Question 1001229
The probability of failure can be modeled as a function of the number of components that fail,
{{{P(x)=C(100,x)*(1-0.0022)^(100-x)(0.0022)^(x)}}}
So for between 1 and 3 components failing calculate {{{P(1)}}}, {{{P(2)}}}, and {{{P(3)}}} and sum the probabilities.
{{{C(100,1)=100}}}
{{{C(100,2)=4950}}}
{{{C(100,3)=161700}}}
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{{{P(1)= 100*(0.9978)^99*(0.0022)^1 =0.1769}}}
{{{P(2)= 4950*(0.9978)^98*(0.0022)^2 =0.01931}}}
{{{P(3)= 161700*(0.9978)^97*(0.0022)^3 =0.001391}}}

So then,
{{{P=0.1769+0.01931+0.001391}}}
{{{P=0.1976}}}