Posted by **anonymous** on Monday, April 12, 2010 at 9:52pm.

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!) -
**drwls**, Monday, April 12, 2010 at 10:18pm
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!) -
**anonymous**, Monday, April 12, 2010 at 10:24pm
thank you so very much! you're my hero! =D

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