【数学A】例題2.1.7：和事象の確率（One More）★★

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% 例題A2.1.7：和事象の確率（One More）★★
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1から99までの番号をつけた99枚のカードがあり，この中から1枚のカードを取り出すとき，その番号が5の倍数または8の倍数である確率を求めよ．

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% 例題A2.1.7の解答（One More）★★
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1枚のカードを取り出す場合の数は，99通り

カードの番号が5の倍数である事象を$A$，8の倍数である事象を$B$とすると，番号が5の倍数または8の倍数である事象は$A\cup B$である．

$A=\{5\cdot 1,5\cdot 2,\ldots,5\cdot 19\}$より，$n(A)=19$

$B=\{8\cdot 1,8\cdot 2,\ldots,8\cdot 12\}$より，$n(B)=12$

したがって，事象$A,B$が起こる確率はそれぞれ，

\[
P(A)=\frac{19}{99},P(B)=\frac{12}{99}
\]

また，事象$A\cap B$は，カードの番号が5の倍数かつ8の倍数，すなわち，40の倍数である事象である．

$A\cap B=\{40,80\}$より，$n(A\cap B)=2$

ゆえに，事象$A\cap B$が起こる確率は，$P(A\cap B)=\frac{2}{99}$

よって，求める確率は，

\[
P(A\cup B)=P(A)+P(B)-P(A\cap B)=\frac{19}{99}+\frac{12}{99}-\frac{2}{99}=\frac{29}{99}
\]

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% 問題A2.1.7：和事象の確率（One More）★★
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1から120までの番号をつけた120枚のカードがあり，この中から1枚のカードを取り出すとき，その番号が6の倍数または7の倍数である確率を求めよ．

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% 問題A2.1.7の解答（One More）★★
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1枚のカードを取り出す場合の数は，120通り

カードの番号が6の倍数である事象を$A$，7の倍数である事象を$B$とすると，番号が6の倍数または7の倍数である事象は$A\cup B$である．

$A=\{6\cdot 1,6\cdot 2,\ldots,6\cdot 20\}$より，$n(A)=20$

$B=\{7\cdot 1,7\cdot 2,\ldots,7\cdot 17\}$より，$n(B)=17$

したがって，事象$A,B$が起こる確率はそれぞれ，

\[
P(A)=\frac{20}{120},P(B)=\frac{17}{120}
\]

また，事象$A\cap B$は，カードの番号が6の倍数かつ7の倍数，すなわち，42の倍数である事象である．

$A\cap B=\{42,84\}$より，$n(A\cap B)=2$

ゆえに，事象$A\cap B$が起こる確率は，$P(A\cap B)=\frac{2}{120}$

よって，求める確率は，

\[
P(A\cup B)=P(A)+P(B)-P(A\cap B)=\frac{20}{120}+\frac{17}{120}-\frac{2}{120}=\frac{7}{24}
\]

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% 例題A2.1.7の側注：和事象の確率（One More）★★
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% 側注1
$5\times 19=95$

% 側注2
$8\times 12=96$

% 側注3
$A\cap B$を忘れないように注意すること．

% 側注4
和事象の確率を考える．

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% 問題A2.1.7の側注：和事象の確率（One More）★★
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% 側注1
$6\times 20=120$

% 側注2
$7\times 17=119$

% 側注3
$A\cap B$を忘れないように注意すること．

% 側注4
和事象の確率を考える．

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% 例題A2.1.7の考え方・補足など：和事象の確率（One More）★★
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% 考え方
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5の倍数である事象と8の倍数である事象は互いに排反ではないので，和事象の確率を用いる．5の倍数である事象と8の倍数である事象をそれぞれ$A,B$とすると，5の倍数かつ8の倍数，すなわち，5と8の最小公倍数である40の倍数は$A\cap B$となり，${P(A\cup B)=P(A)+P(B)-P(A\cap B)}$を考えればよい．

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% 和事象の確率
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$P(A\cup B)=P(A)+P(B)-P(A\cap B)$