5. Given that among five products, three are qualified and two are defective. Each time one is randomly selected for inspection, and it is not put back after inspection. Then the probability of finding two defective items exactly after three inspections is $\qquad$
5. $\frac{3}{10}$.

Exactly three inspections to find two defective items have three scenarios.
(1) The probability of finding defective items on the first and third inspections is
$$
\frac{2}{5} \times \frac{3}{4} \times \frac{1}{3}=\frac{1}{10}
$$
(2) The probability of finding defective items on the second and third inspections is
$$
\frac{3}{5} \times \frac{2}{4} \times \frac{1}{3}=\frac{1}{10} \text {; }
$$
(3) The probability of all three inspections being non-defective is
$$
\frac{3}{5} \times \frac{2}{4} \times \frac{1}{3}=\frac{1}{10} \text {. }
$$

Therefore, the probability of exactly three inspections to find two defective items is
$$
\text { } \frac{1}{10} + \frac{1}{10} + \frac{1}{10}=\frac{3}{10} \text {. }
$$