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0 pointsS4M10y ago0 comments

For 11 I think you're wrong because you forgot the part: "As minutes pass, you begin to vaguely remember lurking in your basement and typing the command ':(){ :|: & };:' in an xterm session just before everything went black.". The fact that you have a machine which has an xterm session means that it's either an OS X or Linux Box, which rules out the possibility of having 3 windows boxes. So you have 1/26 chances of having 3 linux boxes and 3*2/26 chances of having 2 exactly 2 linux boxes, so in total it's 7/26 (or 0.2692307692307692).

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Retr0spectrum10y ago· 2 in thread

Yes, that would make more sense.

oakhaven10y ago

ad 11. I think that all the possible "combinations" are not 27 but 10, because e.g. tuples (Linux, Windows, Linux) and (Linux, Linux, Windows) are identical. I.e you should count combinations not variations.

All the poss combinations: LLL, LLW, LLO, LWW, LWO, LOO, WWW, WWO, OWO, OOO

Then you cross out WWW (xterm constraint), and you have 9 combinations, of which 3 are valid for the condition (at least two "Ls"). So 1/3.

Let me know if I got sth wrong :).

S4MOP10y ago

No, there are 27 combinations(when you still account for the fact that you can have three windows machines).

You can see it in that way:

machine1: L, machine2: W, machine3: O machine1: L, machine2: W, machine3: W

The reason why it matters is that you have more ways to have exactly two linux machines:

m1: L, m2: L, m3: O

m1: L, m2: O, m2: L,

etc.

That gives you 2*3 ways of having exactly two linux machines (you have 3 ways of having exactly one non linux machine, and then this machine can be two different things). If you say that LLO is OLL, you would have only 2 ways of having exactly two linux machines.

Hope this makes sense.

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