Answer:
1/15
Step-by-step explanation:
When we form such three-digit numbers with distinct digits using the digits  1
,
2
,
3
,
5
,
8  and  9  (in all six digits all different), observe that the order of digits does matter. For example, if we make a number using digits  1
,
2
,  and  3
, we can have  123
,
132
,
231
,
213
,
312  or  321
.
Hence we have to find number of  3  digit numbers that can be made from these six digits using permutation and answer is ⁶
P
₃ =  6  ×  5  ×  4  =  120
.
.How haw many of them will have first digit as even, we have two choices  2  and  8
. Once we have chosen  2
 for hundreds place, we can have only  8  in units place and any one of remaining  4  can be used in tens place. Hence four choices, with  2  in hundreds place and another four choices when we have  8  in hundreds place (and  2  in units place) i.e. total 
8  possibilities.
Hence, the probability, that both the first digit and the last digit of the three digit number are even numbers, is  8  /120  =  1
/15
.