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
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.