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1.5(易) 条件付き確率の典型問題

Last updated at Posted at 2021-07-08

方針

「0を送信する」という事象をA,「1を送信する」という事象をA^cと置き,「0を受信する」という事象をB,「1を受信する」という事象をB^cと置き,条件付き確率を計算する.1.の結果を2.で利用する.

答案

「0を送信する」という事象をA,「1を送信する」という事象をA^cと置き,「0を受信する」という事象をB,「1を受信する」という事象をB^cと置く.今,問題文より

P[A]=\frac{1}{3},\ P[A^c]=\frac{2}{3},\ P[B^c|A]=P[B|A^c]=\frac{1}{10}

であることが分かっている.

1.

\begin{align}
P[B]&= P[A\cap B]+P[A^c\cap B]\\
&= P[A]P[B|A]+P[A^c]P[B|A^c]\\
&= \frac{1}{3}\frac{9}{10}+\frac{2}{3}\frac{1}{10}\\
&= \frac{11}{30}
\end{align}

2.

\begin{align}
P[A^c|B]&= \frac{P[A\cap B^c]}{P[B]}\\
&= \frac{P[A^c]P[B|A^c]}{P[B]}\\
&= \frac{\frac{2}{3}\frac{1}{10}}{\frac{11}{30}}\\
&= \frac{2}{11}\\
P[A|B^c]&= \frac{P[A\cap B^c]}{P[B^c]}\\
&= \frac{P[A]P[B^c|A]]}{P[B^c]}\\
&= \frac{\frac{1}{3}\frac{1}{10}}{\frac{19}{30}}\\
&= \frac{1}{19}
\end{align}

参考文献

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