On $0<x<2$, $2x-x^2=x(2-x)>0$. A valid probability density function must integrate to one, soThus $\alpha =3/4$, which is positive and ensures nonnegativity. Its expected value isThereforeThe probability density function is symmetric around $1$, providing a second check on the expected value.

First restrict to integers in the marginal supports: $2\le i\le 12$ and $-5\le j\le 5$. Their marginal probabilities are positive. By part (i), a feasible joint pair identifies a unique ordered pair of rolls, so its joint probability is $1/36$; an infeasible pair has joint probability zero.

Put $a_i=6-|i-7|$ and $b_j=6-|j|$. If the two events were independent, a feasible pair would requireSince $1\le a_i,b_j\le 6$, equality forces $a_i=b_j=6$, hence $(i,j)=(7,0)$. But that pair is infeasible. An infeasible pair cannot give independence either, since its joint probability is zero while the product of its positive marginal probabilities is positive. ThusIf all real values of $i,j$ are allowed literally, an impossible event is independent of every event. The complete answer without a support restriction is therefore

Count the ordered roll pairs producing each value. The probability mass functions areand are zero outside these ranges. Both counts have a unique peak:However, $Y=0$ forces equal rolls, whose sum is even. Hence $\mathbb{P}(X=7,Y=0)=0$, whereas the product of the two marginal probabilities is $1/36>0$. This disproves independence of random variables:

The identity $XY=(S_1+S_2)(S_1-S_2)=S_1^2-S_2^2$ gives $\mathbb{E}[XY]=0$, since the two rolls have the same probability distribution. Also $\mathbb{E}[X]\mathbb{E}[Y]=7\cdot 0=0$. ThereforeThe two random variables are uncorrelated random variables, but that alone does not establish independence of random variables.

Part (a) with $n=6$ gives $\mathbb{E}[S_1]=\mathbb{E}[S_2]=7/2$ and $\mathrm{var}(S_1)=\mathrm{var}(S_2)=35/12$. By linearity of expectation,Independence of random variables and $\mathrm{var}(-S_2)=\mathrm{var}(S_2)$ give the same variance for the sum and difference:

This numbered item specifies the model used below. A convenient sample space is the collection of $36$ ordered pairs $(s_1,s_2)\in \{1,\dots ,6{\}}^2$, each with probability $1/36$. The corresponding random variables are determined by the sum and difference. Inverting the two equations givesThus a candidate pair $(X,Y)=(i,j)$ is possible exactly when $(i+j)/2$ and $(i-j)/2$ are both integers between $1$ and $6$. In particular, $i$ and $j$ must have the same parity. This inverse description will determine the joint probability mass function.

Assume the variances are finite. Expand the square and apply linearity of expectation:Part (i), applied to the identity functions, makes the bracketed covariance zero by independence of random variables. Hence

Independence of random variables factors the joint probability mass function as $\mathbb{P}(X=x,Y=y)=\mathbb{P}(X=x)\mathbb{P}(Y=y)$. ThereforeWith finite expected values, these rearrangements are justified by absolute convergence: the analogous sum with absolute values equals $\mathbb{E}[|f(X)|]\mathbb{E}[|g(Y)|]<\infty$. Thus

Let $B$ count the red balls among the $n$ added balls. By independence of random variables, $B$ has a binomial distribution with parameters $n,1/2$. The total red count is $R=1+B$, and the box has $n+1$ balls. Conditional on $R=r$, sampling without replacement giveswhere $D_1,D_2$ denote red first and second draws. Averaging these conditional probabilities and taking their ratio givesThe binomial distribution moments give $\mathbb{E}[B]=n/2$ and $\mathbb{E}[B^2]=n/4+n^2/4$. ConsequentlyTherefore the red-ball prediction with one certain red ball isIt reduces to $2/3$ at $n=1$ and approaches $1/2$ as the number of randomly coloured extra balls grows. Taking an unweighted average of the remaining red fractions would miss the change in weights caused by taking a conditional probability given the first red draw.

Let $H$ mean the added ball was red, and $D$ mean the first draw was red. Their relevant probabilities are $\mathbb{P}(H)=\mathbb{P}(H^c)=1/2$, $\mathbb{P}(D\mid H)=1$, and $\mathbb{P}(D\mid H^c)=1/2$. The law of total probability gives $\mathbb{P}(D)=3/4$.

Given the observed red draw, the remaining ball is red precisely when $H$ occurred. Bayes' theorem therefore givesThe observation increases the weight of the two-red-ball box, because that box is more likely to yield a red first draw.

The events $A\cap B$ and $A\cap B^c$ are disjoint and their union is $A$. Additivity of probability givesBoth conditioning events have positive probability because $0<\mathbb{P}(B)<1$. Apply the definition of conditional probability to each term to obtain the law of total probability:

The conditional probability rescales the probability of the part of $A$ lying in $B$ by the probability of $B$:The denominator makes $B$ itself have conditional probability one.

For the binomial distribution, the binomial theorem givesDifferentiation of the probability generating function recovers the first expected value and second factorial moment:Since $X^2=X(X-1)+X$, the variance isThusThe formulas also hold for $n=0,1$ and the degenerate choices $p=0,1$, with derivatives read directly from the resulting polynomial.

For a discrete random variable taking nonnegative integer values, its probability generating function isIt is defined at least for $|s|\le 1$. The usual probability generating function is intended for nonnegative integer-valued random variables; a general discrete real-valued random variable need not have this form. The coefficient of $s^k$ recovers $\mathbb{P}(X=k)$.

For $N$ equal intervals, the composite trapezoidal rule isAdding the missing $N^2/N^3$ term therefore gives the exact relationSince $x^2$ is continuous, its Riemann sums, and hence its endpoint-averaged trapezoidal sums, converge to its definite integral. Thus $T_N\to 1/3$ and $1/(2N)\to 0$, givingNo pre-existing formula for the sum of squares is needed.

With step size $h=1/4$, the composite trapezoidal rule gives half weight to the two endpoints:ThereforeIt exceeds the exact definite integral by $1/96$, as expected because $x^2$ is a convex function and each trapezoid lies above the curve.

The power rule for a definite integral gives

For a partial fraction decomposition, writeMultiplying by $(x-1{)}^2(x+1)$ gives $4=A(x^2-1)+B(x+1)+C(x-1{)}^2$. Setting $x=1,-1$ gives $B=2,C=1$, and comparing the coefficient of $x^2$ gives $A=-1$. ThusIntegrating each term gives the antiderivativeThe constant of integration can be chosen independently on each interval separated by $-1$ and $1$.

Differentiate the product $4(x-1{)}^{-2}(x+1{)}^{-1}$ using the product rule and chain rule:Combining terms over a common denominator gives

This is equality of remainders in modular arithmetic. It is reflexive because $3\mid (n-n)=0$, symmetric because $3\mid (n-m)$ implies $3\mid (m-n)$, and transitive because $3\mid (n-m)$ and $3\mid (k-n)$ imply $3\mid (k-m)$ by addition. ThusIts three equivalence classes areIf zero is included in $\mathbb{N}$, it belongs to $C_0$; otherwise that class starts at three.