Question

A system consists of five identical components connected in series as shown:

As soon as one components fails, the entire system will fail. Suppose each component has a lifetime that is exponentially distributed with ? = 0.01 and that components fail independently of one another. Define eventsA_i= {ith component lasts at least t hours}, i = 1, . . . , 5, so that the A_is are independent events. Let X = the time at which the system fails

that is, the shortest (minimum) lifetime among the five components.  The event {X ? t} is equivalent to what event involving A1, . . . , A5?  Using the independence of the A_is, compute P(X ? t).  What type of distribution does X have?  Suppose there are n components, each having exponential lifetime with parameter ? What type of distribution does X have?

Answer 1

Concepts and reason

Exponential distribution: An exponential distribution is used to determine the time related events. In this problem, there is an exponential distribution and it is required obtain some provided probabilities.

The total probability in distribution table is equal to one and each probability is greater than zero.

The cumulative distribution function is the probability that the variable takes a value less than or equal to $x$ . That is, $F\left( x \right) = P\left( {X \le x} \right)$

Independent events: Let A and B be two events. The events A and B are said to be independent if happening of one event does not affect the happening of another event.

Fundamentals

The probability density function of an exponential distribution is provided as follows:

$f\left( x \right) = \left\{ \begin{array}{l}\\\lambda {e^{ - \lambda x}},\,\,\,\,\,\,x \ge 0,\\\\0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x < 0.\\\end{array} \right.$

The cumulative distribution function can be written as follows:

$\begin{array}{c}\\F\left( x \right) = P\left( {X \le x} \right)\\\\ = \left\{ \begin{array}{l}\\1 - \lambda {e^{ - \lambda x}},\,\,\,\,\,\,x \ge 0,\\\\0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x < 0.\\\end{array} \right.\\\end{array}$

The formula for $P\left( {X > a} \right)$ is, $P\left( {X > a} \right) = 1 - P\left( {X \le a} \right)$

The formula for the density function is,

$f\left( t \right) = \frac{d}{{dt}}F\left( t \right)$

If $A$ and $B$ are two independent events then $P\left( {A \cap B} \right) = P\left( A \right) \times P\left( B \right)$

(a)

Given a system consists of five identical components connected in a series.

Since the components are connected in a series, any failure of one component results the failure of the system.

All the five components fail independently and follow exponential distribution with $\lambda = 0.01$ .

The event $\{ X \ge t\}$ is equivalent to the event involving ${A_1},{A_2},{A_3},{A_4},{\rm{ and }}{A_5}$ is,

$\{ X \ge t\} = {A_1} \cap {A_2} \cap {A_3} \cap {A_4} \cap {A_5}$

(b)

The probability of each event can be obtained is as shown below:

$\begin{array}{c}\\P\left( {{A_i}} \right) = P\left( {{\rm{Exponential random varaible with }}\lambda = .01{\rm{ is greater than }}t} \right)\\\\ = 1 - P\left( {{\rm{Exponential random varaible with }}\lambda = .01{\rm{ is greater than }}t} \right)\\\\ = 1 - \left[ {1 - {e^{\lambda t}}} \right]\\\\ = {e^{ - 0.01t}}\\\end{array}$

Since, the entire system fails as soon as one component fails; the system lasts at least t hours if and only if all five of the components last at least t hours.

Compute $P\left( {X \ge t} \right)$

$\begin{array}{c}\\P\left( {X \ge t} \right) = P\left( {{\rm{All five }}{A_i}'s{\rm{ occur}}} \right)\\\\ = P\left( {{A_1}} \right)P\left( {{A_2}} \right)P\left( {{A_3}} \right)P\left( {{A_4}} \right)P\left( {{A_5}} \right){\rm{ }}\left( \begin{array}{l}\\{\rm{Since the components }}\\\\{\rm{independent}}\\\end{array} \right)\\\\ = {\left[ {{e^{ - 0.01t}}} \right]^5}\\\\ = {e^{ - 0.05t}}\\\end{array}$

The distribution function of the random variable $X$ is,

$\begin{array}{c}\\F\left( t \right) = P\left( {X \le t} \right)\\\\ = P\left( {X < t} \right)\\\\ = 1 - P\left( {X \ge t} \right)\\\\ = 1 - {e^{ - 0.05t}}\\\end{array}$

The probability density function of the random variable $X$ is,

$\begin{array}{c}\\f\left( t \right) = \frac{d}{{dt}}F\left( t \right)\\\\ = \frac{d}{{dt}}\left( {1 - {e^{ - 0.05t}}} \right)\\\\ = 0.05{e^{ - 0.05t}}\\\end{array}$

The random variable $X$ follows exponential distribution with parameter $\lambda = 0.05$

(c)

If there are ‘n’ components, each having exponential distributed with $\lambda$ then

$P\left( {X \le t} \right) = 1 - {e^{ - n\lambda t}}$

The density function is,

$\begin{array}{c}\\f\left( t \right) = \frac{d}{{dt}}F\left( t \right)\\\\ = \frac{d}{{dt}}\left( {1 - {e^{ - n\lambda t}}} \right)\\\\ = n\lambda {e^{ - n\lambda t}}\\\end{array}$

The random variable $X$ follows exponential distribution with parameter $n\lambda$

Ans: Part a

The event $\{ X \ge t\}$ is equivalent to the event involving ${A_1},{A_2},{A_3},{A_4},{\rm{ and }}{A_5}$ is,

$\{ X \ge t\} = {A_1} \cap {A_2} \cap {A_3} \cap {A_4} \cap {A_5}$

Part b

The probability density function of the random variable $X$ is, $f\left( t \right) = 0.05{e^{ - 0.05t}}$

The random variable $X$ follows exponential distribution with parameter $\lambda = 0.05$

Part c

The random variable $X$ follows exponential distribution with parameter $n\lambda .$