There are 15 CDs in a box. On 10 of the CDs there are saved data files,(adsbygoogle = window.adsbygoogle || []).push({});

and the other CDs have no data files saved on them.

i) Suppose that 12 CDs are randomly selected. Determine the probability that exactly 9 of these CDs selected have saved data files.

I'm not sure how to do this. First I thought some kind of conditional probability? But I'm confused about that.

P(A|B) = P(A intersection B)/P(B)?

Is that simply [P(A)*P(B)]/P(B) ? Wouldn't the P(B)'s cancel out?

If not, then if 12 are selected. The probability of any of them having data is (10/15)*(12/15) = 8/15

Now,

(9/12)*(8/15) = 40% [probability of 9 of those having any data from the 12 selected]

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# Determining probability

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