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Old 09-18-2002, 03:27 AM   #1
heat23
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Default 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!
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Old 09-18-2002, 03:39 AM   #2
Whitecloak
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Default 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)
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Old 09-18-2002, 06:31 AM   #3
skibum827
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Default 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.
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Old 09-18-2002, 10:45 AM   #4
heat23
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Default 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
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Old 09-18-2002, 10:52 AM   #5
Ameesh
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Default Math: Probability Question

your first answer and whitecloak's answer the same and they both happen to be the ones i agree with it.
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Old 09-18-2002, 11:00 AM   #6
LordSnailz
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Default Math: Probability Question

whitecloak is correct and his answer is worded correctly =)
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Old 09-18-2002, 11:08 AM   #7
HomeBrewerDude
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Default 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. 1
D U U U
  1. 2
U D U U
  1. 3
U U D U
  1. 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.
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Old 09-18-2002, 11:13 AM   #8
HomeBrewerDude
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Default Math: Probability Question

btw,

whitecloak is incorrect
skibum827 is correct

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Old 09-18-2002, 11:27 AM   #9
Ameesh
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Default 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. 1
D U U U
  1. 2
U D U U
  1. 3
U U D U
  1. 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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Old 09-18-2002, 11:41 AM   #10
bmd
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Yamaha & Ski are correct .
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Old 09-18-2002, 01:48 PM   #11
HonkeyDonk
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Default Math: Probability Question

Let me guess:

EE351K w/ Dr. Cogdell?

hated that class...
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Old 09-18-2002, 03:31 PM   #12
heat23
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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
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Old 09-19-2002, 12:55 AM   #13
heat23
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upupup
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Old 09-19-2002, 02:13 AM   #14
heat23
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Default 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?
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Old 09-19-2002, 04:48 AM   #15
DeadHead
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Default Math: Probability Question [New Question for your enjoyment!!]

ahhh don't make me think classes havn't started yet!
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Old 09-19-2002, 05:06 AM   #16
Moonbeam
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Default Math: Probability Question [New Question for your enjoyment!!]

22.16% of the time seems right to me.
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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.
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Old 09-19-2002, 09:06 AM   #17
Krakerjak
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How about the probability of 2 people having the same birthdate, in a room of 40 people??
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Old 09-19-2002, 02:09 PM   #18
heat23
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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.
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Old 09-19-2002, 07:21 PM   #19
heat23
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bump
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