\newfam\rsfs
\newfam\bbold
\def\scr#1{{\fam\rsfs #1}}
\def\bb#1{{\fam\bbold #1}}
\let\oldcal\cal
\def\cal#1{{\oldcal #1}}
\font\rsfsten=rsfs10
\font\rsfssev=rsfs7
\font\rsfsfiv=rsfs5
\textfont\rsfs=\rsfsten
\scriptfont\rsfs=\rsfssev
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\font\bbsev=msbm7
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\textfont\bbold=\bbten
\scriptfont\bbold=\bbsev
\scriptscriptfont\bbold=\bbfiv

\def\Pr{\bb P}
\def\E{\bb E}
\newcount\qnum
\def\q{\afterassignment\qq\qnum=}
\def\qq{\qqq{\number\qnum}}
\def\qqq#1{\bigskip\goodbreak\noindent{\bf#1)}\smallskip}
\def\fr#1#2{{#1\over #2}}
\def\var{\mathop{\rm var}\nolimits}

\q5

The generating function for the number of flowers is $$G_N(s) =
\sum_{s=0}^\infty qp^ns^n = {q\over 1 - ps}.$$
The probability generating function for each flower becoming a fruit is
$x = {1+s\over2}$, and from theorem 4.36, $$G_S(s) = G_N(G_X(s)) =
{q\over 1 - p{1+s\over2}} = {2q\over2-p}{1\over 1 - {p\over2-p}s}.$$
This corresponds to $$\Pr(R=r) = {2q\over2-p}\left({p\over2-p}\right)^r.$$

$$\Pr(N=n|R=r) = {\Pr(R=r|N=n)\over\Pr(R=r)}
= {{n\choose r}(1/2)^r \over
{2q\over2-p}\left({p\over2-p}\right)^r} =
{(2-p)^{r+1}\over2q(2p)^r}{n\choose r}.$$

\q6

$\Pr(X=x) = {n\choose x}p^xq^{n-x}.$
Its probability generating function $G_X$ is $$G_X(s) = (q+ps)^n.$$
$$\E(X) = G_X'(1) = pn(q+ps)^{n-1} = pn.$$
$$G_X''(1) = p^2n(n-1)(q+ps)^{n-2} = p^2n(n-1).$$
$$\var(X) = G_X''(1) + G_X'(1) - G_X'(1)^2 = p^2n(n-1) + pn - p^2n^2
= pn - p^2n = qpn$$

Evenness can be determined as $(G_X(-1)+G_X(1))/2$, because $(-1)^s +
1^s = 2$ iff s is even (else 0). These are in turn $G_X(-1) = (q-p)^n =
(1-2p)^n$ and $G_X(1) = 1,$ giving $(1+(1-2p)^n)/2.$

Divisibility by three is similar, but instead of $1$ and $-1$, the cubic
roots of unity are used: $1,$ $e^{i2\pi/3},$ and $e^{i4\pi/3}.$ $G_X$ on
these values will sum to $3$ for any number divisible by three because
these numbers cubed is $1$ and sum to $0$ for other numbers.
$(G_X(1) + G_X(e^{i2\pi/3}) + G_X(e^{i4\pi/3}))/3
= (1 + (q+pe^{i2\pi/3})^n + (q+pe^{i4\pi/3})^n).%
%(\cos(2\pi/3) + i\sin(2\pi/3)))^n + (q+p(\cos(2\pi
$
% yeah ig you have to do the simplification

\q8

For some value of $N$, $X$ and $Y$ follow binomial distributions.
They are independent,

Because $X$ and $Y$ are independent and $N=X+Y$, $$G_N(s) =
G_X(s)G_Y(s).$$ For a given value of $N$, $X$ and $Y$ both follow
binomial distributions, meaning, for a given value of $N=n$, $$G_X(s) =
G_Y(s) = (\fr12[1+s])^n.$$ Conditioning on $N$ gives the more valid
expressions $$G_X(s) = G_Y(s) = \sum_{i=0}^\infty
\Pr(N=i)(\fr12[1+s])^i = G_N(\fr12[1+s]).$$
Combining this with the first expression gives $$G_N(s) =
G_N(\fr12[1+s])^2.$$ Let $H_N(s) = G_N(1-s)\to G_N(s) = H_N(1-s).$ This
gives an identity $$H_N(1-s) = H_N(1-\fr12[1+s])^2 =
H_N(\fr12[1-s])^2 \to H_N(s) = H_N(\fr12s)^2.$$
There is only one function that fits this description: the exponential
curve $H_N(s) = e^{-\lambda s} \to G_N(s) = e^{-\lambda(1-s)} =
e^{\lambda(s-1)},$ corresponding uniquely to $\sum_{k=0}{1\over
k!}\lambda^ke^{-\lambda}s^k.$ This is a Poisson distribution.

\q9

There is always a $p=1/3$ chance of finding a red token in each
collection. The probability of a string of $j$ collections not having a
red token in the first $j-1$ and a red token in the final collection is
$p(1-p)^{j-1},$ so the generating function is $$\sum_{j=0}^\infty
p(1-p)^{j-1}s^j = ps\sum_{j=0}^\infty (s(1-p))^{j-1} =
{ps\over1-s(1-p)}.$$

In the general case,
let $G_{Y_n}$ be the time to acquire $n$ coupons. $G_{Y_1} = s$ because
it always takes only one collection to acquire a unique coupon. After
$n$ coupons have been collected, there is a $p = (m-n)/m$ chance of
obtaining a new coupon, so the probability of ``time to next coupon''
being $k$ is $pq^{k-1},$ giving a generating function $ps\over1-qs$
after any given previous value. Convolving with $G_{Y_n}$ gives
$$G_{Y_{n+1}}(s) = G_{Y_n}(s){{m-n\over m}s\over1-(n/m)s}
\Longrightarrow G_{Y_n}(s) = {s^n\over m^n}{(m-1)(m-2)\cdots(m-n) \over
(m-s)(m-2s)(m-3s)\cdots(m-ns)/m^n}.$$
$$\Longrightarrow G_Y(s) = {s^m(m-1)!\over(m-s)(m-2s)\cdots(m-ms)}.$$
$$\Longrightarrow \E(Y) = G_Y'(1) = m(m-1!)\cdot
    {{1\over 1}+{1\over2}+\cdots+{1\over m}\over(m-1)!}.$$

This simplifies to the given form.

\q10

The mean value of a discrete random variable $X$ is $\E(X) =
\sum_{i\in X} i\Pr(X=i).$ For a nonnegative integer-valued random
variable, its generating function is $\phi(s) = \sum_{i=0}^\infty
\Pr(X=i)s^i.$

$\phi'(1) = \sum_{i=0}^\infty i\Pr(X=i) = \E(X),$ so $\phi(s) =
p(s)/q(s)$ has mean value $$\phi'(1) = {p'(1)\over q(1)} -
{p(1)q'(1)\over q(1)^2} = {p'(1)\over q(1)} - \phi(1){q'(1)\over q(1)} =
{p'(1)-q'(1)\over q(1)}$$
because $\phi(1) = 1$ because it is a probability generating function.

Duelist A wins the duel on their nth shot with probability $a$ for
making the nth shot and $(1-a)^{n-1}(1-b)^{n-1}$ for it not already
having been made. Let $r = (1-a)(1-b).$
$$\sum_{i=0}^\infty ar^i = {a\over 1-r} = {a\over a+b-ab}$$
This is the probability duelist A wins.

The probability distribution of the number of shots fired is
well-described by a probability generating function
$\phi(s) = {a\over 1-rs^2} + {b(1-a)s\over 1-rs^2} = {a+b(1-a)s\over
1-rs^2}.$ These represent, respectively, the series of chances A and B
have to win the duel, so $$\E(\hbox{shots fired}) = \phi'(1) = {b(1-a) +
2r\over 1-r}$$
by the earlier established identity with $p=a+bs(1-a)$ and $q=1-rs^2.$
This simplifies to $(2-b)(1-a)\over a+b-ab.$

\q11

With $N=n,$
$$G_F(s) = \sum_{i=0}^n{n\choose i}p^i(1-p)^{n-i}s^i = (ps+1-p)^n.$$
Conditioning on $N$ gives
$$G_F(s) = \Pr(N=0) + \Pr(N=1)(ps+1-p) + \Pr(N=2)(ps+1-p)^2 +\cdots
= G_N(ps+1-p).$$

(b) is true by 3.6.14.

(c) is a case of 4.5.8, where from $p=1/2$ (fair coin) and independence
of two variables summing to a larger variable $N$ proves that the larger
variable $N$ is a Poisson distribution.

\bye
