【数学A】例題2.1.3：組合せと確率（One More）★

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% 例題A2.1.3：組合せと確率（One More）★
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赤玉8個と白玉4個の合計12個の玉が入っている袋の中から，3個の玉を同時に取り出すとき，次の確率を求めよ．

(1) 3個とも赤玉である確率

(2) 赤玉が2個，白玉が1個である確率

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% 例題A2.1.3の解答（One More）★
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12個の玉から3個の玉を取り出す場合の数は，${}_{12}\mathrm{C}_3$通り

(1) 赤玉8個から3個の玉を取り出す場合の数は，${}_8\mathrm{C}_3$通り

よって，求める確率は，

\[
\frac{{}_8\mathrm{C}_3}{{}_{12}\mathrm{C}_3}=\frac{8\cdot 7\cdot 6}{3\cdot 2\cdot 1}\times\frac{3\cdot 2\cdot 1}{12\cdot 11\cdot 10}=\frac{14}{55}
\]

(2) 赤玉8個から2個を取り出す場合の数は，${}_8\mathrm{C}_2$通り

そのそれぞれについて，白玉4個から1個を取り出す場合の数は，${}_4\mathrm{C}_1$通り

したがって，赤玉2個，白玉1個を取り出す場合の数は，

\[
{}_8\mathrm{C}_2\times{}_4\mathrm{C}_1(\text{通り})
\]

よって，求める確率は，

\[
\frac{{}_8\mathrm{C}_2\times{}_4\mathrm{C}_1}{{}_{12}\mathrm{C}_3}=\frac{8\cdot 7}{2\cdot 1}\times 4\times\frac{3\cdot 2\cdot 1}{12\cdot 11\cdot 10}=\frac{28}{55}
\]

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% 問題A2.1.3：組合せと確率（One More）★
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赤玉9個と白玉6個の合計15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) 4個とも赤玉である確率

(2) 赤玉が3個，白玉が1個である確率

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15個の玉から4個の玉を取り出す場合の数は，${}_{15}\mathrm{C}_4$通り

(1) 赤玉9個から4個の玉を取り出す場合の数は，${}_9\mathrm{C}_4$通り

よって，求める確率は，

\[
\frac{{}_9\mathrm{C}_4}{{}_{15}\mathrm{C}_4}=\frac{9\cdot 8\cdot 7\cdot 6}{4\cdot 3\cdot 2\cdot 1}\times\frac{4\cdot 3\cdot 2\cdot 1}{15\cdot 14\cdot 13\cdot 12}=\frac{6}{65}
\]

(2) 赤玉9個から3個を取り出す場合の数は，${}_9\mathrm{C}_3$通り

そのそれぞれについて，白玉6個から1個を取り出す場合の数は，${}_6\mathrm{C}_1$通り

したがって，赤玉3個，白玉1個を取り出す場合の数は，${}_9\mathrm{C}_3\times{}_6\mathrm{C}_1(\text{通り})$

よって，求める確率は，

\[
\frac{{}_9\mathrm{C}_3\times{}_6\mathrm{C}_1}{{}_{15}\mathrm{C}_4}=\frac{9\cdot 8\cdot 7}{3\cdot 2\cdot 1}\times 6\times\frac{4\cdot 3\cdot 2\cdot 1}{15\cdot 14\cdot 13\cdot 12}=\frac{24}{65}
\]

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% 側注1
合計12個の玉を$\text{赤}_1,\text{赤}_2,$

$\ldots,$ $\text{赤}_8,$ $\text{白}_1,\text{白}_2,\ldots,\text{白}_4$のように，同じ色の玉を区別して考えている．

% 側注2
$\frac{8\cdot 7\cdot 6}{3\cdot 2\cdot 1}\div\frac{12\cdot 11\cdot 10}{3\cdot 2\cdot 1}$

% 側注3
赤玉8個，白玉4個から，赤玉2個，白玉1個を取り出す場合の数を考える．

% 側注4
積の法則を用いる．

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% 側注1
合計15個の玉を$\text{赤}_1,\text{赤}_2,$

$\ldots,$ $\text{赤}_9,$ $\text{白}_1,\text{白}_2,\ldots,\text{白}_6$のように，同じ色の玉を区別して考えている．

% 側注2
赤玉9個，白玉6個から，赤玉3個，白玉1個を取り出す場合の数を考える．

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% 例題A2.1.3の考え方・補足など：組合せと確率（One More）★
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% 考え方
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確率を考えるときは，同じ色の玉を区別して考える（玉それぞれが同じ確率で選ばれる，同様の確からしさ）．試行の内容によって，根元事象の総数を求める．ここでは，並べる（順列を用いる）のではなく，組を作るので組合せを用いて考える．