91Ó°ÊÓ

The electric system contains 10 components.

The probability of failing each individual component is 0.2.

Given that at least one of the components has failed.

Let, X, denote the number of components.

The number X of components that fail will have the binomial distribution with parameters \(n = 10\)and \(p = 0.2\).

The p.m.f is given by

\(\begin{array}{c}f\left( x \right) = \left( {\begin{array}{*{20}{c}}n\\x\end{array}} \right){p^x}{\left( {1 - p} \right)^{1 - x}}\\ = \left( {\begin{array}{*{20}{c}}{10}\\x\end{array}} \right){\left( {0.2} \right)^x}{\left( {1 - 0.2} \right)^{1 - x}}\end{array}\)

The probability that at least two of the components have failed is \(P\left( {X \ge 2\left| {X \ge 1} \right.} \right)\).

\(\begin{align}P\left( {X \ge 2\left| {X \ge 1} \right.} \right) &= \frac{{P\left( {X \ge 2} \right)}}{{P\left( {X \ge 1} \right)}}\\ &= \frac{{1 - P\left( {X = 0} \right) - P\left( {X = 1} \right)}}{{1 - P\left( {X = 0} \right)}}\end{align}\).

When \(X = 0\),

\(\begin{array}{c}P\left( {X = 0} \right) = \left( {\begin{array}{*{20}{c}}n\\x\end{array}} \right){p^x}{\left( {1 - p} \right)^{1 - x}}\\ = \left( {\begin{array}{*{20}{c}}{10}\\0\end{array}} \right){\left( {0.2} \right)^0}{\left( {1 - 0.2} \right)^{1 - 0}}\\ = 0.1074\end{array}\).

When\(X = 1\),

\(\begin{array}{c}P\left( {X = 1} \right) = \left( {\begin{array}{*{20}{c}}n\\x\end{array}} \right){p^x}{\left( {1 - p} \right)^{1 - x}}\\ = \left( {\begin{array}{*{20}{c}}{10}\\1\end{array}} \right){\left( {0.2} \right)^1}{\left( {1 - 0.2} \right)^{1 - 1}}\\ = 0.2684\end{array}\)

Now, the probability is

\(\begin{align}P\left( {X \ge 2\left| {X \ge 1} \right.} \right) &= \frac{{1 - 0.1074 - 0.2684}}{{1 - 0.1074}}\\ &= \frac{{0.6242}}{{0.8926}}\\ &= 0.6993\end{align}\)

Hence, the probability that at least two of the components have failed is 0.6993.