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Chapter 5: Q 8E (page 280)

A certain electronic system contains 10 components. Suppose that the probability that each individual component will fail is 0.2 and that the components fail independently of each other. Given that at least one of the components have failed, what is the probability that at least two of the components have failed?

Short Answer

Expert verified

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

Step by step solution

01

Given information

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.

02

Define the p.m.f

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

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

03

Calculate the probability

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.

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Most popular questions from this chapter

Prove that the p.f. of the negative binomial distribution can be written in the following alternative form:

\[{\bf{f}}\left( {{\bf{x|r,p}}} \right){\bf{ = }}\left\{ \begin{array}{l}\left( \begin{array}{l}{\bf{ - r}}\\{\bf{x}}\end{array} \right){{\bf{p}}^{\bf{r}}}{\left( {{\bf{ - }}\left[ {{\bf{1 - p}}} \right]} \right)^{\bf{x}}}\;\;{\bf{for}}\,{\bf{x = 0,1,2}}...\\{\bf{0}}\;\;\;{\bf{otherwise}}{\bf{.}}\end{array} \right.\]Hint: Use Exercise 10 in Sec. 5.3.

Suppose that a pair of balanced dice are rolled 120 times, and let X denote the number of rolls on which the sum of the two numbers is 12. Use the Poisson approximation to approximate \({\bf{Pr}}\left( {{\bf{X = 3}}} \right){\bf{.}}\)

Let\({X_1},{X_2},{X_3}\)be a random sample from the exponential distribution with parameter\(\beta \). Find the probability that at least one of the random variables is greater than t, where\(t > 0\)

Suppose that the random variables \({{\bf{X}}_{\bf{1}}}{\bf{,}}{{\bf{X}}_{\bf{2}}}{\bf{ \ldots }}{{\bf{X}}_{\bf{k}}}\)are independent

and that \({{\bf{X}}_{\bf{i}}}\) has the negative binomial distribution with parameters \({{\bf{r}}_{\bf{i}}}\) and\({\bf{p}}\left( {{\bf{i = 1 \ldots k}}} \right)\). Prove that the sum \({{\bf{X}}_{\bf{1}}}{\bf{,}}{{\bf{X}}_{\bf{2}}}{\bf{ \ldots }}{{\bf{X}}_{\bf{k}}}\)has the negative binomial distribution with parameters \({\bf{r = }}{{\bf{r}}_{\bf{1}}}{\bf{ + \ldots + }}{{\bf{r}}_{\bf{k}}}\)and p.

Let \({{\bf{X}}_{{\bf{1,}}}}{\bf{ }}{\bf{. }}{\bf{. }}{\bf{. , }}{{\bf{X}}_{\bf{n}}}\)be i.i.d. random variables having thenormal distribution with mean \({\bf{\mu }}\) and variance\({{\bf{\sigma }}^{\bf{2}}}\). Define\(\overline {{{\bf{X}}_{\bf{n}}}} {\bf{ = }}\frac{{\bf{1}}}{{\bf{n}}}\sum\limits_{{\bf{i = 1}}}^{\bf{n}} {{{\bf{X}}_{\bf{i}}}} \) , the sample mean. In this problem, weshall find the conditional distribution of each \({{\bf{X}}_{\bf{i}}}\)given\(\overline {{{\bf{X}}_{\bf{n}}}} \).

a.Show that \({{\bf{X}}_{\bf{i}}}\)and\(\overline {{{\bf{X}}_{\bf{n}}}} \) have the bivariate normal distributionwith both means \({\bf{\mu }}\), variances\({{\bf{\sigma }}^{\bf{2}}}{\rm{ }}{\bf{and}}\,\,\frac{{{{\bf{\sigma }}^{\bf{2}}}}}{{\bf{n}}}\),and correlation\(\frac{{\bf{1}}}{{\sqrt {\bf{n}} }}\).

Hint: Let\({\bf{Y = }}\sum\limits_{{\bf{j}} \ne {\bf{i}}} {{{\bf{X}}_{\bf{j}}}} \).

Nowshow that Y and \({{\bf{X}}_{\bf{i}}}\) are independent normals and \({{\bf{X}}_{\bf{n}}}\)and \({{\bf{X}}_{\bf{i}}}\) are linear combinations of Y and \({{\bf{X}}_{\bf{i}}}\) .

b.Show that the conditional distribution of \({{\bf{X}}_{\bf{i}}}\) given\(\overline {{{\bf{X}}_{\bf{n}}}} {\bf{ = }}\overline {{{\bf{x}}_{\bf{n}}}} \)\(\) is normal with mean \(\overline {{{\bf{x}}_{\bf{n}}}} \) and variance \({{\bf{\sigma }}^{\bf{2}}}\left( {{\bf{1 - }}\frac{{\bf{1}}}{{\bf{n}}}} \right)\).
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