【数学A】2章：確率（節末問題・章末問題）

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% 節末問題A2.1.1：（One More）★★
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1から10までの番号が1つずつ書かれた10枚のカードから1枚取り出し，その数字を記録して元に戻す．この操作を3回繰り返し，記録した数を順に$x,y,z$とする．このとき，次の確率を求めよ．

(1) $\frac{y}{x}$が整数になる確率

(2) $x<y<z$になる確率

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% 節末問題A2.1.1の解答（One More）★★
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(1) $x,y$の取り出し方は$10^2=100(\text{通り})$

(i) $x=1$のとき，$y$は$1,2,\ldots,10$の$10$通り

(ii) $x=2$のとき，$y$は2の倍数であり，$5$通り

(iii) $x=3$のとき，$y$は3の倍数であり，$3$通り

(iv) $x=4$のとき，$y$は4の倍数であり，$2$通り

(v) $x=5$のとき，$y$は5の倍数であり，$2$通り

(vi) $x=6$のとき，$y=6$の$1$通り

(vii) $x=7,8,9,10$のとき，$y$は$y=x$の$1$通り

(i)〜(vii)より，$\frac{y}{x}$が整数となる場合の数は，$10+5+3+2+2+1+4\times 1=27(\text{通り})$

よって，求める確率は，$\frac{27}{100}$

(2) $x,y,z$の取り出し方は$10^3=1000(\text{通り})$

$x<y<z$となる場合の数は，1から10までの10個の数字から異なる3個を選び，それらを小さい方から順に$x,y,z$と定めればよいから，

\[
{}_{10}\mathrm{C}_3=\frac{10\cdot 9\cdot 8}{3\cdot 2\cdot 1}=120(\text{通り})
\]

よって，求める確率は，

\[
\frac{{}_{10}\mathrm{C}_3}{10^3}=\frac{120}{1000}=\frac{3}{25}
\]

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% 節末問題A2.1.2：（One More）★★
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12人が円形に座るとき，次の確率を求めよ．

(1) 特定の2人$\mathrm{X},\mathrm{Y}$が1人おいて隣り合う確率

(2) 特定の3人$\mathrm{X},\mathrm{Y},\mathrm{Z}$が1人ずつおいて隣り合う確率

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% 節末問題A2.1.2の解答（One More）★★
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すべての場合の数は，12人の円順列であるから，$(12-1)!=11!(\text{通り})$

(1) X，Yとその間に座る1人をまとめて1組と考えると，残りの9人と合わせた10個の円順列より，

\[
(10-1)!=9!(\text{通り})
\]

$\mathrm{X},\mathrm{Y}$の2人の並び方は，$2!$通り

$\mathrm{X}$と$\mathrm{Y}$の間に座る1人は残りの10人から選ぶので，${}_{10}\mathrm{C}_1$通り

したがって，$\mathrm{X},\mathrm{Y}$が1人おいて隣り合う座り方の総数は，$9!\times 2!\times{}_{10}\mathrm{C}_1(\text{通り})$

よって，求める確率は，$\frac{9!\times 2!\times{}_{10}\mathrm{C}_1}{11!}=\frac{2}{11}$

(2) $\mathrm{X},\mathrm{Y},\mathrm{Z}$とその間に座る2人をまとめて1組と考えると，残りの7人と合わせた8個の円順列より，$(8-1)!=7!(\text{通り})$

$\mathrm{X},\mathrm{Y},\mathrm{Z}$の3人の並び方は，$3!$通り

間に座る2人は残りの9人から選んで並べるので，その場合の数は，${}_9\mathrm{P}_2$通り

したがって，$\mathrm{X},\mathrm{Y},\mathrm{Z}$が1人ずつおいて隣り合う座り方の総数は，

\[
7!\times 3!\times{}_9\mathrm{P}_2(\text{通り})
\]

よって，求める確率は，$\frac{7!\times 3!\times{}_9\mathrm{P}_2}{11!}=\frac{3}{55}$

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% 節末問題A2.1.3：（One More）★★★
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赤玉5個，白玉3個，青玉4個の合計12個の玉が入っている袋の中から，3個の玉を同時に取り出すとき，次の確率を求めよ．

(1) 3個の玉がすべて同じ色である確率

(2) 3個とも色が異なる確率

(3) 少なくとも1個は青玉である確率

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% 節末問題A2.1.3の解答（One More）★★★
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12個の玉から3個の玉を取り出す方法の総数は，${}_{12}\mathrm{C}_3=220(\text{通り})$

(1) 3個とも赤玉のとき，赤玉5個から3個を取り出す場合の数は，${}_5\mathrm{C}_3=10(\text{通り})$

3個とも白玉のとき，白玉3個から3個を取り出す場合の数は，${}_3\mathrm{C}_3=1(\text{通り})$

3個とも青玉のとき，青玉4個から3個を取り出す場合の数は，${}_4\mathrm{C}_3=4(\text{通り})$

よって，求める確率は，$\frac{10}{220}+\frac{1}{220}+\frac{4}{220}=\frac{3}{44}$

(2) 赤玉，白玉，青玉をそれぞれ1個ずつ選ぶ場合の数は，

\[
{}_5\mathrm{C}_1\times{}_3\mathrm{C}_1\times{}_4\mathrm{C}_1=5\times 3\times 4=60(\text{通り})
\]

よって，求める確率は$\frac{60}{220}=\frac{3}{11}$

(3) 少なくとも1個は青玉である事象を$A$とすると，余事象$\overline{A}$は3個のうち青玉が1個もない事象である．

青玉ではない玉の個数は，8個あるから，$\overline{A}$の場合の数は，${}_8\mathrm{C}_3=\frac{8\cdot 7\cdot 6}{3\cdot 2\cdot 1}=56(\text{通り})$

よって，求める確率は，$P(A)=1-P(\overline{A})=1-\frac{56}{220}=\frac{41}{55}$

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% 節末問題A2.1.4：（One More）★★
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箱の中に赤玉5個，白玉2個，青玉4個が入っている．この箱から同時に3個の玉を取り出すとき，次の確率を求めよ．

(1) 玉の色が少なくとも2種類ある

(2) 取り出した玉の色がちょうど2種類になる

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% 節末問題A2.1.4の解答（One More）★★
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11個の玉から3個の玉を取り出す場合の数は，${}_{11}\mathrm{C}_3$通り

(1) 取り出される3個の玉の色が少なくとも2種類ある事象を$A$とすると，余事象$\overline{A}$はすべて同じ色の玉が取り出される事象である．

事象$\overline{A}$が起こる場合の数は，${}_5\mathrm{C}_3+{}_4\mathrm{C}_3(\text{通り})$

よって，求める確率は，

\[
P(A)=1-P(\overline{A})=1-\frac{{}_5\mathrm{C}_3+{}_4\mathrm{C}_3}{{}_{11}\mathrm{C}_3}=1-\frac{14}{165}=\frac{151}{165}
\]

(2) 取り出される3個の玉の色がすべて異なる確率は，

\[
\frac{{}_5\mathrm{C}_1\times{}_2\mathrm{C}_1\times{}_4\mathrm{C}_1}{{}_{11}\mathrm{C}_3}=\frac{40}{165}
\]

玉の色が2種類である事象の余事象は，玉の色が1種類または3種類である事象であるから，玉の色がちょうど2種類になる確率は，

\[
1-(\frac{14}{165}+\frac{40}{165})=\frac{111}{165}=\frac{37}{55}
\]

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% 節末問題A2.1.5：（One More）★★★
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$n$人でじゃんけんを1回行うとき，次の確率を求めよ．ただし，$n\geqq 5$とする．

(1) ちょうど4人が勝つ

(2) あいこになる

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% 節末問題A2.1.5の解答（One More）★★★
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$n$人のじゃんけんの手の出し方は，$3^n$通り

(1) 勝つ4人の選び方は，${}_n\mathrm{C}_4$通りであり，その勝つ4人の手の出し方は，${}_3\mathrm{C}_1$通りであるから，その場合の数は，

\[
{}_n\mathrm{C}_4\times{}_3\mathrm{C}_1={}_n\mathrm{C}_4\times 3(\text{通り})
\]

よって，求める確率は，

\[
\frac{{}_n\mathrm{C}_4\times 3}{3^n}=\frac{n(n-1)(n-2)(n-3)\times 3}{4\cdot 3\cdot 2\cdot 1\times 3^n}=\frac{n(n-1)(n-2)(n-3)}{8\cdot 3^n}
\]

(2) あいこになる事象は，勝敗が決まる事象の余事象である．勝敗が決まるのは，ちょうど2種類の手が出る場合である．

2種類の手の選び方は，${}_3\mathrm{C}_2$通りであり，その手の出し方は，$(2^n-2)$通り

よって，求める確率は，

\[
1-\frac{{}_3\mathrm{C}_2\cdot(2^n-2)}{3^n}=\frac{3^n-3\cdot 2^n+6}{3^n}
\]

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% 節末問題A2.2.1：（One More）★★
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X，Yの2人があるゲームで繰り返し対戦し，先に4ゲーム勝った方を優勝者とする．各ゲームにおいてXが勝つ確率は$\frac{3}{4}$で，引き分けはないものとする．このとき，次の問いに答えよ．

(1) 4ゲーム目で優勝が決まる確率を求めよ．

(2) 5ゲーム目でXが優勝する確率を求めよ．

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% 節末問題A2.2.1の解答（One More）★★
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(1) Xが4連勝で勝つ確率は，$(\frac{3}{4})^4=\frac{81}{256}$

1回のゲームでYが勝つ確率は，$1-\frac{3}{4}=\frac{1}{4}$であるから，Yが4連勝で勝つ確率は，

\[
(\frac{1}{4})^4=\frac{1}{256}
\]

よって，求める確率は，$\frac{81}{256}+\frac{1}{256}=\frac{82}{256}=\frac{41}{128}$

(2) Xが4ゲーム目までに3勝1敗となり，5ゲーム目に勝つ確率であるから，

\[
{}_4\mathrm{C}_3(\frac{3}{4})^3(\frac{1}{4})^1\times\frac{3}{4}=\frac{81}{256}
\]

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% 節末問題A2.2.2：（One More）★★★
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1個のさいころを4回投げるとき，1の目と6の目が同じ回数だけ出る確率を求めよ．

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% 節末問題A2.2.2の解答（One More）★★★
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さいころを1回投げるとき，1の目，6の目が出る確率は，それぞれ$\frac{1}{6}$であり，1，6以外の目が出る確率は，$\frac{2}{3}$である．

1の目と6の目が出る回数が同じであるのは，1の目と6の目の出る回数が0回，1回，2回の3つの場合がある．

(i) 1の目と6の目が1回も出ないとき

\[
(\frac{2}{3})^4=\frac{16}{81}
\]

(ii) 1の目と6の目が1回ずつ出るとき

\[
\frac{4!}{2!1!1!}\times(\frac{1}{6})^1(\frac{1}{6})^1(\frac{2}{3})^2=\frac{4}{27}
\]

(iii) 1の目と6の目が2回ずつ出るとき

\[
\frac{4!}{2!2!}\times(\frac{1}{6})^2(\frac{1}{6})^2=\frac{1}{216}
\]

よって，(i)〜(iii)より，求める確率は，

\[
\frac{16}{81}+\frac{4}{27}+\frac{1}{216}=\frac{227}{648}
\]

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% 節末問題A2.2.3：（One More）★★★★
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12本のくじの中に3本の当たりくじがある．当たりくじを2回引くまで繰り返しくじを引くとき，$n$回目で終わる確率$p_n$を最大にする$n$の値を求めよ．ただし，引いたくじは毎回もとに戻すものとする．

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% 節末問題A2.2.3の解答（One More）★★★★
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このくじから1本を引くとき，当たりくじを引く確率は$\frac{1}{4}$であり，$n\geqq 2$である．$n$回目で終わるのは，$n-1$回目までに当たりくじを1回引き，$n$回目で当たりくじを引くときであるから，

\[
p_n={}_{n-1}\mathrm{C}_1(\frac{1}{4})(\frac{3}{4})^{n-2}\times\frac{1}{4}=\frac{3^{n-2}(n-1)}{4^n}
\]

$n\geqq 2$において，$p_{n+1}$と$p_n$の比を求めると，

\[
\frac{p_{n+1}}{p_n}=\frac{3^{n-1}n}{4^{n+1}}\div\frac{3^{n-2}(n-1)}{4^n}=\frac{n}{n-1}\cdot\frac{3^{n-1}\cdot 4^n}{3^{n-2}\cdot 4^{n+1}}=\frac{3n}{4(n-1)}
\]

(i) $\frac{p_{n+1}}{p_n}\geqq 1$のとき

$\frac{3n}{4(n-1)}\geqq 1$より，$3n\geqq 4(n-1)$であるから，$n\leqq 4$

したがって，$n=2,3$のとき，$\frac{p_{n+1}}{p_n}>1$より，$p_n<p_{n+1}$

$n=4$のとき，$p_4=p_5$

(ii) $\frac{p_{n+1}}{p_n}<1$のとき

$\frac{3n}{4(n-1)}<1$より，$3n<4(n-1)$であるから，$n>4$

したがって，$n=5,6,7,\ldots$のとき，$\frac{p_{n+1}}{p_n}<1$より，$p_n>p_{n+1}$

(i)，(ii)より$p_2<p_3<p_4,p_4=p_5,p_5>p_6>\cdots>p_{11}>p_{12}>\cdots$

よって，$p_n$を最大にする$n$の値は，$n=4,5$

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% 節末問題A2.2.4：（One More）★★
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あるコンテストで，$a$が優勝する確率は$70\%$である．$a$の結果にかかわらず，確率$\frac{1}{4}$でうそをつく$b$が，$a$の結果を知ったうえで「$a$が優勝した」と発言した．このとき，$a$が本当に優勝した確率を求めよ．

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% 節末問題A2.2.4の解答（One More）★★
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このコンテストで，$a$が優勝する事象を$E$，$b$が「$a$が優勝した」と発言する事象を$F$とする．

\begin{align*}
&P(E\cap F)=P(E)\times P_E(F)=\frac{70}{100}\times\frac{3}{4}=\frac{21}{40},\\
&P(\overline{E}\cap F)=P(\overline{E})\times P_{\overline{E}}(F)=\frac{30}{100}\times\frac{1}{4}=\frac{3}{40}
\end{align*}

したがって，

\[
P(F)=P(E\cap F)+P(\overline{E}\cap F)=\frac{21}{40}+\frac{3}{40}=\frac{24}{40}=\frac{3}{5}
\]

よって，求める確率は，

\[
P_F(E)=\frac{P(E\cap F)}{P(F)}={\frac{21}{40}}\div{\frac{3}{5}}=\frac{7}{8}
\]

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% 節末問題A2.2.5：（One More）★★★★
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原点$\mathrm{O}$から出発して，数直線上を動く点$\mathrm{P}$がある．$\mathrm{P}$は，1枚の硬貨を投げて表が出た場合には$+5$，裏が出た場合は$+3$移動する．硬貨を続けて投げていき，点$\mathrm{P}$の座標が初めて$18$以上になるまでの投げた回数を$X$とする．

(1) $X=4$となる確率を求めよ．

(2) $X$の期待値を求めよ．

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% 節末問題A2.2.5の解答（One More）★★★★
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(1) 4回投げて$\mathrm{P}$の座標が初めて$18$以上になるのは，4回とも表が出る場合と，4回のうち3回表，1回裏が出る場合があるから，その確率は，

\[
(\frac{1}{2})^4+{}_4\mathrm{C}_3(\frac{1}{2})^3(\frac{1}{2})^1=\frac{5}{16}
\]

(2) 硬貨を投げる回数が3回以下のとき，$\mathrm{P}$の座標は$18$以上にならない．また，硬貨を6回投げるまでに必ず$\mathrm{P}$の座標は$18$以上になる．つまり，$X$の値は$4,5,6$の3つの場合がある．

(i) $X=4$のとき

(1) より，その確率は，$\frac{5}{16}$

(ii) $X=5$のとき

2回表，2回裏で，5回目に表が出る確率は，${}_4\mathrm{C}_2(\frac{1}{2})^2(\frac{1}{2})^2\times\frac{1}{2}=\frac{3}{16}$

2回表，3回裏が出る確率は，${}_5\mathrm{C}_2(\frac{1}{2})^2(\frac{1}{2})^3=\frac{5}{16}$

したがって，$X=5$になる確率は，

\[
\frac{3}{16}+\frac{5}{16}=\frac{1}{2}
\]

(iii) $X=6$のとき

6回とも裏が出る確率は，$(\frac{1}{2})^6=\frac{1}{64}$

1回表，4回裏で，6回目に表が出る確率は，${}_5\mathrm{C}_1(\frac{1}{2})^1(\frac{1}{2})^4\times\frac{1}{2}=\frac{5}{64}$

1回表，5回裏が出る確率は，${}_6\mathrm{C}_1(\frac{1}{2})^1(\frac{1}{2})^5=\frac{3}{32}$

したがって，$X=6$になる確率は，

\[
\frac{1}{64}+\frac{5}{64}+\frac{3}{32}=\frac{3}{16}
\]

(i)〜(iii)より，$X$のとりうる値と，それぞれの値をとる確率は，次の表のようになる．

\begin{tabular}{|c||c|c|c|c|}
\hline
$X$&4&5&6&計\\
\hline
確率&$\frac{5}{16}$&$\frac{1}{2}$&$\frac{3}{16}$&1\\
\hline
\end{tabular}

よって，求める期待値は，

\[
4\times\frac{5}{16}+5\times\frac{1}{2}+6\times\frac{3}{16}=\frac{39}{8}
\]

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% 章末問題A2.1：（One More）★★★
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正六角形の頂点を反時計回りに$1$から$6$までの番号を付ける．$1$個のさいころを$3$回投げて，出た目の番号に対応する頂点を線分で結び図形を作るとき，次の確率を求めよ．

(1) 三角形ができる確率

(2) 正三角形ができる確率

(3) 直角三角形ができる確率

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% 章末問題A2.1の解答（One More）★★★
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さいころを$3$回投げるとき，目の出方の総数は，$6^3=216(\text{通り})$

(1) $3$点がすべて異なる場合の数は，${}_6\mathrm{P}_3=120(\text{通り})$

よって，求める確率は，$\frac{120}{216}=\frac{5}{9}$

(2) 正三角形ができるには，選ばれる$3$個の頂点が$\{1,3,5\}$または$\{2,4,6\}$のときであり，それぞれの目の出方は$3!$通りあるので，正三角形ができる目の出方は，

\[
2\times 3!=12(\text{通り})
\]

よって，求める確率は，$\frac{12}{216}=\frac{1}{18}$

(3) 向かい合う$2$個の頂点と残りの$4$個の頂点から$1$個の頂点を選ぶと，$1$個の直角三角形ができる．向かい合う$2$個の頂点の選び方は$3$組あるので，直角三角形は全部で，

\[
3\times 4=12(\text{個})
\]

それぞれの3個の番号の目の出方は$3!$通り

したがって，直角三角形ができる目の出方は，$12\times 3!=72\text{(通り)}$

よって，求める確率は，$\frac{72}{216}=\frac{1}{3}$

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% 章末問題A2.2：（One More）★★★
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座標平面上の原点$\mathrm{O}$から出発して，毎回確率$\frac{1}{4},\frac{1}{2},\frac{1}{4}$でそれぞれ左，上，右へ1ずつ移動する点$\mathrm{Q}$がある．8回の移動後に点$(2,4)$にいる確率を求めよ．

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% 章末問題A2.2の解答（One More）★★★
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左へ$x$回，上へ$y$回，右へ$z$回進むとすると，$x+y+z=8\cdots(\mathrm{i})$

このとき，$x$軸方向には$-x+z$，$y$軸方向には$y$動くので，移動後の座標は$(-x+z,y)$であるから，$-x+z=2,y=4\cdots(\mathrm{ii})$

(i)，(ii)より，$x=1,y=4,z=3$

よって，求める確率は，

\[
\frac{8!}{1!3!4!}(\frac{1}{4})^1(\frac{1}{2})^4(\frac{1}{4})^3=\frac{35}{512}
\]

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% 章末問題A2.3：（One More）★★★★
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3個のさいころ$\mathrm{A},\mathrm{B},\mathrm{C}$を同時に振り，出た目の最小値が3であったとき，最大値が5である条件付き確率を求めよ．

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% 章末問題A2.3の解答（One More）★★★★
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出た目の最小値が3であるという事象を$A$，最大値が5であるという事象を$B$とする．

$A$は，出た目がすべて3以上である場合から，出た目がすべて4以上である場合を除いた場合であるから，

\[
P(A)=(\frac{4}{6})^3-(\frac{3}{6})^3=\frac{64-27}{216}=\frac{37}{216}
\]

また，最小値が3，最大値が5となる目の組合せは，$\{3,3,5\},\{3,4,5\},\{3,5,5\}$

したがって，それぞれの確率を考えると，

\[
P(A\cap B)=\frac{2\times{}_3\mathrm{C}_1+3!}{6^3}=\frac{12}{216}
\]

よって，求める確率は，

\[
P_A(B)=\frac{P(A\cap B)}{P(A)}=\frac{12}{216}\div\frac{37}{216}=\frac{12}{37}
\]

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% 章末問題A2.4：（One More）★★★
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箱の中に4個の白玉と$n$個の赤玉が入っている．この箱から同時に2個の玉を取り出したとき，赤玉の数を$X$とする．$X$の期待値が1.5であるとき，$n$の値を求めよ．ただし，$n\geqq 2$であるとする．

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% 章末問題A2.4の解答（One More）★★★
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$X=k$である確率を$P(X=k)$で表すとする．

玉の取り出し方の総数は，${}_{n+4}\mathrm{C}_2$通り

$X=1$となるのは，白玉と赤玉を1個ずつ取り出す場合であり，その確率は，

\[
P(X=1)=\frac{{}_4\mathrm{C}_1\cdot{}_n\mathrm{C}_1}{{}_{n+4}\mathrm{C}_2}=\frac{8n}{(n+4)(n+3)}
\]

$X=2$となるのは，赤玉を2個取り出す場合であり，その確率は，

\[
P(X=2)=\frac{{}_n\mathrm{C}_2}{{}_{n+4}\mathrm{C}_2}=\frac{n(n-1)}{(n+4)(n+3)}
\]

したがって，$X$の期待値は，

\begin{align*}
&1\times\frac{8n}{(n+4)(n+3)}+2\times\frac{n(n-1)}{(n+4)(n+3)}\\
&=\frac{8n+2n(n-1)}{(n+4)(n+3)}=\frac{2n(n+3)}{(n+4)(n+3)}=\frac{2n}{n+4}
\end{align*}

よって，$\frac{2n}{n+4}=1.5$であるとき，$n=12$

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% 章末問題A2.5：（One More）★★★★
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2つのチームA，Bが繰り返し試合をして，先に4勝した方を優勝チームとする．各試合においてAが勝つ確率は$\frac{2}{3}$で，引き分けはないとする．このとき，優勝チームが決まるまでの試合数の期待値を求めよ．

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% 章末問題A2.5の解答（One More）★★★★
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(i) 4試合目にAが優勝するとき

Aが4勝0敗する場合であるから，その確率は，$(\frac{2}{3})^4=\frac{16}{81}$

(ii) 4試合目にBが優勝するとき

Bが4勝0敗する場合であるから，その確率は，$(\frac{1}{3})^4=\frac{1}{81}$

(iii) 5試合目にAが優勝するとき

4試合目までにAが3勝1敗となり，5試合目にAが勝つ場合であるから，その確率は，

\[
{}_4\mathrm{C}_3(\frac{2}{3})^3(\frac{1}{3})^1\times\frac{2}{3}=\frac{64}{243}
\]

(iv) 5試合目にBが優勝するとき

4試合目までにBが3勝1敗となり，5試合目にBが勝つ場合であるから，その確率は，

\[
{}_4\mathrm{C}_3(\frac{1}{3})^3(\frac{2}{3})^1\times\frac{1}{3}=\frac{8}{243}
\]

(v) 6試合目にAが優勝するとき

5試合目までにAが3勝2敗となり，6試合目にAが勝つ場合であるから，その確率は，

\[
{}_5\mathrm{C}_3(\frac{2}{3})^3(\frac{1}{3})^2\times\frac{2}{3}=\frac{160}{729}
\]

(vi) 6試合目にBが優勝するとき

5試合目までにBが3勝2敗となり，6試合目にBが勝つ場合であるから，その確率は，

\[
{}_5\mathrm{C}_3(\frac{1}{3})^3(\frac{2}{3})^2\times\frac{1}{3}=\frac{40}{729}
\]

(vii) 7試合目に優勝チームが決まるとき

6試合目までにAとBが3勝3敗となり，このとき，7試合目はどちらのチームが勝っても優勝チームが決まるから，その確率は，

\[
{}_6\mathrm{C}_3(\frac{2}{3})^3(\frac{1}{3})^3\times 1=\frac{160}{729}
\]

\begin{tabular}{|c||c|c|c|c|c|}
\hline 試合数&4&5&6&7&計\\
\hline 確率&$\frac{17}{81}$&$\frac{8}{27}$&$\frac{200}{729}$&$\frac{160}{729}$&1\\
\hline
\end{tabular}

よって，(i)〜(vii)より，求める試合数の期待値は，

\[
4\times\frac{17}{81}+5\times\frac{8}{27}+6\times\frac{200}{729}+7\times\frac{160}{729}=\frac{4012}{729}
\]

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% 節末問題A2.1.1の側注：（One More）★★
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% 側注1
分母である$x$の値で場合分けをする．

% 側注2
$y$は，3，6，9の3通り

% 側注3
和の法則を用いる．

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% 節末問題A2.1.2の側注：（One More）★★
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% 側注1
間に入る人をAとすると，下の図のようになる．

% 側注2
間に入る人をA，Bとすると，下の図のようになる．

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% 節末問題A2.1.3の側注：（One More）★★★
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% 側注1
${}_{12}\mathrm{C}_3=\frac{12\cdot 11\cdot 10}{3\cdot 2\cdot 1}=220$

% 側注2
赤玉5個と白玉3個を合わせた，合計8個の玉から3個の玉を取り出す．

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% 節末問題A2.1.4の側注：（One More）★★
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% 側注1
すべてが赤玉，またはすべてが青玉のときである．

% 側注2
余事象の確率を考える．

% 側注3
赤玉，白玉，青玉をそれぞれ1個ずつ取り出す．

% 側注4
余事象の確率を考える．

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% 節末問題A2.1.5の側注：（One More）★★★
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% 側注1
グー，チョキ，パーの3通りを$n$人が出す（重複順列）．

% 側注2
$2^n$の手の出し方から，$n$人すべてが同じ手を出す場合（あいこになる場合）の2通りを除く．

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% 節末問題A2.2.1の側注：（One More）★★
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% 側注1
$\underbrace{\bigcirc\bigcirc\times\bigcirc}_{\text{3勝1敗}}\underbrace{\bigcirc}_{\text{Xが勝つ}}$

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% 節末問題A2.2.2の側注：（One More）★★★
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% 側注1
4回とも1，6以外の目が出る（1の目0回，6の目0回）．

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% 節末問題A2.2.3の側注：（One More）★★★★
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% 側注1
当たりくじが出ない確率は，$1-\frac{1}{4}=\frac{3}{4}$

% 側注2
反復試行の確率を考える．また，${}_{n-1}\mathrm{C}_1=n-1(n\geqq 2)$を用いる．

% 側注3
$p_{n+1}$は$p_n$の$n$に$n+1$を代入して求められる．

% 側注4
$4(n-1)>0$である．

% 側注5
$n=2$のとき，$p_2<p_3$，$n=3$のとき，$p_3<p_4$，$n=4$のとき，$p_4=p_5$

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% 節末問題A2.2.4の側注：（One More）★★
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% 側注1
求める確率は，

\begin{align*}
P_F(E)&=\frac{P(F\cap E)}{P(F)}\\
&=\frac{P(E\cap F)}{P(F)}
\end{align*}

% 側注2
aが優勝しなかったとき，bが「$a$は優勝した」とうそをつくときの確率は，$P_{\overline{E}}(F)=\frac{1}{4}$

% 側注3
$E\cap F$と$\overline{E}\cap F$は互いに排反である．

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% 節末問題A2.2.5の側注：（One More）★★★★
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% 側注1
$X=4,5$のときの確率の和を，1から引くことで求めてもよい．

% 側注2
4回とも表のとき，Pの座標は20となり，3回表，1回裏のとき，Pの座標は18となる．

% 側注3
3回とも表のとき，Pの座標は15，6回とも裏のとき，Pの座標は18である．

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% 章末問題A2.1の側注：（One More）★★★
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% 側注1
三角形ができるのは，3点がすべて異なるときである．

% 側注2
向かい合う2個の頂点を結ぶと，直径となる．

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% 章末問題A2.2の側注：（One More）★★★
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% 側注1
3つの方向に移動するので，未知数を3つ設定する．

% 側注2
左の方向に1回，右の方向に3回，上の方向に4回移動する反復試行である．

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% 章末問題A2.3の側注：（One More）★★★★
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% 側注1
最小値が$k$となる確率は，最小値が$k$以上の確率から，$k+1$以上の確率を引く．

% 側注2
$\{3,3,5\},\{3,5,5\}$のとき，${}_3\mathrm{C}_1$通りあり，$\{3,4,5\}$のとき，$3!$通りある．

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% 章末問題A2.4の側注：（One More）★★★
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% 側注1
${}_{n+4}\mathrm{C}_2=\frac{(n+4)(n+3)}{2\cdot 1}$

% 側注2
$n+3\neq 0$で分母・分子を割る．

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% 章末問題A2.5の側注：（One More）★★★★
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% 側注1
試合数は$4,5,6,7$の4つの場合がある．

% 側注2
例えば，Aが勝つことを$\bigcirc$，Aが負けることを$\times$で表し，左から試合順に並べると，
\[
\underbrace{\bigcirc\bigcirc\times\bigcirc}_{\text{3勝1敗}}\underbrace{\bigcirc}_{\text{Aが勝つ}}
\]

% 側注3
(i)と(ii)，(iii)と(iv)，(v)と(vi)は同じ試合数であるから，その確率をそれぞれ足し合わせて1つの枠にまとめている．