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?

1 Answer
R. Nguyen
Jan 11, 2018

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#.

Explanation:

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#.

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