Exercise
ABy Admin
Apr 29'23
Answer
Solution: B
[[math]]
\begin{align*}
\operatorname{P}[ D] &= \operatorname{P}[ H ]\operatorname{P}[ D | H ] + \operatorname{P}[ M ]\operatorname{P}[ D | M ] + \operatorname{P}[ L]\operatorname{P}[ D | L] \\
0.009 &= \operatorname{P}[ H ]\operatorname{P}[ D | H ] + \operatorname{P}[M] \left( \frac{1}{2} \operatorname{P}[D | H] \right) + \operatorname{P}[L] \left( \frac{1}{2} \frac{1}{3} \operatorname{P}[D | H] \right) \\
0.009 &= 0.20 \operatorname{P}[ D | H ] + 0.35 \left( \operatorname{P}[ D | H ] \right) + 0.45 \left( \operatorname{P}[ D | H ] \right) = 0.45 \operatorname{P}[ D | H ] \\
\operatorname{P}[ D | H ] &= 0.009/ 0.45 = 0.02
\end{align*}
[[/math]]