Question

Assume that 3 digits are selected at random from the set { 1, 3, 4, 6, 9 } and are arranged in random order.
What is the probability that the resulting 3-digit number is less than 900?

Answer 1

i think this is the same as asking what is the probability that the first digit is not a 9

Answer 2

I don't think the double randomization matters, because there will be still be random choices for each of the digit positions. If that holds true:

There are 60 possible 3-digit numbers that could be chosen, because 5 * 4 * 3 = 60.
There are 12 possible 3-digit numbers beginning with 9, because 4 * 3 = 12.

So (60 - 12) / 60 = 48 / 60 = 0.8 = 80% chance.