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discrete probability distribution (help please!)

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a machine has 7 identical components which function independently. the probability that a component will fail is 0.2. the machine will stop working if more than three components fail. find the probability that the machine will be working.

  • discrete probability distribution (help please!) - ,

    Add the probabilities that 0, 1, 2 or 3 components have failed. That sum will be the probability that the machine works.

    The probability that none have failed is
    P(0) = 0.8^7 = 0.210

    The probability that one part has failed is
    P(1) = 0.8^6*0.2*7 = 0.367

    The probability that two parts have failed is P(2) (0.8)^5*(0.2)^2*[7!/(5!2!)] = 0.275

    The probability that three parts have failed is P(3) = (0.8)^4*(0.2)^30*[7!/4!3!)] = 0.115

    The sum of these probabilities is 0.967.

  • discrete probability distribution (help please!) - ,

    thank you so very much! you're my hero! =D

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