Berkeley CSUA MOTD:Entry 26854
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2025/05/25 [General] UID:1000 Activity:popular
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2002/12/19-20 [Computer/SW/Languages/Java, Computer/Theory] UID:26854 Activity:very high
12/18   Math help!  I'm a software engineer who forgot how to do problems
        like this: I have a bag with 60% black marbles and 40% white.
        Drawing with replacement, how many marbles do I have to draw to
        have a 90% chance of drawing 5 black marbles (among any number
        of white)?  I'd appreciate a formula or explanation because I might
        need to change those numbers.  THANKS!
        \_ Stat 2, kids.
        \_ Couldn't you figure this out using your own logical reasoning?
           \_ No, that is why I am asking.
        \_ Likely wrong:
        * Drawing 5 black marbles on the first try is .6**5 = .07776
        You want to make the chances of 5 black marbles .9, ie,
        * .007776 * numDraws = .9
        ==> numDraws = .9 / .07776 = 11.57
        ==> rounding up, you need 12 draws.
        \_ When you can't find the right forumale, the easy backup plan
           is to write a program that will do X trials for you and find it
           empiracally.
        \_ Real answer:
           The answer is the smallest N, such that the definite integral
           of the binomial distrubution function with parameters n = N,
                \_ too bad you can't take a definite integral
                   of a binomial distribution -- it's discrete.
                   \_ What the hell are you talking about?  Of course you can.
                      Just take the sum.  You are an idiot.  Go away and kill
                      yourself now.
                   \_ So do a discrete summation instead
        \_ Too bad there are only five marbles in the bag (3 black and 2
           white) making a solution impossible. Unless you know the number
           of marbles in the bag, you can't give a reasonable answer. As it
           approaches a very high number (not infinity per se), the above
           solution becomes more valid.
           p = 0.6, taken from 5 to N >= .9.  Feel free to play around with
           many java applets of the binomial distribution to find out what
           N is.
           \_ you can also use the chernov bound to bound this integral from
              above. if you're interested in these specfific parameters,
              you should implement what the above says.
              \_ It's a poly time algorithm to find N, no need to bound.
        \_ The question is asking drawing exactly 5 out of N, so should the
           answer be just: 90% = C(N,5)*0.6^5*0.4^(N-5)?
ERROR, url_link recursive (eces.Colorado.EDU/secure/mindterm2) 2025/05/25 [General] UID:1000 Activity:popular
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