4. The morning class schedule needs to arrange Chinese, Math, English, and two self-study periods for a total of 5 periods. The probability that the first period is not Math and the last period is not Chinese is $\qquad$
The total number of arrangements is $A_{5}^{3}=60$, let $A=$ "the first period is not Math, and the last period is not Chinese", then the number of arrangements for event $A$ is $A_{5}^{3}-2 A_{4}^{2}+A_{3}^{1}=39$, so $P(A)=\frac{39}{60}=\frac{13}{20}$.