Component fail

There is a 90 percent chance that a particular type of component will perform adequately under high temperature conditions. If the device involved has four such components, determine the probability that the device is inoperable because exactly one of the four components fails

Correct result:

p =  70.84 %

Solution:

p1=0.90  C1(4)=(41)=4!1!(41)!=41=4  p2=(41) p141 (1p1)1=4 0.941 (10.9)1=7292500=0.2916  p=100 (1p2)=100 (10.2916)=177125=70.84%p_{1}=0.90 \ \\ \ \\ C_{{ 1}}(4)=\dbinom{ 4}{ 1}=\dfrac{ 4! }{ 1!(4-1)!}=\dfrac{ 4 } { 1 }=4 \ \\ \ \\ p_{2}={ { 4 } \choose 1 } \cdot \ p_{1}^{ 4-1 } \cdot \ (1-p_{1})^1=4 \cdot \ 0.9^{ 4-1 } \cdot \ (1-0.9)^1=\dfrac{ 729 }{ 2500 }=0.2916 \ \\ \ \\ p=100 \cdot \ (1-p_{2})=100 \cdot \ (1-0.2916)=\dfrac{ 1771 }{ 25 }=70.84 \%



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