths as all eight edges of the square pyramid. The equal lateral edges put its apex above the base centre. The base-centre-to-vertex distance is half the diagonal, $1/\sqrt{2}\mathrm{m}$. The Pythagorean theorem therefore givesUsing the pyramid volume formula with a $1{\mathrm{m}}^2$ base,

The sphere has surface area $4\pi R^2=400\pi {\mathrm{m}}^2$. Dividing the given surface-patch area by that total gives the dimensionless fractionor about $0.796\%$ of the surface. The patch area is already measured on the surface, so no planar projection or spherical-cap conversion is needed.

Put the equation into standard ellipse form:The centre is $(1,0)$, with horizontal semiaxis $\sqrt{3}$ and vertical semiaxis $\sqrt{3/2}$. Its four axis vertices are $(1\pm \sqrt{3},0)$ and $(1,\pm \sqrt{3/2})$; the ellipse meets the vertical coordinate axis at $(0,\pm 1)$.

Substituting the line into the hyperbola givesThe two polynomial roots give all intersections:Both satisfy the original line as well as the curve.

The general solution of this first-order linear ordinary differential equation is $y=C{e}^{3x}$. Its initial condition gives $C=3$, soDifferentiation verifies $y^{\prime }=3y$ and substitution at zero gives the required value.