(7) There are 5 pairs of shoes with different sizes. If 4 shoes are taken out, the probability of being able to form exactly one pair is $\qquad$
(7) $\frac{4}{7}$ Hint: $P=\frac{\mathrm{C}_{5}^{1} \mathrm{C}_{4}^{2} \mathrm{C}_{2}^{1} \mathrm{C}_{2}^{1}}{\mathrm{C}_{10}^{4}}=\frac{120}{210}=\frac{4}{7}$.