【数学A】例題2.1.6：確率の加法定理（One More）★

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% 例題A2.1.6：確率の加法定理（One More）★
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赤玉7個と白玉5個の合計12個の玉が入っている袋の中から，4個の玉を同時に取り出すとき，次の確率を求めよ．

(1) 赤玉が3個以上取り出される確率

(2) 4個の玉がすべて同じ色である確率

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% 例題A2.1.6の解答（One More）★
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12個の玉から4個の玉を取り出す方法の総数は，${}_{12}\mathrm{C}_4=495(\text{通り})$

(1) 赤玉7個から3個，白玉5個から1個を取り出す場合の数は，
\[
{}_7\mathrm{C}_3\times{}_5\mathrm{C}_1=35\times 5=175(\text{通り})
\]

赤玉7個から4個を取り出す場合の数は，
\[
{}_7\mathrm{C}_4=35(\text{通り})
\]

よって，求める確率は，

\[
\frac{175}{495}+\frac{35}{495}=\frac{210}{495}=\frac{14}{33}
\]

(2) 白玉5個から4個を取り出す場合の数は，${}_5\mathrm{C}_4=5(\text{通り})$

(1)より，赤玉7個から4個を取り出す場合の数は，35通り

よって，求める確率は，

\[
\frac{5}{495}+\frac{35}{495}=\frac{40}{495}=\frac{8}{99}
\]

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% 問題A2.1.6：確率の加法定理（One More）★
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赤玉6個と白玉5個の合計11個の玉が入っている袋の中から，4個の玉を同時に取り出すとき，次の確率を求めよ．

(1) 赤玉が2個以上取り出される確率

(2) 4個の玉がすべて同じ色である確率

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% 問題A2.1.6の解答（One More）★
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11個の玉から4個の玉を取り出す方法の総数は，${}_{11}\mathrm{C}_4=330(\text{通り})$

(1) 赤玉6個から2個，白玉5個から2個を取り出す場合の数は，
\[
{}_6\mathrm{C}_2\times{}_5\mathrm{C}_2=15\times 10=150(\text{通り})
\]

赤玉6個から3個，白玉5個から1個を取り出す場合の数は，
\[
{}_6\mathrm{C}_3\times{}_5\mathrm{C}_1=20\times 5=100(\text{通り})
\]

赤玉6個から4個を取り出す場合の数は，${}_6\mathrm{C}_4=15(\text{通り})$よって，求める確率は，

\[
\frac{150}{330}+\frac{100}{330}+\frac{15}{330}=\frac{265}{330}=\frac{53}{66}
\]

(2) 白玉5個から4個を取り出す場合の数は，${}_5\mathrm{C}_4=5(\text{通り})$

(1)より，赤玉6個から4個を取り出す場合の数は，15通り

よって，求める確率は，

\[
\frac{5}{330}+\frac{15}{330}=\frac{20}{330}=\frac{2}{33}
\]

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% 例題A2.1.6の側注：確率の加法定理（One More）★
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% 側注1
\begin{align*}
{}_{12}\mathrm{C}_4&=\frac{12\cdot 11\cdot 10\cdot 9}{4\cdot 3\cdot 2\cdot 1}\\
&=495
\end{align*}

% 側注2
\begin{align*}
&(\text{赤玉が3個以上の確率})\\
=&(\text{赤玉3個と白玉1個の確率})\\
&+(\text{赤玉4個の確率})
\end{align*}

% 側注3
「すべて白玉である」と「すべて赤玉である」は互いに排反である．

% 側注4
$P(A\cup B)=P(A)+P(B)$

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% 問題A2.1.6の側注：確率の加法定理（One More）★
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% 側注1
白玉を3個取り出す場合と，白玉を4個取り出す場合を考えて，余事象を用いて求めてもよい．

% 側注2
\begin{align*}
&(\text{赤玉が2個以上の確率})\\
=&(\text{赤玉2個と白玉2個の確率})\\
&+(\text{赤玉3個と白玉1個の確率})\\
&+(\text{赤玉4個の確率})
\end{align*}

% 側注3
「すべて白玉である」と「すべて赤玉である」は互いに排反である．

% 側注4
$P(A\cup B)=P(A)+P(B)$

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% 例題A2.1.6の考え方・補足など：確率の加法定理（One More）★
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% 考え方
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事象$A,B$が互いに排反（$A\cap B=\varnothing$）であるとき，確率の加法定理${P(A\cup B)=P(A)+P(B)}$を考える．

(1) 「赤玉が3個以上」の場合は，「赤玉3個と白玉1個」と「赤玉4個」の場合を合わせたものである．これらは互いに排反であることから，それぞれ確率を求め，確率の加法定理によって足し合わせる．

(2) 「すべて白玉である」と「すべて赤玉である」の2つの場合をそれぞれ考え，確率の加法定理によって足し合わせる．

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% 確率の加法定理
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事象$A,B$が互いに排反（$A\cap B=\varnothing$）であるとき，$P(A\cup B)=P(A)+P(B)$