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Binomial distribution

  1. Apr 2, 2009 #1
    Hi Guys,
    I have been given the probability that a drill strikes oil in a region = 0.2.

    I know that if I wanted to find the probabilty of say striking oil 3 times out of 5 wells

    It would be 5Choose3 = 5!/((2!)(3!)) * (1/5)3* (4/5)2 = 0.0512


    My question is how would I go about finding the probability that the third strike occurs on the fifth well?

    regards
    Brendan
     
  2. jcsd
  3. Apr 2, 2009 #2

    HallsofIvy

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    That would be the probability that two strikes were found in four wells times the probability that the fifth well is a strike.
     
  4. Apr 2, 2009 #3
    So,
    It would be 4Choose2 * Px=0.2 = 4!/((2!)(2!)) * (1/5)2* (4/5)2 *(1/5) = 0.03072
    regards
    Brendan
     
  5. Apr 6, 2009 #4
    Is the Expected value of striking oil 3 successive times just 0.2 * 0.2 * 0.2

    = 0.008 ?
    regards
    Brendan
     
  6. Apr 6, 2009 #5

    CRGreathouse

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    That's the probability of striking oil three successive times out of three tries. If you have five tries, the chance is higher:
    SSSnn
    SSSnS
    SSSSn
    SSSSS
    nSSSn
    nSSSS
    nnSSS

    where n is nothing and S is a strike. Your probability covers the first four cases, but the following three cases are also possible.
     
  7. Apr 6, 2009 #6
    So,
    Let say I have 5 tries there are 5P3 = 20 ways of getting 3 in a row.
    and there are 5! = 120 combinations. so would that make the probability 5P3 divide 5! = 1/6 ?

    regards
    Brendan
     
  8. Apr 7, 2009 #7
    Sorry guys,
    I should have said how many wells would have to be drilled to strike oil 3 times in in succession.

    I should be using the Negative binomial for that calculation.

    It is the number of successes k divided by the probabilty of success p

    so k/p = 3/0.2 = 15 wells drilled before success
     
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