discrete probability distribution (help please!)
posted by anonymous .
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.

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