8 There are 5 table tennis balls, among which 3 are new balls, and 2 are old balls (i.e., balls that have been used at least once). For each match, 2 balls are taken out to use, and after use, they are all put back. Let the number of new balls taken in the second match be $\xi$, then the mathematical expectation $E \xi=$ $\qquad$
$8 \frac{18}{25}$ Hint: Assume the number of new balls taken in the first game is $m$. Then
$$
\begin{array}{l}
P(m=0)=\frac{\mathrm{C}_{2}^{2}}{\mathrm{C}_{5}^{2}}=\frac{1}{10}, \\
P(m=1)=\frac{\mathrm{C}_{3}^{1} \mathrm{C}_{2}^{1}}{\mathrm{C}_{5}^{2}}=\frac{6}{10}, \\
P(m=2)=\frac{\mathrm{C}_{3}^{2}}{\mathrm{C}_{5}^{2}}=\frac{3}{10} \text {. } \\
P(\xi=0)=P(m=0) \times \frac{\mathrm{C}_{2}^{2}}{\mathrm{C}_{5}^{2}}+P(m=1) \times \frac{\mathrm{C}_{3}^{2}}{\mathrm{C}_{5}^{2}}+P(m=2) \times \frac{\mathrm{C}_{4}^{2}}{\mathrm{C}_{5}^{2}} \\
=\frac{1}{10} \times \frac{1}{10}+\frac{6}{10} \times \frac{3}{10}+\frac{3}{10} \times \frac{6}{10}=\frac{37}{100}, \\
P(\xi=1)=P(m=0) \times \frac{\mathrm{C}_{3}^{1} \mathrm{C}_{2}^{1}}{\mathrm{C}_{5}^{2}}+P(m=1) \times \frac{\mathrm{C}_{2}^{1} \mathrm{C}_{3}^{1}}{\mathrm{C}_{5}^{2}}+ \\
P(m=2) \times \frac{\mathrm{C}_{1}^{1} \mathrm{C}_{4}^{1}}{\mathrm{C}_{5}^{2}} \\
=\frac{1}{10} \times \frac{6}{10}+\frac{6}{10} \times \frac{6}{10}+\frac{3}{10} \times \frac{4}{10}=\frac{54}{100} \text {, } \\
P(\xi=2)=P(m=0) \times \frac{\mathrm{C}_{3}^{2}}{\mathrm{C}_{5}^{2}}+P(m=1) \times \frac{\mathrm{C}_{2}^{2}}{\mathrm{C}_{5}^{2}} \\
=\frac{1}{10} \times \frac{3}{10}+\frac{6}{10} \times \frac{1}{10}=\frac{9}{100} \text {. } \\
\end{array}
$$

Therefore,
$$
E \xi=0 \times \frac{37}{100}+1 \times \frac{54}{100}+2 \times \frac{9}{100}=\frac{18}{25} .
$$