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1.6(易) 条件付き確率の典型問題2

Last updated at Posted at 2021-07-08

方針

「陽性反応が出る」という事象をA,「陰性反応が出る」という事象をA^cとし,「疾患がある」という事象をB,「疾患がない」という事象をB^cと置き,条件付き確率を計算する.

答案

「陽性反応が出る」という事象をA,「陰性反応が出る」という事象をA^cとし,「疾患がある」という事象をB,「疾患がない」という事象をB^cと置き,条件付き確率を計算する.今,問題文より

P[A|B^c]=\frac{2}{10},\ P[A^c|B]=\frac{1}{10},\ P[B]=\frac{1}{10}

であるから,

\begin{align}
P[B|A]&=\frac{P[A\cap B]}{P[A]}\\
&= \frac{P[A\cap B]}{P[A\cap B]+P[A\cap B^c]}\\
&= \frac{P[B]P[A|B]}{P[B]P[A|B]+P[B^c]P[A|B^c]}\\
&= \frac{P[B](1-P[A^c|B])}{P[B](1-P[A^c|B])+P[B^c]P[A|B^c]}\\
&= \frac{\frac{1}{10}(1-\frac{1}{10})}{\frac{1}{10}(1-\frac{1}{10})+\frac{2}{10}\frac{9}{10}}\\
&= \frac{1}{3}
\end{align}

参考文献

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