The lifetime (in hours) of an electronic part is a RVdistribution with the follo
ID: 2913506 • Letter: T
Question
The lifetime (in hours) of an electronic part is a RVdistribution with the following density function:F(x) = (1/100)e-x/100 , x > 0
0, elsewhere
Three of these parts operate independently inside a complexelectronic equipment. The
equipment fails if at least two of the parts fail. Find theprobability that the equipment
operates for at least 200 hours without failure.
Explanation / Answer
The lifetime (inhours) of an electronic part is a RV distribution with thefollowing density function: F(x) = (1/100)e-x/100 , x > 0 0, elsewhere Three of these parts operate independently inside a complexelectronic equipment. The equipment fails if at least two of the parts fail. Find theprobability that the equipment operates for at least 200 hours withoutfailure. {Probability Each Device Works 200hours Or More} = p = = 200inff(x) dx = 200inf(1/100)*exp(-x/100)dx = [ -exp(-x/100) ]200inf = (0)+ exp(-(200)/100) = exp(-2) = 0.1353 Prob{Equipment Operates200 hours Or More} = = Prob{2 Or More Devices Operate 200hours Or More} = Binomial{N = 3; k = 2 to3; p = 0.1353} = C(3,2)*((0.1353)^2)*(1- 0.1353)^1 + C(3, 3)*((0.1353)^3)*(1 - 0.1353)^0 = 0.04996 .
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