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Independent Events Probability

  1. Jan 29, 2012 #1
    1. The problem statement, all variables and given/known data
    During a winter season at one ski resort, there are two roads from Area A to Area B and two roads from Area B to Area C. Each of the four roads is blocked by snow with a probability p=.25 independently of the other roads.

    What is the probability that there exists an open route from Area A to Area C

    It was decided to add a new path connecting areas A and C directly. It is also blocked by snow with probability p=.25 independently of all the other paths. Now what is the probability that there is an open route from area A to area C.




    2. Relevant equations



    3. The attempt at a solution
    For the first part I assumed that at least one road had to be open from A to B so I got .4375 which is the same for at least one road from B to C. And then that both A to B and B to C had to be open so I got .191 which was wrong.

    Any help would be appreciated
     
  2. jcsd
  3. Jan 29, 2012 #2
    Well, I'm sure there is probably an easier way but there is nothing wrong with drawing a good old probability tree. It will be quite a big one (as there are four roads and so 4 "rounds" to the tree). But once you've drawn the tree, figure out which branches correspond to it being possible to get from A to C, then figure out their probabilities and you should get the answer.

    James
     
  4. Jan 29, 2012 #3

    LCKurtz

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    Remember that the probability of a road between two points being open is 1 minus the probability that both are blocked.
     
  5. Jan 29, 2012 #4
    Right so what I did to find at least one road being open between A and B or B and C was
    1-(1-.25)(1-.25)= .4375
    and then to find the probability of both being open I did
    .4375*.4375=.191
     
  6. Jan 29, 2012 #5
    But the probability of being blocked is .25, so you've worked out the opposite!
     
  7. Jan 29, 2012 #6
    Ohh so what I would do is 1-(1-.75)(1-.75)= .5
    and then
    .5*.5=.25 which would be the final answer for the first part?
     
  8. Jan 29, 2012 #7
    I agree with the logic, not the answer ;)
     
  9. Jan 29, 2012 #8
    sorry :)
    it would be 1-(1-.75)(1-.75)= .9375
    and then
    .9375*.9375=.8789 right?

    For the second part how would I approach it?
     
  10. Jan 29, 2012 #9
    Sounds good.

    Well now you have the probability that there is a route to C through B that is open, and you also know the probability of getting directly from A to C. Can you think of a way of combining these to get the second answer?
     
  11. Jan 29, 2012 #10
    my though was addition but then the probabilities would be over 1
     
  12. Jan 29, 2012 #11
    Probabilities over 1 are never really a good sign!

    Well imagine you've got two coins, what would be the probability of at least 1 head?
     
  13. Jan 29, 2012 #12
    1-(1-.5)(1-.5)=.75
     
  14. Jan 29, 2012 #13
    Good, so can you see what the probability of at least one route being open is?
     
  15. Jan 29, 2012 #14
    1-(1-.8789)(1-.75)=.9697
     
  16. Jan 29, 2012 #15
    Yep, I would agree
     
  17. Jan 29, 2012 #16
    thank you!
     
  18. Jan 29, 2012 #17
    No problem, hope they're right!
     
  19. Jan 29, 2012 #18

    HallsofIvy

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    That is correct but you did it the hard way. If P(A)= .25 and P(B)= .25, and A and B are independent, [itex]P(A and B)= .25^2= 0.0625[/itex] the probability of both roads being blocked is 0.0625 so the probability that at least one of the roads is not is [itex]1- P(A)P(B)= 1- .25^2= 1- 0.0625= 0.9375[/itex], as you say.
     
  20. Jan 29, 2012 #19

    Ray Vickson

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    I get something different from all of you!

    In the first problem, {AC blocked} = {both AB blocked} or {both BC blocked}, so P{AC blocked} = (1/4)(1/4) + (1/4)(1/4) - (1/4)^4 = 31/256 = .12109375, so P{AC open} = 225/256 = .87890625 . This uses P{A or B} = P{A} + P{B} - P{A & B}.

    RGV
     
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