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Probability is the likelihood that an event will occur and is calculated by dividing the number of favourable outcomes by the total number of possible outcomes. The simplest example is a coin flip. When you flip a coin there are only two possible outcomes, the result is either heads or tails.

(a)

It is assumed that \(X\)represents the time at which the system fails, i.e. the component with the shortest (minimum) lifetime among the five.

Let \(E\) be the event \(\{ X \ge t\} \), or in other words :

The system's lifetime is at least \(E\).

Events \({A_i}\)for \(i = \{ 1,2,3,4,5\} \)are defined as :

\({A_i} = \left\{ {{i^{{\rm{th }}}}{\rm{ component lasts at least t hours }}} \right\}\)

Since the components are connected in series, Hence

\(E = \{ X \ge t\} = {A_1} \cap {A_2} \cap {A_3} \cap {A_4} \cap {A_5}\)

(b)

The chance that the \({i^{{\rm{th }}}}\)component lasts at least \(t\)hours is \(P\left( {{A_i}} \right)\). With the parameter \(\lambda = 0.01\), each component has an exponentially distributed lifetime. Let's use a random variable \(Y\) to represent the lifetime of a single component.

\(P\left( {{A_i}} \right) = P(Y \ge t)\)

Using the cdf of exponential distribution we can write cdf of \(Y\)as : :

\(F(y;\lambda ) = \left\{ {\begin{array}{*{20}{l}}0&{y < 0}\\{1 - {e^{ - (0.01)y}}}&{y \ge 0}\end{array}} \right.\)

Hence

\(P\left( {{A_i}} \right) = P(Y \ge t)\)

\({\rm{ = 1 - F(t)}}\)

\( = 1 - \left( {1 - {e^{ - (0.01)t}}} \right)\)

\(P\left( {{A_i}} \right) = {e^{ - (0.01)t}}\)

Since all the events are independent, Hence

\(P(X \ge t) = P\left( {{A_1}} \right) \cdot P\left( {{A_2}} \right) \cdot P\left( {{A_3}} \right) \cdot P\left( {{A_4}} \right) \cdot P\left( {{A_5}} \right)\)

\( = {\left( {{e^{ - (0.01)t}}} \right)^5}\)

\(P(X \ge t) = {e^{ - (0.05)t}}\)

Let \({\rm{Z}}\)be a continuous rv with cdf \(F(z)\)as the cdf. After that, for any number a,

\(P(Z \ge a) = 1 - \phi (a)\)

\(P(Z \le a) = \phi (a)\)

The \(X\) cdf can thus be written as:

\(F(x) = P(X \le t)\)

\( = 1 - P(X \ge t)\)

\(F(x) = 1 - {e^{ - (0.05)t}}\)

Then, by differentiating\(X\)'s cdf for \(t \ge 0\), we can get a pdf of it.

\(f(x) = {F^\prime }x\)

\(f(x) = (0.05){e^{ - (0.05)t}}\;\;\;{\rm{ for }}t \ge 0\)

We may deduce that \(X\)has an exponential distribution with parameter \(\lambda = 0.05\)by looking at the pdf. Proposition: If \(X\)is a continuous rv with pdf \(f(x)\)and cdf \(F(x)\), then the derivative \({F^\prime }(x)\) exists at every \(x\).

\(f(x) = {F^\prime }(x)\)

(c)

So, we get that

From observing the last part we can say that \(X\)will have exponential distribution with parameter with parameter \(n\lambda \).