Chapter 11 - Counting Methods and Probability Theory - 11.5 Probability with the Fundamental Counting Principle, Permutations, and Combinations - Exercise Set 11.5 - Page 724: 30

$0.1$

We know that if we want to select $r$ objects out of $n$ regarding the order, we can do this in $_nP_r=\frac{n!}{(n-r)!}$ ways. Here we have $_5P_3$ options for the numbers: $_{5}C_3=\frac{5!}{2!}=60$ Out of these the following are even and greater than $500$: $\{512,514,524,532,534,542\}$ hence the probability is: $\frac{6}{60}=0.1$