2. A box contains 12 good items and 3 defective items. Each time an item is drawn without replacement, the number of defective items $\xi$ drawn before a good item is obtained. The mathematical expectation $\mathrm{E} \xi=$ $\qquad$ .
2. $\frac{24}{91}$.
$\xi$ takes values $0, 1, 2, 3$, and
$$
\begin{array}{l}
P(\xi=0)=\frac{\mathrm{C}_{12}^{1}}{\mathrm{C}_{15}^{1}}=\frac{4}{5}, \\
P(\xi=1)=\frac{\mathrm{C}_{3}^{1} \mathrm{C}_{12}^{1}}{2 \mathrm{C}_{15}^{2}}=\frac{6}{35}, \\
P(\xi=2)=\frac{\mathrm{C}_{3}^{2} \mathrm{C}_{12}^{1}}{2 \mathrm{C}_{15}^{3}}=\frac{18}{455}, \\
P(\xi=3)=\frac{\mathrm{C}_{3}^{3} \mathrm{C}_{12}^{1}}{2 \mathrm{C}_{15}^{4}}=\frac{2}{455} . \\
\text { Therefore, } \mathrm{E} \xi=0 \times \frac{4}{5}+1 \times \frac{6}{35}+2 \times \frac{18}{455}+3 \times \frac{2}{455} \\
=\frac{24}{91} .
\end{array}
$$