A five digit number is formed by the digits 1,2,3,4, 5,6 and 8. The probability that the number has even digit at both ends is

$\frac{2}{7}$

Total number of 5 digit numbers formed with the given digits = ${^7C}_5 × 5!$

There are 4 even digits out of which two even digits (for both ends) can be chosen in ${^4C}_2$ ways. Therefore, the two ends can be filled in $2× {^4C}_2$ ways.

Now, remaining three places can be filled in ${^5C}_2$ × 3!$ ways.

∴ Number of 5 digit, numbers having even digits at both ends

$= 2 × {^4C}_2 ×{^5C}_2 ×3!$

Required probability $=\frac{2 × {^4C}_2 ×{^5C}_2 ×3!}{^7C_5×5!}=\frac{2}{7}$