【数学A】例題2.2.9：確率の最大値（One More）★★★★

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% 例題A2.2.9：確率の最大値（One More）★★★★
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1個のさいころを14回投げるとき，1の目が出る回数が何回のとき，その確率が最も大きくなるか．

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% 例題A2.2.9の解答（One More）★★★★
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さいころを1回投げたとき，1の目が出る確率は$\frac{1}{6}$であるから，さいころを14回投げたときに1の目が$n$回（$0\leqq n\leqq 14$）出る確率$p_n$は，

\[
p_n={}_{14}\mathrm{C}_n(\frac{1}{6})^n(\frac{5}{6})^{14-n}=\frac{14!}{n!(14-n)!}\cdot\frac{5^{14-n}}{6^{14}}
\]

$n=0,1,2,\ldots,13$において，$p_{n+1}$と$p_n$の比を求めると，

\begin{align*}
\frac{p_{n+1}}{p_n}&=\{\frac{14!}{(n+1)!(13-n)!}\cdot\frac{5^{13-n}}{6^{14}}\}\div\{\frac{14!}{n!(14-n)!}\cdot\frac{5^{14-n}}{6^{14}}\}\\
&=\frac{n!(14-n)!}{(n+1)!(13-n)!}\cdot\frac{5^{13-n}}{5^{14-n}}\\
&=\frac{14-n}{5(n+1)}
\end{align*}

(i) $\frac{p_{n+1}}{p_n}\geqq 1$のとき

$\frac{14-n}{5(n+1)}\geqq 1$より，$14-n\geqq 5(n+1)$であるから，$n\leqq\frac{3}{2}$

したがって，$n=0,1$のとき，$\frac{p_{n+1}}{p_n}>1$より，$p_n<p_{n+1}$

(ii) $\frac{p_{n+1}}{p_n}<1$のとき

$\frac{14-n}{5(n+1)}<1$より，$14-n<5(n+1)$であるから，$n>\frac{3}{2}$

したがって，$n=2,3,\ldots,13$のとき，$\frac{p_{n+1}}{p_n}<1$より，$p_n>p_{n+1}$

(i)，(ii)より，$p_0<p_1<p_2>p_3>p_4>\cdots>p_{13}>p_{14}$

よって，1の目が2回出る確率が最も大きい．

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% 問題A2.2.9：確率の最大値（One More）★★★★
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1個のさいころを18回投げるとき，2の目が出る回数が何回のとき，その確率が最も大きくなるか．

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% 問題A2.2.9の解答（One More）★★★★
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さいころを1回投げたとき，2の目が出る確率は$\frac{1}{6}$であるから，さいころを18回投げたときに2の目が$n$回（$0\leqq n\leqq 18$）出る確率$p_n$は，

\[
p_n={}_{18}\mathrm{C}_n(\frac{1}{6})^n(\frac{5}{6})^{18-n}=\frac{18!}{n!(18-n)!}\cdot\frac{5^{18-n}}{6^{18}}
\]

$n=0,1,2,\ldots,17$において，$p_{n+1}$と$p_n$の比を求めると，

\begin{align*}
\frac{p_{n+1}}{p_n}&=\{\frac{18!}{(n+1)!(17-n)!}\cdot\frac{5^{17-n}}{6^{18}}\}\div\{\frac{18!}{n!(18-n)!}\cdot\frac{5^{18-n}}{6^{18}}\}\\
&=\frac{n!(18-n)!}{(n+1)!(17-n)!}\cdot\frac{5^{17-n}}{5^{18-n}}=\frac{18-n}{5(n+1)}
\end{align*}

(i) $\frac{p_{n+1}}{p_n}\geqq 1$のとき

$\frac{18-n}{5(n+1)}\geqq 1$より，$18-n\geqq 5(n+1)$であるから，$n\leqq\frac{13}{6}$

したがって，$n=0,1,2$のとき，$\frac{p_{n+1}}{p_n}>1$より，$p_n<p_{n+1}$

(ii) $\frac{p_{n+1}}{p_n}<1$のとき

$\frac{18-n}{5(n+1)}<1$より，$18-n<5(n+1)$であるから，$n>\frac{13}{6}$

したがって，$n=3,4,\ldots,17$のとき，$\frac{p_{n+1}}{p_n}<1$より，$p_n>p_{n+1}$

(i)，(ii)より，$p_0<p_1<p_2<p_3>p_4>p_5>\cdots>p_{17}>p_{18}$

よって，2の目が3回出る確率が最も大きい．

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% 例題A2.2.9の側注：確率の最大値（One More）★★★★
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% 側注1
$1$の目が出ない確率は，$1-\frac{1}{6}=\frac{5}{6}$

% 側注2
反復試行の確率を考える．また，${}_n\mathrm{C}_r=\frac{n!}{r!(n-r)!}$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
$(n+1)!=(n+1)\cdot n!,(14-n)!=(14-n)\cdot(13-n)!,5^{14-n}=5^{13-n}\cdot 5$より，約分する．

% 側注5
$5(n+1)>0$である．

% 側注6
$n=0$のとき，$p_0<p_1$，$n=1$のとき，$p_1<p_2$

% 側注7
$n=2$のとき，$p_2>p_3$，$n=3$のとき，$p_3>p_4,\cdots$，$n=13$のとき，$p_{13}>p_{14}$

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% 問題A2.2.9の側注：確率の最大値（One More）★★★★
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% 側注1
反復試行の確率を考える．また，${}_n\mathrm{C}_r=\frac{n!}{r!(n-r)!}$を用いる．

% 側注2
$p_{n+1}$は$p_n$の$n$に$n+1$を代入して求められる．

% 側注3
$(n+1)!=(n+1)\cdot n!,(18-n)!=(18-n)\cdot(17-n)!,5^{18-n}=5^{17-n}\cdot 5$より，約分する．

% 側注4
$n=0$のとき，$p_0<p_1$，$n=1$のとき，$p_1<p_2$，$n=2$のとき，$p_2<p_3$

% 側注5
$n=3$のとき，$p_3>p_4$，$n=4$のとき，$p_4>p_5,\cdots$，$n=17$のとき，$p_{17}>p_{18}$

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% 例題A2.2.9の考え方・補足など：確率の最大値（One More）★★★★
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% 考え方
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さいころを14回投げたときに1の目が$n$回出る確率$p_n$を$n$の式で表す．すると，$p_n=\frac{14!}{n!(14-n)!}\cdot\frac{5^{14-n}}{6^{14}}$とすることができるが，関数と見て最大値を求めるのは難しい．そこで，$p_n,p_{n+1}$の大小関係$(p_n>p_{n+1},p_n<p_{n+1})$を調べることを考える．$A>B$を示すには，$A-B>0$を示す方法などもあるが，ここでは比を考えて1と比べる方法を用いるとよい．

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% One Point
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$n$回起こる確率$p_n$の最大は，$\frac{p_{n+1}}{p_n}$と1の大小を比べるとよい．

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% 余談
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$p_n$をグラフで表すと，右の図のようになり，$n=2$で確率が最大となっている（1の目が2回出る確率が最も大きい）．