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Probability Problem

  1. Nov 29, 2009 #1
    Problem:
    The probability of defaulting on the nth payment is 0.017n - 0.013 where n is a whole number, n = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
    find:
    1) the probability of making all 10 payments
    2) the probability of making only the first 5 payments
    Note: each payment made are equal

    My idea:

    I was thinking that defaulting means I won't make the payment, so
    is this the right answer to the first question?
    P = ( 1 - (0.017 - 0.013)) * ( 1 - (0.017(2) - 0.013)) * ( 1 - (0.017(3) - 0.013)) * ... * ( 1 - (0.017(10) - 0.013))

    and similarly for the second one:
    P = P = ( 1 - (0.017 - 0.013)) * ( 1 - (0.017(2) - 0.013)) * ( 1 - (0.017(3) - 0.013)) * ( 1 - (0.017(4) - 0.013)) * ( 1 - (0.017(5) - 0.013))

    Does that make sense? Taking the probability for each payment by 1 - failure and then multiply each of them. Am I doing this right? I'm not good at probability yet.
     
  2. jcsd
  3. Nov 30, 2009 #2
    You have the right idea. They are independent events. My TI-86 gets 42.6% possibility of not defaulting on ten payments.

    A simple problem is to consider the possibility of test failure to be 1/4 and they are to be three tests. Then the possibiility of passing all tests is (3/4)^3.
     
    Last edited: Nov 30, 2009
  4. Nov 30, 2009 #3
    It often helps to draw the probability tree - try this and you might spot the term missing from one of your expressions.
     
  5. Nov 30, 2009 #4
    ??? I checked it over with a tree diagram. How am I missing a term?

    (edit): would I have to include the probability of defaulting in the second question? i.e. keep multiplying my answer by (0.017n - 0.013) where n = 6, ... 10 ?

    (edit #2): I got it now. Thanks guys!
     
    Last edited: Nov 30, 2009
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