

09182002, 03:27 AM

#1

Elite Member
Join Date: Oct 1999
Location: San Diego
Posts: 3,944

Math: Probability Question
ANOTHER NEW QUESTION! Check my last post
NEW QUESTION!
A peculiar sixsided 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!



09182002, 03:39 AM

#2

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.



09182002, 06:31 AM

#3

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.



09182002, 10:45 AM

#4

Elite Member
Join Date: Oct 1999
Location: San Diego
Posts: 3,944

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



09182002, 10:52 AM

#5

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.



09182002, 11:00 AM

#6

Diamond Member
Join Date: Nov 1999
Posts: 4,822

Math: Probability Question
whitecloak is correct and his answer is worded correctly =)



09182002, 11:08 AM

#7

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) = 1p(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 U  2
U D U U  3
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.



09182002, 11:13 AM

#8

Lifer
Join Date: Jan 2001
Posts: 14,466

Math: Probability Question
btw,
whitecloak is incorrect
skibum827 is correct



09182002, 11:27 AM

#9

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) = 1p(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 U 2
U D U U 3
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



09182002, 11:41 AM

#10

Golden Member
Join Date: Feb 2001
Posts: 1,043

Math: Probability Question
Yamaha & Ski are correct .



09182002, 01:48 PM

#11

Diamond Member
Join Date: Oct 2001
Posts: 4,020

Math: Probability Question
Let me guess:
EE351K w/ Dr. Cogdell?
hated that class...



09182002, 03:31 PM

#12

Elite Member
Join Date: Oct 1999
Location: San Diego
Posts: 3,944

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



09192002, 12:55 AM

#13

Elite Member
Join Date: Oct 1999
Location: San Diego
Posts: 3,944

Math: Probability Question [New Question for your enjoyment!!]
upupup



09192002, 02:13 AM

#14

Elite Member
Join Date: Oct 1999
Location: San Diego
Posts: 3,944

Math: Probability Question [New Question for your enjoyment!!]
A peculiar sixsided 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?



09192002, 04:48 AM

#15

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!



09192002, 05:06 AM

#16

Elite Member
Join Date: Nov 1999
Posts: 53,934

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.



09192002, 09:06 AM

#17

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??



09192002, 02:09 PM

#18

Elite Member
Join Date: Oct 1999
Location: San Diego
Posts: 3,944

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.



09192002, 07:21 PM

#19

Elite Member
Join Date: Oct 1999
Location: San Diego
Posts: 3,944

Math: Probability Question [New Question for your enjoyment!!]
bump



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