Probability that the resulting 3 digit number is less than 500

Question

Rework problem 12 from section 2.4 of your text, involving the construction of a number from a set of digits. Assume that 3 digits are selected at random from the set  and are arranged in random order.

What is the probability that the resulting 3-digit number is less than 500? 

Answer 1

Although it is not explicitly stated, I am going to assume the set of digits referred to in this problem is the set of 10 digits with exactly one copy of each digit in the set = {0,1,2,3,4,5,6,7,8,9}.

The total number of three digit numbers that can be created at random from 3 digits out of this set = 1000 minus the numbers in 000-999 inclusive, which contain multiple copies of the same digit. 10 of these numbers are triplets of the same digit. 10 possible pairs of the same digit x 9 possible different digits = 90 numbers that need to be removed fo r each of 3 locations of the different digit in the 3-digit number = 270 numbers, for a total of 10 + 270 = 280 removed numbers. This means that a total of 1000-280=720 numbers can be formed by three different digits. This can also be calculated as 10 choices for the first digit x 9 choices for the second digit x 8 choices for the 3rd digit = 720.

Numbers that are < 500 need to begin with 0, 1, 2, 3 or 4 in the hundreds' place, which is 5 choices. Picking one of these digits, leaves 9 possibilities for the tens' place and 8 possibilities for the ones' place for a total of 5x9x8 = 360 numbers.

So the probability that the resulting number is < 500 is 360/720 = 0.5 or 50%.