Stats

Fortunately, plane crashes are rare events. Suppose that commercial crashes occur on an average (mean) of 1.1 per year (Hint: Poisson).
a. What is the probability that this year will be crash free?
b. If there is a crash, what is the probability that there will be more than two?

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asked by MoJo
  1. What is the question? Are they asking for the probabilities of 0, 1, 2, 3 etc per year?

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    posted by drwls
  2. a. What is the probability that this year will be crash free?

    b. If there is a crash, what is the probability that there will be more than 2?

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    posted by MoJo
  3. Probability of zero = 1.1^0 e^-1.1 / 0!
    = (1/1 )e^-1.1 = .332
    of 1 = 1.1^1 (.332)/1! = .365
    of 2 = 1.1^2 (.332)/2! = .201

    The sum of the probabilities of zero, 1 and 2 is
    .898
    SO, the probability of more than two is
    1 - .898 = .102
    ten percent. That is scary.

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    posted by Damon

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