Solution
= Solution
The <law of total probability> over the two student groups gives
$$
P(\text{absent})=\frac14\frac LN+\frac1{100}\left(1-\frac LN\right),
\qquad\boxed{P(\text{absent})=\frac{24L+N}{100N}.}
$$
= Solution
The <law of total probability> over the two student groups gives
$$
P(\text{absent})=\frac14\frac LN+\frac1{100}\left(1-\frac LN\right),
\qquad\boxed{P(\text{absent})=\frac{24L+N}{100N}.}
$$