【数学A】例題2.2.3：反復試行の確率1（One More）★★

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 例題A2.2.3：反復試行の確率1（One More）★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

1個のさいころを5回投げるとき，次の確率を求めよ．

(1) 6の目がちょうど3回出る確率

(2) 6の目が出る回数が4回以上である確率

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 例題A2.2.3の解答（One More）★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

(1) 1個のさいころを1回投げるとき，6の目が出る確率は，$\frac{1}{6}$

6の目が出ない確率は，$\frac{5}{6}$

よって，求める確率は，

\[
{}_5\mathrm{C}_3(\frac{1}{6})^3(\frac{5}{6})^2=\frac{10\times 1^3\times 5^2}{6^5}=\frac{125}{3888}
\]

(2)

(i) 6の目が4回出るとき

\[
{}_5\mathrm{C}_4(\frac{1}{6})^4(\frac{5}{6})=\frac{5\times 1^4\times 5}{6^5}=\frac{25}{7776}
\]

(ii) 6の目が5回出るとき

\[
(\frac{1}{6})^5=\frac{1}{7776}
\]

よって，(i)，(ii)は互いに排反であるから，求める確率は，

\[
\frac{25}{7776}+\frac{1}{7776}=\frac{13}{3888}
\]

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 問題A2.2.3：反復試行の確率1（One More）★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

1個のさいころを4回投げるとき，次の確率を求めよ．

(1) 1の目がちょうど3回出る確率

(2) 1の目が出る回数が1回以下である確率

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 問題A2.2.3の解答（One More）★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

(1) 1個のさいころを1回投げるとき，1の目が出る確率は，$\frac{1}{6}$

1の目が出ない確率は，$\frac{5}{6}$

よって，求める確率は，

\[
{}_4\mathrm{C}_3(\frac{1}{6})^3(\frac{5}{6})^1=\frac{4\times 1^3\times 5}{6^4}=\frac{5}{324}
\]

(2)

(i) 1の目が0回出るとき

\[
(\frac{5}{6})^4=\frac{625}{1296}
\]

(ii) 1の目が1回出るとき

\[
{}_4\mathrm{C}_1(\frac{1}{6})^1(\frac{5}{6})^3=\frac{4\times 1\times 125}{6^4}=\frac{500}{1296}
\]

よって，(i)，(ii)は互いに排反であるから，求める確率は，

\[
\frac{625}{1296}+\frac{500}{1296}=\frac{125}{144}
\]

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 例題A2.2.3の側注：反復試行の確率1（One More）★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% 側注1
5回のうち6の目が4回出る場合の数は，${}_5\mathrm{C}_4$通り

% 側注2
${}_5\mathrm{C}_5(\frac{1}{6})^5(1-\frac{1}{6})^0$としてもよい．${}_5\mathrm{C}_5=1,(1-\frac{1}{6})^0=1$である．

% 側注3
独立な反復試行である．

% 側注4
$1-\frac{1}{6}=\frac{5}{6}$

% 側注5
${}_5\mathrm{C}_3\underbrace{(\frac{1}{6})^3}_{\text{6が3回}}\underbrace{(\frac{5}{6})^{5-3}}_{\text{6以外が2回}}$

% 側注6
排反な事象であるから，足し合わせる．

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 問題A2.2.3の側注：反復試行の確率1（One More）★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% 側注1
${}_4\mathrm{C}_0(\frac{1}{6})^0(1-\frac{1}{6})^4$としてもよい．${}_4\mathrm{C}_0=1,(\frac{1}{6})^0=1$である．

% 側注2
4回のうち1の目が1回出る場合の数は，${}_4\mathrm{C}_1$通り

% 側注3
独立な反復試行である．

% 側注4
$1-\frac{1}{6}=\frac{5}{6}$

% 側注5
${{}_4\mathrm{C}_3}\underbrace{(\frac{1}{6})^3}_{\text{1が3回}}\underbrace{(\frac{5}{6})^{4-3}}_{\text{1以外が1回}}$

% 側注6
排反な事象であるから，足し合わせる．

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 例題A2.2.3の考え方・補足など：反復試行の確率1（One More）★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

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

さいころを投げる試行は，毎回独立な試行である．6の目が出ることを$\bigcirc$，6の目が出ないことを$\times$で表すと，6の目がちょうど3回出るのは右の表のような場合である．6の目が出る確率は$\frac{1}{6}$，6の目が出ない確率は$1-\frac{1}{6}=\frac{5}{6}$であることを利用する．

(1)は，5回の試行のうち，$\bigcirc$となる3回を選び(${}_5\mathrm{C}_3$通り)，反復試行を考える．(2)は，6の目がそれぞれ4回，5回出るときに分けて考えるとよい．

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 反復試行の確率
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

事象$A$の起こる確率を$p$とすると，試行を$n$回繰り返し行うとき，事象$A$が$r$回起こる確率は，

\[
{}_n\mathrm{C}_rp^r(1-p)^{n-r}
\]