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

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heat23

heat23

Elite Member
Oct 9, 1999
3,998
9
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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!
 
Whitecloak

Whitecloak

Diamond Member
May 4, 2001
6,074
2
0
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)
 
S

skibum827

Junior Member
Sep 11, 2002
11
0
0
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.
 
heat23

heat23

Elite Member
Oct 9, 1999
3,998
9
81
www.heatware.com
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
 
A

Ameesh

Lifer
Apr 3, 2001
23,686
0
0
your first answer and whitecloak's answer the same and they both happen to be the ones i agree with it.
 
HomeBrewerDude

HomeBrewerDude

Lifer
Jan 18, 2001
14,465
1
0
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.
 
A

Ameesh

Lifer
Apr 3, 2001
23,686
0
0
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.
Click to expand...

ohh your correct! it could have been any 1 of those 4 permutations

 
heat23

heat23

Elite Member
Oct 9, 1999
3,998
9
81
www.heatware.com
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 :)
 
heat23

heat23

Elite Member
Oct 9, 1999
3,998
9
81
www.heatware.com
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?
 
heat23

heat23

Elite Member
Oct 9, 1999
3,998
9
81
www.heatware.com
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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