probability that at least 2 must be selected to find a blank

For the life of me, I can't figure out how to do this probability problem. I have tried it over and over, but I know I am missing something. Because everytime my answer is not the same as the correct answer. What am I not doing?

A student has a box containing 25 computer disks, of which 15 are blank and the other 10 are not. If she randomly selects disks one by one, what is the probability that at least two must be selected in order to find one that is blank? (I know the answer is .4, I just don't know how to get it)

This is what I have done so far:
A=15, B=10
P(A)= 15/25 P(B|A)= 14/24
P(A)= 3/5 P(B|A)= 7/12
P(A)= .6 P(B|A)= .58

P(A)*P(B/A) = P(.6)*P(.58) = .35 = not right!!!

I have also tried:
P(B)*P(A|B) = P(.4)*P(.625) = .25 = not right!!!

I know I am not on the right path, obviously it isn't as straight forward as I think its supposed to be.

Re: Urgent Help!

Find the probability that the first selection results in a non-blank disk.

Re: Urgent Help!

selects disks one by one, what is the probability that at least
two must be selected in order to find one that is blank?

Yeah, and that's .4, but remember, two must be selected in order to find one that is blank. My classmate and I tried that, and the professor said it wasn't the right answer.

Re: Urgent Help!

Look at the statement of the problem again. If the first disk is non-blank, then you must select a second disk. Therefore, if you don't get a blank disk the first time, you will select at least two disks. So .4 is the correct answer.

Re: Urgent Help!

I think, the term "at least two" includes cases - N1-B, N1-N2-B, N1-N2-N3-B,...N1-N2...N10-B, AND ALL ADDED UP TOGETHER.

Re: Urgent Help!

I just thought about it, and I finally reasoned out why it was .4. I get it now, thanks, for the help.