35. Put 4 identical withered jujubes and 6 different apples into 5 different boxes. What is the probability that each box contains 2 fruits?
35. Putting 4 identical oranges into 5 different boxes has $\mathrm{C}_{4+5-1}^{5-1}=\mathrm{C}_{8}^{4}$ ways, and putting 6 different apples into 5 different boxes has $5^{6}$ ways. In total, there are $\mathrm{C}_{8}^{4} \cdot 5^{6}$ ways.
Each box contains 2 fruits, with the following scenarios:
(1) (Apple, Apple), (Apple, Apple), (Apple, Apple), (Orange, Orange), (Orange, Orange), first choose 2 out of 5 different boxes to place the oranges, since the oranges are identical, just place two in one box, and then place the 6 different apples into the remaining 3 different boxes, with 2 apples in each box. The number of such methods is $\mathrm{C}_{5}^{2} \cdot \frac{6!}{2!2!2!}$.
(2) (Apple, Apple), (Apple, Apple), (Apple, Orange), (Apple, Orange), (Orange, Orange), first choose 1 out of 5 boxes to place 2 oranges, then choose 2 out of the remaining 4 different boxes to place one orange each, and place the 6 apples into the remaining 4 different boxes, with the number of apples in these 4 boxes being $2,2,1,1$. Therefore, there are $\mathrm{C}_{5}^{1} \cdot \mathrm{C}_{4}^{2} \cdot \frac{6!}{2!2!1!1!}$ ways.
(3) (Apple, Apple), (Apple, Orange), (Apple, Orange), (Apple, Orange), (Apple, Orange), similarly, the number of methods is $\mathrm{C}_{5}^{4} \frac{6!}{2!(1!)^{4}}$. In summary, the probability that each box has two fruits is
$$
P=\frac{\mathrm{C}_{5}^{2} \cdot \frac{6!}{2^{3}}+\mathrm{C}_{5}^{1} \cdot \mathrm{C}_{4}^{2} \cdot \frac{6!}{2^{2}}+\mathrm{C}_{5}^{4} \cdot \frac{6!}{2}}{\mathrm{C}_{8}^{4} \cdot 5^{6}} \approx 7.4 \% .
$$