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 09-18-2002, 03:27 AM #1 heat23 Elite Member     Join Date: Oct 1999 Location: San Diego Posts: 3,983 Math: Probability Question ANOTHER NEW QUESTION! Check my last post NEW QUESTION! A peculiar six-sided die has uneven faces. In particular, the faces showing 1 or 6 are 1 x 1.5 inches, the faces showing 2 or 5 are 1 x 0.4 inches, and the faces showing 3 or 4 are 0.4 x 1.5 inches. Assume that the probability of a particular face coming up is proportional to its area. We independently roll the die twice. What is the probability that we get doubles? So calculating the probabilities for each set of numbers you have... P(1, 6)=.6 P(2, 5)=.16 P(3, 4)=.24 EDIT: Would you do 2(.3)(.3)+2(.08)(.08)+2(.12)(.12) = .2216? Does this mean that the Probility of getting just a 1 is .6?? or is it .3? If you take it as .6, then the sum of the probabilities is 2.0. Im confused about that and everything after that point if someone can help me out I would really appreciate it!! THANKS! Q:An internet access provider (IAP) owns two servers. Each server has a 50% chance of being ?down? independently of the other. Fortunately, only one server is necessary to allow the IAP to provide service to its customers, i.e., only one server is needed to keep the IAP?s system up. Suppose a customer tries to access the internet on four different occasions, which are sufficiently spaced apart in time, so that we may assume that the states of the system corresponding to these four occasions are independent. What is the probability that the customer will only be able to access the internet on 3 out of the 4 occasions? Is the answer (3x3x3x1)/(4x4x4x4) = .105 or (3x3x3x1)x4/(4x4x4x4) = .4218 or neither any help and explanation would be appreciated thanks! __________________ My HeatWare My Blog - Computer Help From A to Z
 09-18-2002, 03:39 AM #2 Whitecloak Diamond Member     Join Date: May 2001 Posts: 6,074 Math: Probability Question well, here's my attempt the probability of each server failing is independent of the other = 0.5 now the ISP is up on exactly 3 out of 4 occasions. = this means that atleast one server was up on those 3 occasions. probabilty that at least one server was up = 1 - probability that both servers were down = 1- (1/2)*(1/2) = 3/4 so probability that the customer is able to access the ISP on exactly 3 out of 4 occasions = probability that atleast one server was up on 3 occasions * probability that both servers were down on 1 occasion= [(3/4)*(3/4)*(3/4)]*(1/4) __________________ hmfg.
 09-18-2002, 06:31 AM #3 skibum827 Junior Member   Join Date: Sep 2002 Posts: 11 Math: Probability Question I think we all agree that the probability of the ISP working properly is 3/4. Now the question is what is the probability that it is working 3 out 4 times? We have 4 situations where this arises. 1. Doesn't work 1st time, but next 3 ( 1/4 * 3/4 * 3/4 * 3/4) 2. Doesn't work 2nd time, but all others ( 3/4 * 1/4 * 3/4 * 3/4) 3. 3rd try didn't work ( 3/4 * 3/4 * 1/4 * 3/4) 4. last try doesn't work ( 3/4 * 3/4 * 3/4 * 1/4) We add all of these together and get (3/4)^3 * (1/4) * 4 = .42 I think this is pretty much what you had, but the question was, where does the last 4 come from. It comes from the fact, that there are 4 scenarios where it could work 3 times and fail one of them. Hope this helps.
 09-18-2002, 10:45 AM #4 heat23 Elite Member     Join Date: Oct 1999 Location: San Diego Posts: 3,983 Math: Probability Question Im getting conflicting answers from both of you Dont know which one to believe if anyone else wants to take a stab at it, i would appreciate it __________________ My HeatWare My Blog - Computer Help From A to Z
 09-18-2002, 10:52 AM #5 Ameesh Lifer   Join Date: Apr 2001 Posts: 23,685 Math: Probability Question your first answer and whitecloak's answer the same and they both happen to be the ones i agree with it. __________________ Xbox Live Gamercard
 09-18-2002, 11:00 AM #6 LordSnailz Diamond Member   Join Date: Nov 1999 Posts: 4,822 Math: Probability Question whitecloak is correct and his answer is worded correctly =)
 09-18-2002, 11:08 AM #7 HomeBrewerDude Lifer     Join Date: Jan 2001 Posts: 14,466 Math: Probability Question You can work this out by hand, step by step: 1) p(at least 1 server is up) = 1-p(no server up) = 1-(p(server1 down) x p(server2 down)) = 1 - .25 = .75 2) p(1 server is up 3 of 4 trials) there are four possible sequences that meet this condition: 1 D U U U2 U D U U3 U U D U 4 U U U D p (D U U U) = .25 x .75 x .75 x .75 = .105 p (U D U U) = .75 x .25 x .75 x .75 = .105 p (U U D U) = .75 x .75 x .25 x .75 = .105 p (U U U D) = .75 x .75 x .75 x .25 = .105 p(any of the above sequences) = 4 x .105 = .420 <--rounding error -------------------------- OR solve this using a binomial distribution where p(success) = .75 (choose 3 of 4) x p(success)^3 x p(fail)^1 = 4!/3!-1! x (.75)^3 x (.25)^1 = 4 x .422 x .25 = . 422 EDIT fixed a rounding error.
 09-18-2002, 11:13 AM #8 HomeBrewerDude Lifer     Join Date: Jan 2001 Posts: 14,466 Math: Probability Question btw, whitecloak is incorrect skibum827 is correct
09-18-2002, 11:27 AM   #9
Ameesh
Lifer

Join Date: Apr 2001
Posts: 23,685
Math: Probability Question

Quote:
 Originally posted by: yamahaXS You can work this out by hand, step by step: 1) p(at least 1 server is up) = 1-p(no server up) = 1-(p(server1 down) x p(server2 down)) = 1 - .25 = .75 2) p(1 server is up 3 of 4 trials) there are four possible sequences that meet this condition: 1 D U U U2 U D U U3 U U D U 4 U U U D p (D U U U) = .25 x .75 x .75 x .75 = .105 p (U D U U) = .75 x .25 x .75 x .75 = .105 p (U U D U) = .75 x .75 x .25 x .75 = .105 p (U U U D) = .75 x .75 x .75 x .25 = .105 p(any of the above sequences) = 4 x .105 = .420 <--rounding error -------------------------- OR solve this using a binomial distribution where p(success) = .75 (choose 3 of 4) x p(success)^3 x p(fail)^1 = 4!/3!-1! x (.75)^3 x (.25)^1 = 4 x .422 x .25 = . 422 EDIT fixed a rounding error.
ohh your correct! it could have been any 1 of those 4 permutations

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 09-18-2002, 11:41 AM #10 bmd Golden Member   Join Date: Feb 2001 Posts: 1,043 Math: Probability Question Yamaha & Ski are correct .
 09-18-2002, 01:48 PM #11 HonkeyDonk Diamond Member   Join Date: Oct 2001 Posts: 4,020 Math: Probability Question Let me guess: EE351K w/ Dr. Cogdell? hated that class...
 09-18-2002, 03:31 PM #12 heat23 Elite Member     Join Date: Oct 1999 Location: San Diego Posts: 3,983 Math: Probability Question nope got dr. baldick...hes alright..its just that i really suck at probability thanks for all the help guys....so it seems like .422 got the most votes __________________ My HeatWare My Blog - Computer Help From A to Z
 09-19-2002, 12:55 AM #13 heat23 Elite Member     Join Date: Oct 1999 Location: San Diego Posts: 3,983 Math: Probability Question [New Question for your enjoyment!!] upupup __________________ My HeatWare My Blog - Computer Help From A to Z
 09-19-2002, 02:13 AM #14 heat23 Elite Member     Join Date: Oct 1999 Location: San Diego Posts: 3,983 Math: Probability Question [New Question for your enjoyment!!] A peculiar six-sided die has uneven faces. In particular, the faces showing 1 or 6 are 1 x 1.5 inches, the faces showing 2 or 5 are 1 x 0.4 inches, and the faces showing 3 or 4 are 0.4 x 1.5 inches. Assume that the probability of a particular face coming up is proportional to its area. We independently roll the die twice. What is the probability that we get doubles? So calculating the probabilities for each set of numbers you have... P(1, 6)=.6 P(2, 5)=.16 P(3, 4)=.24 Would you do 2(.3)(.3)+2(.08)(.08)+2(.12)(.12) = .2216? __________________ My HeatWare My Blog - Computer Help From A to Z
 09-19-2002, 04:48 AM #15 DeadHead Senior Member   Join Date: Jun 2002 Posts: 243 Math: Probability Question [New Question for your enjoyment!!] ahhh don't make me think classes havn't started yet!
 09-19-2002, 05:06 AM #16 Moonbeam Elite Member     Join Date: Nov 1999 Posts: 56,949 Math: Probability Question [New Question for your enjoyment!!] 22.16% of the time seems right to me. __________________ The above is probably just my usual sarcasm and in no way reflects my real opinion (and,or) may include subtleties of sufficient rarity as to appear to the unsuspecting like total gibberish. It may not be so much a matter that I'm far out, but rather that you have never been anywhere.
 09-19-2002, 09:06 AM #17 Krakerjak Senior Member   Join Date: Jul 2001 Posts: 768 Math: Probability Question [New Question for your enjoyment!!] How about the probability of 2 people having the same birthdate, in a room of 40 people??
 09-19-2002, 02:09 PM #18 heat23 Elite Member     Join Date: Oct 1999 Location: San Diego Posts: 3,983 Math: Probability Question [New Question for your enjoyment!!] man this class is really pissin me off..i cant do any of these problems and its due today...btw heres another A parking lot contains 100 cars that all look quite nice from the outside. However, K of these cars happen to be lemons. The number K is known to lie in the range {0, 1, . . . , 9}, with all values equally likely. (a) We testdrive 20 distinct cars chosen at random, and to our pleasant surprise, none of them turns out to be a lemon. Given this knowledge, what is the probability that K = 0? (b) Repeat part (a) when the 20 cars are chosen with replacement; that is, at each testdrive, each car is equally likely to be selected, including those that were selected earlier. __________________ My HeatWare My Blog - Computer Help From A to Z
 09-19-2002, 07:21 PM #19 heat23 Elite Member     Join Date: Oct 1999 Location: San Diego Posts: 3,983 Math: Probability Question [New Question for your enjoyment!!] bump __________________ My HeatWare My Blog - Computer Help From A to Z