13. Place 5 small balls, red, yellow, blue, white, and black, into 5 boxes, red, yellow, blue, white, and black, respectively, with 1 ball in each box. The probability that the red ball is not in the red box and the yellow ball is not in the yellow box is $\qquad$ .
13.0.65.

Placing 5 balls into 5 boxes has a total of \( N = 5! = 120 \) (ways).
The number of ways where the red ball is not in the red box and the yellow ball is not in the yellow box can be divided into two categories:
(1) The red ball is in the yellow box, in which case there are
$$
n_{1} = 4! = 24 \text{ (ways); }
$$
(2) The red ball is neither in the red box nor in the yellow box, in which case there are
$$
n_{2} = \mathrm{C}_{3}^{1} \mathrm{C}_{3}^{1} \mathrm{~A}_{3}^{3} = 54 \text{ (ways); }
$$

The total number of ways where the red ball is not in the red box and the yellow ball is not in the yellow box is
$$
n = n_{1} + n_{2} = 24 + 54 = 78 \text{ (ways). }
$$

Therefore, the probability that the red ball is not in the red box and the yellow ball is not in the yellow box is
$$
P = \frac{n}{N} = \frac{78}{120} = 0.65 \text{. }
$$