Let's see whether I can convince you -- of cause, feel free to point out where I'm wrong. Let's use a simplified version, as yingying suggested, let's consider the case where there are 2(instead of 5) different CD-ROM games, and 3 (instead of 12) boxes. Let's label the 2 games as A, B. Since each box can hold either A or B, so there are totally 2^3 = 8 possibilities. Of these, 2 configurations consists of all A's or B's, hence does not satisfy the requirement. Formally, the number of possibilities that have all 2 games is: 2^3-C(2,1)*1^3 = 6. So the probability to get all 2 games is 6/8 = 0.75.www.ddhw.com You can also use yingying's formula: 1-C(2,1)( 1/2)^3 = 0.75 Or you can simply list all possibilities: AAA BBB AAB ABA BAA ABB BAB BBA Each of these 8 possibilities are equally possible (you agree, right? if not, can you provide your assumption?), and you can see 6 out of 8 satisfy the requirement (except the first 2), so the probability is, again, 6/8 = 0.75. www.ddhw.com So, this means yingying's formula is right at least for this simple case, of cause this does not necessarily mean that it is right for other cases. But before we go further, how about show your method (using ball/hole), and if we get the same answer, we can then proceed to more complicated cases, if not, we can see where the difference came from. www.ddhw.com
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