Digit d is randomly selected from the set {1,2,3,4,5,6,7,8,9}. Without replacement of d, another digit, e, is selected. What is the probability that the two-digit number de is a multiple of 3?

Recall that for any multiple of 3, the sum of its digits is also divisible by 3. Now we may reframe this question to be, "what is the probability that #d + e# is divisible by #3#?"

Let's say #d# is #1#. Out of the remaining #9# options, #2#, #5#, and #8# will all give us a multiple of #3#.

Similary, #2# has #1#, #4#, and #7#.

This pattern holds for all integers from 1-9 with the exception of #3#, #6#, and #9# which all have only #2# options as you cannot add the same number.

So, six values for #d# have #3# working values for #e#, and three values have #2# working values. Calculating the number of ways the sum of #d# and #e# is a multiple of 3:

#6 * 3 + 3 * 2#

#= 18 + 6#

#= 24#

And finally, take this as a fraction of the total number of permutations, which is the number of different numbers you can pick for #d#, times the number of ways you can pick #e#. This is #9*8# as we do not replace #d#.

#24/(9*8)#

#=24/72#

#=1/3#

So, the probability that a two-digit number formed from the set of digits #{1, 2, 3, 4, 5, 6, 7, 8, 9}# without replacement happens to be a multiple of #3# is #1/3#.