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Conditional probability problem - help need

  1. Apr 5, 2007 #1
    hi

    I got a stats problem infornt of me. I figured out that it is abaut conditional probability. But I am stuck :confused: .

    # hurricanes 0 1 2 3 4 5 6
    probability .25 .33 .24 .11 .04 .02 .01

    prob >6 is 0
    questions are independent.
    a.) what is prob 2< X <5?
    b.) 2 hurr are already observed. what is the prob for 4 or more?
    c.) 1 hurr is already observed. what is the prob for at least 1 or more?
    thanks
     
  2. jcsd
  3. Apr 5, 2007 #2

    arildno

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    Let us take b) as an example:

    Now, we must have P(4)=P(2)*P(4|2)
    P(4|2) is the one you are looking after; that is:

    P(4|2)=P(4)/P(2)=0.04/0.24=4/24=1/6.
     
  4. Apr 6, 2007 #3
    thanks arildno

    but one doubt.
    as the question asks for prob of 4 or more, do i have to find for 5 and 6 the same way and add the probs?
     
  5. Apr 7, 2007 #4

    arildno

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    Dear oh dear, I made a grievous mistake here!

    In the conditional probability, I should have used the "probability of getting a first hurricane", P(>=1), rather than "the prob of getting exactly 1 hurricane", P(1).
    Sorry about that!

    Now, P(>=1) equals the sum of the probabilities of non-zero events, i.e P(>=1)=0.75.
    In addition, I misreadread in b) the probability to be that of EXACTLY 4, rather than 4 or more.

    Thus, in b) P(>=4|>=1)=P(>=4)/P(>=1), P(>=4)=0.07

    And in c)
    P(>=2)|>=1)=P(>=2)/P(>=1), P(>=2)=0.42.
     
  6. Apr 8, 2007 #5
    thanks a lot arildno
     
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