【数学A】例題2.2.4：反復試行の確率2（One More）★★

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 例題A2.2.4：反復試行の確率2（One More）★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

A，Bの2人が繰り返し卓球の試合をして，先に3勝した方を優勝者とする．各試合においてAが勝つ確率は$\frac{1}{4}$で，引き分けはないものとする．このとき，Aが優勝する確率を求めよ．

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 例題A2.2.4の解答（One More）★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

(i) Aが3勝0敗で優勝する確率は，$(\frac{1}{4})^3=\frac{1}{64}$

(ii) Aが3勝1敗で優勝する確率は，3試合目までに2勝1敗となり，4試合目に勝つ確率であるから，

\[
{}_3\mathrm{C}_2(\frac{1}{4})^2(\frac{3}{4})^1\times\frac{1}{4}=\frac{9}{4^4}=\frac{9}{256}
\]

(iii) Aが3勝2敗で優勝する確率は，4試合目までに2勝2敗となり，5試合目に勝つ確率であるから，

\[
{}_4\mathrm{C}_2(\frac{1}{4})^2(\frac{3}{4})^2\times\frac{1}{4}=\frac{54}{4^5}=\frac{27}{512}
\]

よって，(i)〜(iii)より，求める確率は，

\[
\frac{1}{64}+\frac{9}{256}+\frac{27}{512}=\frac{53}{512}
\]

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 問題A2.2.4：反復試行の確率2（One More）★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

A，Bの2人が繰り返しカードゲームで対戦し，先に3勝した方を優勝者とする．各試合においてAが勝つ確率は$\frac{2}{5}$で，引き分けはないものとする．このとき，Aが優勝する確率を求めよ．

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 問題A2.2.4の解答（One More）★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

(i) Aが3勝0敗で優勝する確率は，$(\frac{2}{5})^3=\frac{8}{125}$

(ii) Aが3勝1敗で優勝する確率は，3試合目までに2勝1敗となり，4試合目に勝つ確率であるから，

\[
{}_3\mathrm{C}_2(\frac{2}{5})^2(\frac{3}{5})^1\times\frac{2}{5}=\frac{72}{5^4}=\frac{72}{625}
\]

(iii) Aが3勝2敗で優勝する確率は，4試合目までに2勝2敗となり，5試合目に勝つ確率であるから，

\[
{}_4\mathrm{C}_2(\frac{2}{5})^2(\frac{3}{5})^2\times\frac{2}{5}=\frac{432}{5^5}=\frac{432}{3125}
\]

よって，(i)〜(iii)より，求める確率は，

\[
\frac{8}{125}+\frac{72}{625}+\frac{432}{3125}=\frac{992}{3125}
\]

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 例題A2.2.4の側注：反復試行の確率2（One More）★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% 側注1
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，

\[
\bigcirc\bigcirc\underbrace{\bigcirc}_{\text{Aが勝つ}}
\]

% 側注2
$\underbrace{\bigcirc\times\bigcirc}_{\text{2勝1敗}}\underbrace{\bigcirc}_{\text{Aが勝つ}}$

% 側注3
$\underbrace{\bigcirc\times\times\bigcirc}_{\text{2勝2敗}}\underbrace{\bigcirc}_{\text{Aが勝つ}}$

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 問題A2.2.4の側注：反復試行の確率2（One More）★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% 側注1
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，

\[
\bigcirc\bigcirc\underbrace{\bigcirc}_{\text{Aが勝つ}}
\]

% 側注2
$\underbrace{\bigcirc\times\bigcirc}_{\text{2勝1敗}}\underbrace{\bigcirc}_{\text{Aが勝つ}}$

% 側注3
$\underbrace{\bigcirc\times\times\bigcirc}_{\text{2勝2敗}}\underbrace{\bigcirc}_{\text{Aが勝つ}}$

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 例題A2.2.4の考え方・補足など：反復試行の確率2（One More）★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 考え方
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Aが優勝するのは，3勝0敗，3勝1敗，3勝2敗の場合があり，優勝する人は最後の試合で勝つことに注意する．例えば，3勝1敗の場合，右のように3試合目までに2勝1敗で，4試合目に勝つ確率を考える．

[width=5.5cm]$\underbrace{\text{1試合目},\text{2試合目},\text{3試合目}}_{\text{2勝1敗}},\underbrace{\text{4試合目}}_{\text{Aが勝つ}}$

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 注意
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

例えば(iii)において，3勝2敗であることから，${}_5\mathrm{C}_3(\frac{1}{4})^3(\frac{3}{4})^2$としないように注意すること．右のように，4試合で優勝が決まる場合なども含めて考えてしまうので誤りである．

\[
\underbrace{\bigcirc\times\bigcirc\bigcirc}_{\text{4試合で優勝が決まる}}\times
\]