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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