Solution
= Solution
The <prior probability> is $1/10$. Using <Bayes' theorem>, the <posterior probability> after two absences is
$$
\boxed{P(\text{lazy}\mid\text{absent})=\frac{(1/4)(1/10)}{(1/4)(1/10)+(1/100)(9/10)}=\frac{25}{34}\approx0.7353.}
$$
Thus the probability rises from \b[10% to about 73.5%] within the assumed attendance model. Two absences are 25 times as likely under the lazy model, overcoming the prior odds of $1:9$; they still do not make the classification certain.