Answer:
Step-by-step explanation:
There are 6 digits given: 1,2,3,5,8 and 9
But we are only taking three at a time.
= 6!/3!
= 6*5*4*3*2/3*2
= 6*5*4
= 120
The first and last digits can only be even if the number takes the one of the forms "2 _ 8" or "8 _ 2". The middle number can be any of the remaining four.
This means the probability of getting a number beginning and ending with an even digits is:
= 8/120
= 1/15 ....
