From: rusin@washington.math.niu.edu (Dave Rusin) Newsgroups: sci.math Subject: Re: Probability question Date: 5 Sep 1995 18:51:12 GMT In article <424tkj$bf9@acme.freenet.columbus.oh.us>, Kreig Babbert wrote: >I could use some help with calculating some probabilities. This is the >situation. >I have three automated tape silos numbered D, E, and F. >Silo D has 12 tape drives, silo E has 12 tape drives, and silo F has 8 >tape drives, for a total of 32 drives. > >I want to have at least one tape drive empty in each silo at any given >point in time. How many of the 32 tape drives should I keep empty to >give me a 50% and/or 75% chance that there will be an empty tape drive >in each silo. The use of tape drives across all 32 drives is random. >I can set the system to use a maximum of say 27 drives at any one time. >Is that sufficient (or too much) to provide a probability that there will >be an empty drive in each silo 75% of the time? Let's see if I understand what you want. You hope to choose a number N of drives which will be used at once, with the following property: if up to N drives out of the 32 are selected at random, then the probabilty that (at least one of the first 12 drives is not selected) AND (at least one of the next 12 drives is not selected) AND (at least one of the last 8 drives is not selected) be greater than 50% (or 75%). Let's try N=27 as you suggest. I hope you'll be able to adapt this argument to consider other similar possibilities and select the one that's best for you. The most critical situation for you is if indeed the maximum 27 selections are made. Then you have only 5 unselected drives. There are 32 choose 5 = 201376 ways of marking these 5 drives from among the 32. Which among these ways bother you? You're concerned about having, say, 3 unselected drives in the first silo and 2 in the last, since that means all drives in the middle silo are selected. This particular arrangement happens in (12 choose 3)*(8 choose 2)= (220)*(28)=6160 ways. But there are other worrisome combinations too. Here's the list of "bad" arrangements showing the numbers of unselected drives in the three silos, together with the count of ways such a selection can be made: 5, 0, 0 792 4, 1, 0 5940 4, 0, 1 3960 3, 2, 0 14520 3, 0, 2 6160 2, 3, 0 14520 2, 0, 3 3696 1, 4, 0 5940 1, 0, 4 840 0, 5, 0 792 0, 4, 1 3960 0, 3, 2 6160 0, 2, 3 3960 0, 1, 4 840 0, 0, 5 56 which, if I haven't slipped with the pencil, sum to 71872. Thus the likelihood that one of these undesirable arrangements shows up is 71872/201376=.3569..., i.e., you'll be safe 64.31% of the time. (I hope the 8-drive silo is the most resilient, given your description of the system usage, since it will have all its drives operating at once with greater likelihood than the other two silos.) dave