【数学A】例題2.2.11：確率の乗法定理1（One More）★★

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% 例題A2.2.11：確率の乗法定理1（One More）★★
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当たりくじが3本入っている10本のくじがある．$\mathrm{a},\mathrm{b}$がこの順にくじを1本ずつ引くとき，次の確率を求めよ．ただし，引いたくじは戻さないものとする．

(1) a，bがともに当たりくじを引く確率

(2) bが当たりくじを引く確率

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% 例題A2.2.11の解答（One More）★★
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$\mathrm{a},\mathrm{b}$が当たりくじを引く事象をそれぞれ$A,B$とする．

(1) $P(A)=\frac{3}{10},P_A(B)=\frac{2}{9}$であるから，乗法定理より，求める確率$P(A\cap B)$は，

\[
P(A\cap B)=P(A)\times P_A(B)=\frac{3}{10}\times\frac{2}{9}=\frac{1}{15}
\]

(2)

(i) aもbも当たりくじを引くとき

(1)より，$P(A\cap B)=\frac{1}{15}$

(ii) aがはずれくじを引き，bが当たりくじを引くとき

$P(\overline{A})=\frac{7}{10},P_{\overline{A}}(B)=\frac{3}{9}$であるから，乗法定理より，その確率は，

\[
P(\overline{A}\cap B)=P(\overline{A})\times P_{\overline{A}}(B)=\frac{7}{10}\times\frac{3}{9}=\frac{7}{30}
\]

よって，(i)，(ii)は互いに排反であるから，求める確率は，

\[
P(B)=P(A\cap B)+P(\overline{A}\cap B)=\frac{1}{15}+\frac{7}{30}=\frac{3}{10}
\]

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% 問題A2.2.11：確率の乗法定理1（One More）★★
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当たりくじが4本入っている13本のくじがある．$\mathrm{a},\mathrm{b}$がこの順にくじを1本ずつ引くとき，次の確率を求めよ．ただし，引いたくじは戻さないものとする．

(1) a，bがともに当たりくじを引く確率

(2) bが当たりくじを引く確率

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% 問題A2.2.11の解答（One More）★★
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$\mathrm{a},\mathrm{b}$が当たりくじを引く事象をそれぞれ$A,B$とする．

(1) $P(A)=\frac{4}{13},P_A(B)=\frac{3}{12}$であるから，乗法定理より，求める確率$P(A\cap B)$は，

\[
P(A\cap B)=P(A)\times P_A(B)=\frac{4}{13}\times\frac{3}{12}=\frac{1}{13}
\]

(2)

(i) aもbも当たりくじを引くとき

(1)より，$P(A\cap B)=\frac{1}{13}$

(ii) aがはずれくじを引き，bが当たりくじを引くとき

$P(\overline{A})=\frac{9}{13},P_{\overline{A}}(B)=\frac{4}{12}$であるから，乗法定理より，その確率は，

\[
P(\overline{A}\cap B)=P(\overline{A})\times P_{\overline{A}}(B)=\frac{9}{13}\times\frac{4}{12}=\frac{3}{13}
\]

よって，(i)，(ii)は互いに排反であるから，求める確率は，

\[
P(B)=P(A\cap B)+P(\overline{A}\cap B)=\frac{1}{13}+\frac{3}{13}=\frac{4}{13}
\]

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% 例題A2.2.11の側注：確率の乗法定理1（One More）★★
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% 側注1
$P_{\overline{A}}(B)$は，aが引いたくじがはずれであるとき，残るくじは9本あり，その中に当たりくじは3本含まれるので，
\[
P_{\overline{A}}(B)=\frac{3}{10-1}=\frac{3}{9}
\]

% 側注2
$P_A(B)$は，aが引いたくじが当たりであるとき，残るくじは9本あり，その中に当たりくじは2本含まれるので，
\[
P_{A}(B)=\frac{3-1}{10-1}=\frac{2}{9}
\]

% 側注3
$P(A)=P(B)$であることがわかる．

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% 問題A2.2.11の側注：確率の乗法定理1（One More）★★
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% 側注1
$P_{\overline{A}}(B)$は，aが引いたくじがはずれであるとき，残るくじは12本あり，その中に当たりくじは4本含まれるので，
\[
P_{\overline{A}}(B)=\frac{4}{13-1}=\frac{4}{12}
\]

% 側注2
$P_A(B)$は，aが引いたくじが当たりであるとき，残るくじは12本あり，その中に当たりくじは3本含まれるので，
\[
P_{A}(B)=\frac{4-1}{13-1}=\frac{3}{12}
\]

% 側注3
$P(A)=P(B)$であることがわかる．

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% 例題A2.2.11の考え方・補足など：確率の乗法定理1（One More）★★
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% 考え方
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順列の考え方を用いても解くことができるが，確率の乗法定理$P(A\cap B)=P(A)P_{A}(B)$を用いて解くことを考える．a，bの順にくじを引き，引いたくじは戻さないことから，くじの残りの本数に注目して確率を計算する．

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% 余談
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$a$が当たりくじを引く確率は$\frac{3}{10}$であり，これは(2)で求めた$b$が当たりくじを引く確率と等しい．一般に，当たりくじを引く確率はくじを引く順番に関係なく一定であり，$n$本のくじの中に$a$本の当たりくじが入っているとき，$k$番目$(k=1,2,\cdots,n)$にくじを引く人が当たりくじを引く確率は常に$\frac{a}{n}$である．