Here's the question you clicked on:
xEnOnn
Suppose there is a set of numbers: \(\{1,2,3,4,5\}\). I pick from the set of numbers 3 times. Each pick is independent. So there is replacement. Then, what is the probability of me getting a 1, a 3 and a 5? And, what is the probability of getting a 2, two 4s?
This is for the first part or the second part of the question?
why would it be the same for the second one? There is a repeated number in there ie, two 4s.
but there is a replacement. So it doesn't mater....
Actually, I was thinking why wouldn't it be \(\frac{6}{125}\) for the first part, and \(\frac{3}{125}\) for the second part?
you should let people know if the order that you pick then numbers matters.
@Zarkon I think the orders do matter in this case since it is a combination because I just need to get a 1, a 3, and a 5. Similarly for the other combination of 2,4,4.
if the order matter than myko's answer is correct. If the order does not matter then you are correct. whether the order matters or not depends on what the problem is asking for. to me, it sounds like the order does not matter...though the problem could have been written better
oh yea you are right that the order does not matter. I said it wrongly in my previous comment. Thanks for the help! :)