Total number of 5 digit numbers obtained by the digits 1, 2, 3, 4, 5, 6 and 8.
=7P5=7.6.5.4.3=2520There are 4 even digits (2, 4, 6 and 8).
∴ 2 even digits can be selected in
4C2=6 ways
∴ The two ends can be filled in 2 × 6 =12 ways
Remaining 3 places from remaining 5 digits can be filled in
5P3=60 ways.
Hence, the required probability
==.