現在,One Moreの数学II・B版を作成中です!

【数学A】例題2.2.5:3つの事象に関する反復試行の確率(One More)★★★

【数学A】例題2.2.5:3つの事象に関する反復試行の確率(One More)
【数学A】例題2.2.5:3つの事象に関する反復試行の確率の例題ページ
問題の解答

【数学A】問題2.2.5:3つの事象に関する反復試行の確率の解答
検索用コード(LaTeX)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 例題A2.2.5:3つの事象に関する反復試行の確率(One More)★★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

赤玉1個,白玉2個,青玉3個が入っている袋の中から,1個の玉を取り出し,色を調べてからもとに戻すことを5回行うとき,次の確率を求めよ.

(1) 赤玉が1回,白玉が2回,青玉が2回出る確率

(2) 赤玉と白玉が出る回数が同じである確率

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 例題A2.2.5の解答(One More)★★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

この袋から玉を1個取り出すとき,赤玉,白玉,青玉が出る確率は,それぞれ$\frac{1}{6},\frac{1}{3},\frac{1}{2}$である.

(1) 求める確率は,

\[
\frac{5!}{1!2!2!}\times(\frac{1}{6})^1(\frac{1}{3})^2(\frac{1}{2})^2=\frac{5}{36}
\]

(2) 赤玉と白玉が出る回数が同じであるのは,赤玉と白玉の出る回数が0回,1回,2回の3つの場合がある.

(i) 赤玉と白玉が1回も出ないとき

\[
(\frac{1}{2})^5=\frac{1}{32}
\]

(ii) 赤玉と白玉が1回ずつ出るとき

\[
\frac{5!}{1!1!3!}\times(\frac{1}{6})^1(\frac{1}{3})^1(\frac{1}{2})^3=\frac{5}{36}
\]

(iii) 赤玉と白玉が2回ずつ出るとき

\[
\frac{5!}{2!2!1!}\times(\frac{1}{6})^2(\frac{1}{3})^2(\frac{1}{2})^1=\frac{5}{108}
\]

よって,(i)〜(iii)より,求める確率は,

\[
\frac{1}{32}+\frac{5}{36}+\frac{5}{108}=\frac{187}{864}
\]

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 問題A2.2.5:3つの事象に関する反復試行の確率(One More)★★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

赤玉1個,白玉2個,青玉2個が入っている袋の中から,1個の玉を取り出し,色を調べてからもとに戻すことを5回行うとき,次の確率を求めよ.

(1) 赤玉が1回,白玉が2回,青玉が2回出る確率

(2) 赤玉が出る回数が白玉が出る回数よりも1回だけ多くなる確率

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 問題A2.2.5の解答(One More)★★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

この袋から玉を1個取り出すとき,赤玉,白玉,青玉が出る確率は,それぞれ$\frac{1}{5},\frac{2}{5},\frac{2}{5}$である.

(1) 求める確率は,

\[
\frac{5!}{1!2!2!}\times(\frac{1}{5})^1(\frac{2}{5})^2(\frac{2}{5})^2=\frac{96}{625}
\]

(2) 赤玉が出る回数が白玉が出る回数よりも1回だけ多くなるのは,以下の3つの場合がある.

(i) 赤玉が1回,白玉が0回,青玉が4回出るとき

\[
\frac{5!}{1!4!}\times(\frac{1}{5})^1(\frac{2}{5})^4=\frac{16}{625}
\]

(ii) 赤玉が2回,白玉が1回,青玉が2回出るとき

\[
\frac{5!}{2!1!2!}\times(\frac{1}{5})^2(\frac{2}{5})^1(\frac{2}{5})^2=\frac{48}{625}
\]

(iii) 赤玉が3回,白玉が2回,青玉が0回出るとき

\[
\frac{5!}{3!2!}\times(\frac{1}{5})^3(\frac{2}{5})^2=\frac{8}{625}
\]

よって,(i)〜(iii)より,求める確率は,

\[
\frac{16}{625}+\frac{48}{625}+\frac{8}{625}=\frac{72}{625}
\]

あわせて読みたい
【数学A】2章:確率(基本事項) 検索用コード(LaTeX) 本文・解答側注 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % 基本事項A2.1.1:事象と確率(One More) %%%%%%%%%%%%%%%%%%%%%%%%%%%%...
あわせて読みたい
【数学A】2章:確率(節末問題・章末問題) 節末A2.1.1〜A2.1.5の解答 節末A2.1.1節末A2.1.2節末A2.1.3節末A2.1.4節末A2.1.5 リンク(関連例題) https://onemath.net/onemorea-reidai2-1-2 https://onemath.net/o...
目次