Question

For this problem, assume the balls in the box are numbered 1 through 9, and that an experiment consists of randomly selecting 3 balls one after another without replacement.

What probability should be assigned to the event that at least one ball has an odd number?

Answer 1

So what we DON"T want is the case of "all even"
Let's find the prob of that.
prob of 3 all even = (4/9)(3/8)(2/7) = 1/27


So prob of at least one odd = 1 - 1/27 = 26/27

Answer 2

It says that answer is wrong :(

the probability to each outcome is 1/504
I tried 5/504 for my question and it came out wrong

Answer 3

It comes to the same thing!

Each individual outcome is 1/504 (=(1/9)*(1/8)*(1/7)).

There are 4!=24 ways of picking the even numbers, so the probability of picking all even is
24/504=1/21

This is the same as Reiny's number of 2*3*4/(9*8*7), there was a transcription error at the end and became 1/27 instead.

Proceeding with Reiny's logic, the probability of picking at least one odd ball is therefore 1-1/21=20/21 (or 480/504).