# Probability of getting a 3 digit number without repetition.

• Jan 11th 2011, 04:53 AM
FailCalculus
Probability of getting a 3 digit number without repetition.
guys i encountered this problem and i'm kinda having a hard time answering it. can anybody help me if im on the right path and explain what to do next .here's the problem:

given the #'s 0-9

what is the probability of getting a 3 digit number without repetition? how many of these numbers are odd? even? how many of these numbers are greater than 342? and how many of the numbers greater than 342 are even?

so im guessing that the 1st number cant be zero so there can be 9 possible outcomes
the next number will be anything except the 1st number so 9 possible outcomes
and the last number will be anything but the 1st and second so 8 possible outcomes

9 x 9 x 8 = 648

and i'm stuck with these. anyone know what to do next?
• Jan 11th 2011, 05:04 AM
DrSteve
You've computed the total number of 3 digit positive integers without repetition (correctly). What is the total number of 3 digit positive integers (possibly with repetition)? See if this helps get you to the next step.
• Jan 11th 2011, 05:34 AM
FailCalculus
if w/ repetition it will be 9 x 10 x 10 right?

so if i will be needing even the number of even numbers from 648 i must make it

648 - (9 x 8 x 5) ?