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Another question!

  1. Aug 12, 2009 #1
    I think this is correct. I would appreciate any input.

    Question:
    The potential buyer of a particular engine requires that the engine start successfully 10 consecutive times. Suppose that the probability of a successful start is .990. Consider that the outcomes of attempted starts is independent. What is the probability that the engine is accepted after just 10 starts?

    My solution:

    P(x<=10) = p(2) + p(3) +.....+ p(10)

    where:
    p(2) = P(X = 2) = P(S on #1 and S on #2) = p2
    p(3) = P(S on #3 and S on #2 and F on #1) = (1 – p)p2
    p(4) = P(S on #4 and S on #3 and F on #2) = (1 – p)p2
    p(5) = P(S on #5 and S on #4 and F on #3 and no 2 consecutive S’s on trials prior to #3) = [ 1 – p(2) ](1 – p)p2
    p(6) = P(S on #6 and S on #5 and F on #4 and no 2 consecutive S’s on trials prior to #4) = [ 1 – p(2) – p(3)](1 – p)p2

    In general, for x = 5, 6, 7, …: p(x) = [ 1 – p(2) - … – p(x - 3)](1 – p)p2

    Does this make any sense?
    AJ
     
  2. jcsd
  3. Aug 12, 2009 #2

    mathman

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    I can't make sense of what you wrote. However the probability of 10 successes in a row is simply .9910.
     
  4. Aug 12, 2009 #3
    But what is the question asking for, in terms of probability? That's what I don't understand.
     
  5. Aug 12, 2009 #4

    jgens

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    The buyer will accept the engine if and only if it starts successfully ten consecutive times. This means that the engine is "accepted" if it can start ten times in a row. The question is essentially asking what is the probability of this event occuring? That's why the probability is simply 0.9910 as mathman wrote.
     
  6. Aug 13, 2009 #5
    hmm ok. Thank you for your help.
     
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