A probability question

Question

This is a conditional probability question I am struggling to solve and understand. Any insight would be so very useful.  Here is the problem: A five digit number is formed from the digits 1,2,3,4,5. What is the probability that the number ends in the digits 52, given that it is even? Assume no repetition of digits.

Answer 1

the # of ways that ANY EVEN 5 digit number can be formed is 2(4!)--i.e. two choices for units position, then permutation for remaining leading digits.

 

The number of ways that you can form a number as described is the product of all of these:

a) rightmost digit can be 2 and units position is 5 -- only one choice here!

b) the remaining digits can be populated in 3! ways.

 

I think, therefore, that this probability P(---52 | even) = 3! / [2(4!)] = 1/6.

 

Do you agree?