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\(\begin{array}{*{20}{c}}{{\rm{P}}\left( {{\rm{\{ \;system lifetime exceeds\;}}{{\rm{t}}_{{\rm{ - 0\} }}}}{\rm{)}}} \right.{\rm{ = P}}\left( {\left( {{{\rm{A}}_{\rm{1}}}{\rm{C}}{{\rm{A}}_{\rm{2}}}} \right){\rm{E }}\left( {{{\rm{A}}_{\rm{3}}}{\rm{C}}{{\rm{A}}_{\rm{4}}}} \right)} \right)}\\{\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {\rm{P}}\left( {{{\rm{A}}_{\rm{1}}}{\rm{C}}{{\rm{A}}_{\rm{2}}}} \right){\rm{ + P}}\left( {{{\rm{A}}_{\rm{3}}}{\rm{C}}{{\rm{A}}_{\rm{4}}}} \right)}{{\rm{ - P}}\left( {\left( {{{\rm{A}}_{\rm{1}}}{\rm{C}}{{\rm{A}}_{\rm{2}}}} \right){\rm{C}}\left( {{{\rm{A}}_{\rm{3}}}{\rm{C}}{{\rm{A}}_{\rm{4}}}} \right)} \right)}\\ {\mathop {\rm{ = }}\limits^{{\rm{(2)}}} {\rm{P}}\left( {{{\rm{A}}_{\rm{1}}}} \right){\rm{*P}}\left( {{{\rm{A}}_{\rm{2}}}} \right){\rm{ + P}}\left( {{{\rm{A}}_{\rm{3}}}} \right){\rm{*P}}\left( {{{\rm{A}}_{\rm{4}}}} \right)}{{\rm{ - P}}\left( {{{\rm{A}}_{\rm{1}}}} \right){\rm{*P}}\left( {{{\rm{A}}_{\rm{2}}}} \right){\rm{*P}}\left( {{{\rm{A}}_{\rm{3}}}} \right){\rm{*P}}\left( {{{\rm{A}}_{\rm{4}}}} \right)}\\{{\rm{ = }}{{\rm{p}}^{\rm{2}}}{\rm{ + }}{{\rm{p}}^{\rm{2}}}{\rm{ - }}{{\rm{p}}^{\rm{4}}}{\rm{ = 2}}{{\rm{p}}^{\rm{2}}}{\rm{ - }}{{\rm{p}}^{\rm{4}}}{\rm{.}}}\end{array}\)

(1): we utilize the statement given below,

(2): we derive the multiplication property from the proposition given below (since the occurrences are unrelated).

the proposition given below (since the occurrences are unrelated).

Proposition: There should be two events for every two events. \({\rm{A}}\)and \({\rm{B}}\)are two different types of

\({\rm{P(AÉ B) = P(A) + P(B) - P(AÇB)}}\)

Two occurrences \({\rm{A}}\)and \({\rm{B}}\)are independent if and only if they have the same multiplication property. \({\rm{P(AÇB) = P(A)*P(B)}}\)

The chance of a system lifetime surpassing \({{\rm{t}}_{\rm{0}}}\)is supplied to us.

\({\rm{2}}{{\rm{p}}^{\rm{2}}}{\rm{ - }}{{\rm{p}}^{\rm{4}}}{\rm{ = 0}}{\rm{.99}}\)

Set \({\rm{x = }}{{\rm{p}}^{\rm{2}}}\)in order to answer this problem.

\({\rm{2x - }}{{\rm{x}}^{\rm{2}}}{\rm{ = 0}}{\rm{.99}}\)

or, alternatively

\({{\rm{x}}^{\rm{2}}}{\rm{ - 2x + 0}}{\rm{.99 = 0}}{\rm{.}}\)

where \({\rm{a = 1,b = - 2}}\)and \({\rm{c = 0}}{\rm{.99}}\)are the values. This indicates that the quadratic equation's solutions are

\({{\rm{x}}_{{\rm{12}}}}{\rm{ = }}\frac{{{\rm{ - b \pm }}\sqrt {{{\rm{b}}^{\rm{2}}}{\rm{ - 4ac}}} }}{{{\rm{2a}}}}{\rm{ = }}\frac{{{\rm{2 \pm }}\sqrt {{{\rm{2}}^{\rm{2}}}{\rm{ - 4*1*0}}{\rm{.99}}} }}{{{\rm{2*1}}}}\)

The first option is

\(\begin{matrix}{{\text{x}}_{\text{1}}}\text =&\frac{\text{2+}\sqrt{{{\text{2}}^{\text{2}}}\text{-4*0}\text{.99}}}{\text{2}}\text{ }\!\!\hat{\mathrm{U}}\!\!\text{ } \\ {{\text{x}}_{\text{1}}}\text =& \frac{\text{2+}\sqrt{\text{0}\text{.04}}}{\text{2}}\text{ }\!\!\hat{\mathrm{U}}\!\!\text{ } \\ {{\text{x}}_{\text{1}}}\text =&1\text{.1}\text{.} \\ \end{matrix}\)

and the second alternatives

\(\begin{matrix}{{\text{x}}_{\text{2}}}\text=&\frac{\text{2-}\sqrt{{{\text{2}}^{\text{2}}}\text{-4*0}\text{.99}}}{\text{2}}\text{ }\!\!\hat{\mathrm{U}}\!\!\text{ } \\ {{\text{x}}_{\text{2}}}\text =&0 \text{.9}\text{.} \\ \end{matrix}\text{ }\!\!\hat{\mathrm{U}}\!\!\text{ }\)

From the \({\rm{x = }}{{\rm{p}}^{\rm{2}}}\),\({\rm{p > 0}}\)

\({\rm{p}}_{\rm{1}}^{\rm{2}}{\rm{ = 1}}{\rm{.1}}\;\;\;{\rm{Û}}\;\;\;{{\rm{p}}_{\rm{1}}}{\rm{ = }}\sqrt {{\rm{1}}{\rm{.1}}} \;\;\;{\rm{Û}}\;\;\;{{\rm{p}}_{\rm{1}}}{\rm{ = 1}}{\rm{.049}}\)

And \({\rm{p}}_{\rm{2}}^{\rm{2}}{\rm{ = 0}}{\rm{.9}}\;\;\;{\rm{Û }}\;\;\;{{\rm{p}}_{\rm{2}}}{\rm{ = }}\sqrt {{\rm{0}}{\rm{.9}}} \;\;\;{\rm{Û }}\;\;\;{{\rm{p}}_{\rm{2}}}{\rm{ = 0}}{\rm{.949}}{\rm{.}}\).

Because \({\rm{p}}\) is a probability, and a probability cannot be greater than \({\rm{1}}\), the only solution is \({{\rm{p}}_{\rm{2}}}{\rm{ = 0}}{\rm{.949}}{\rm{.}}\)