Reliability of a “one-shot” device. A “one-shot” device can be used only once; after use, the device (e.g., a nuclear weapon, space shuttle, automobile air bag) is either destroyed or must be rebuilt. The destructive nature of a one-shot device makes repeated testing either impractical or too costly. Hence, the reliability of such a device must be determined with minimal testing. Consider a one-shot device that has some probability, p, of failure. Of course, the true value of p is unknown, so designers will specify a value of p that is the largest defective rate they are willing to accept. Designers will conduct n tests of the device and determine the success or failure of each test. If the number of observed failures, x, is less than or equal to some specified value, K, then the device is considered to have the desired failure rate. Consequently, the designers want to know the minimum sample size n needed so that observing K or fewer defectives in the sample will demonstrate that the true probability of failure for the one-shot device is no greater than p.

a. Suppose the desired failure rate for a one-shot device is p = .10. Also, suppose designers will conduct n = 20 tests of the device and conclude that the device is performing to specifications if K = 1 (i.e., if 1 or no failures are observed in the sample). Find$\mathit{P}\left(x\le 1\right)$

b. In reliability analysis,$\mathbf{1}\mathbf{-}\mathbf{P}\left(x\le K\right)$ is often called the level of confidence for concluding that the true failure rate is less than or equal to p. Find the level of confidence for the one-shot device described in part a. In your opinion, is this an acceptable level? Explain.

c. Demonstrate that the confidence level can be increased by either (1) increasing the sample size n or (2) decreasing the number K of failures allowed in the sample.

d. Typically, designers want a confidence level of .90, .95, or .99. Find the values of n and K to use so that the designers can conclude (with at least 95% confidence) that the failure rate for the one-shot device of part a is no greater than p = .10.

Let X is the number of observed failures. The probability of failure is p.

It is given that

$p=0.10$and $n=20$

Consider,

$$\begin{gathered} P\left(x\le 1\right)=P\left(x=0\right)+P\left(x=1\right) \\ =\left[\left(\begin{array}{c}20\\ 0\end{array}\right){\left(0.10\right)}^0{\left(0.90\right)}^{20-0}\right]+\left[\left(\begin{array}{c}20\\ 1\end{array}\right){\left(0.10\right)}^1{\left(0.90\right)}^{20-1}\right] \\ =\left(1\times 1\times 0.1216\right)+\left(20\times 0.10\times 0.{90}^{19}\right) \\ =\left(0.1216\right)+\left(0.2702\right) \\ =0.3918 \\ \approx 0.392 \end{gathered}$$

Thus, the probability of$P\left(x\le 1\right)$ is 0.392.

b.

By the reliability analysis, the level of the confidence for concluding that the true failure rate is less than or equal to p is defined as follows:

Level of confidence$\mathbf{=}\mathbf{1}\mathbf{-}\mathit{P}\left(x\le K\right)$

From the part a, the level of confidence is as follows:

Level of confidence

$$\begin{gathered} =1-P\left(x\le K\right) \\ =1-P\left(x\le K\right) \\ =1-0.392 \\ =0.608 \end{gathered}$$

Thus, the level of confidence for the part a. is 0.608.

Here, the level of confidence is not close to the probability value 1. This clearly indicates that this cannot be an acceptable level.

c.

From the information, it is clear that the sample size is n=20.

Increase the sample size n for 20 to 25 with

Consider,

$$\begin{gathered} P\left(x\le 1\right)=P\left(x=0\right)+P\left(x=1\right) \\ =\left[\left(\begin{array}{c}25\\ 0\end{array}\right){\left(0.10\right)}^0{\left(0.90\right)}^{25-0}\right]+\left[\left(\begin{array}{c}25\\ 1\end{array}\right){\left(0.10\right)}^1{\left(0.90\right)}^{25-1}\right] \\ =\left(1\times 10.0718\right)+\left(20\times 0.10\times 0.{90}^{24}\right) \\ =\left(0.0718\right)+\left(0.1994\right) \\ =0.2712 \\ \approx 0.271 \end{gathered}$$

Thus, the probability of $P\left(x\le 1\right)$is 0.271.

The level of the confidence with$\mathit{p}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{10}$and$\mathit{n}\mathbf{=}\mathbf{25}$can be obtained as follows:

Level of confidence is:

$$\begin{gathered} 1-P\left(x\le K\right)=1-P\left(x\le K\right) \\ =1-0.271 \\ =0.729 \end{gathered}$$

Thus, the level of confidence is 0.729.

From the information, it is clear that the sample size is n=20

Decrease the number of failures 1 to 0.

Consider,

$$\begin{gathered} P\left(x\le 0\right)=P\left(x=0\right) \\ =\left(\begin{array}{c}25\\ 0\end{array}\right){\left(0.10\right)}^0{\left(0.90\right)}^{20-0} \\ =0.1216 \\ \approx 0.122 \end{gathered}$$

Thus, the probability of$P\left(x\le 0\right)$is 0.122.

The level of confidence with$\mathit{p}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{10}$ and$\mathit{n}\mathbf{=}\mathbf{20}$ can be obtained as follows:

Level of confidence$\mathbf{=}\mathbf{1}\mathbf{-}\mathit{P}\left(x\le K\right)$

That is

$$\begin{gathered} 1-P\left(X\le K\right)=1-P\left(x\le 0\right) \\ =1-0.122 \\ =0.878 \end{gathered}$$

Thus, the level of confidence is 0.878.

Here, the level of confidence by increasing the sample size from 20 to 25 is 0.729 whereas by decreasing the number of failure 1 to 0 is 0.878.

Hence, it is clear that the confidence level is increased by decreasing the number of K failures that has been allowed in the sample.

d.

Consider,

K=0,

It is necessary to obtain the value of n. Thus, the level of confidence should be at least 0.95. That is,

$$\begin{gathered} P\left(x=0\right)\le 0.95 \\ P\left(x=0\right)=\left\{\left(\begin{array}{c}n\\ 0\end{array}\right){\left(p\right)}^x{\left(q\right)}^{n-x}\right\}\le 0.05 \\ =\left\{\frac{n!}{0!n!}{\left(0.10\right)}^0{\left(0.90\right)}^{n-0}\right\}\le 0.05 \\ =\left\{1\times 1\times {\left(0.90\right)}^n\right\}\le 0.05 \\ =\left\{{\left(0.90\right)}^n\right\}\le 0.05 \end{gathered}$$

Taking logarithm of both sides,

$$\begin{gathered} \mathrm{In}\left[{\left(0.90\right)}^{\mathrm{n}}\right]\le \mathrm{In}\left[0.05\right] \\ \mathrm{n}\mathrm{In}\left(0.90\right)\le \mathrm{In}\left(0.05\right) \\ \mathrm{n}\ge \frac{\mathrm{In}\left(0.05\right)}{\mathrm{In}\left(0.90\right)} \\ \mathrm{n}\ge \frac{-2.9957}{-0.1054} \\ \mathrm{n}\ge 28.4 \\ \mathrm{n}\approx 29 \end{gathered}$$

Thus, the value of n when $K=0$is 29 in order to get the level of confidence at least 0.95.

Consider, K=1.

It is necessary to obtain the value of n. Thus, the level of confidence should be at least 0.95. That is,$P(x=0)\le 0.95$

role="math" localid="1660716120882" $$\begin{gathered} P\left(x\le 1\right)=\left\{\left(\begin{array}{c}n\\ 0\end{array}\right){\left(p\right)}^x{\left(q\right)}^{n-0}+\left(\begin{array}{c}n\\ 0\end{array}\right){\left(p\right)}^x{\left(q\right)}^{n-1}\right\}\le 0.05 \\ =\left\{\frac{n!}{0!n!}{\left(0.10\right)}^0{\left(0.90\right)}^{n-0}+\frac{n!}{1!\left(n-1\right)!}{\left(0.10\right)}^1{\left(0.90\right)}^{n-1}\right\}\le 0.05 \\ =\left\{1\times 1\times {\left(0.90\right)}^n+n{\left(0.10\right)}^1{\left(0.90\right)}^{n-1}\right\}\le 0.05 \\ ={\left(0.90\right)}^{n-1}\left\{0.90+n\left(0.10\right)\right\}\le 0.05\dots \dots \dots \dots \dots \dots \dots \dots ..\left(1\right) \end{gathered}$$

By the trail, and error,

Consider, n=30 and substitute in the equation (1)

$$\begin{gathered} {\left(0.90\right)}^{\mathrm{n}-1}\left\{0.90+\mathrm{n}\left(0.10\right)\right\}={\left(0.90\right)}^{30-1}\left\{0.90+30\left(0.10\right)\right\} \\ =0.{90}^{29}\left(0.90+3\right) \\ =0.{90}^{29}\left(3.90\right) \\ =0.1837 \end{gathered}$$

Consider, n=40and substitute in the equation (1),

$$\begin{gathered} {\left(0.90\right)}^{\mathrm{n}-1}\left\{0.90+\mathrm{n}\left(0.10\right)\right\}={\left(0.90\right)}^{40-1}\left\{0.90+40(0.10\right\} \\ =0.{90}^{39}\left(0.90+4\right) \\ =0.{90}^{39}\left(4.90\right) \\ =0.0805 \end{gathered}$$

Consider,n=45 and substitute in the equation (1)

$$\begin{gathered} {\left(0.90\right)}^{\mathrm{n}-1}\left\{0.90+\mathrm{n}\left(0.10\right)\right\}={\left(0.90\right)}^{45-1}\left\{0.90+45(0.10\right\} \\ =0.{90}^{44}\left(0.90+4.5\right) \\ =0.{90}^{44}\left(5.4\right) \\ =0.0524 \end{gathered}$$

Consider, and substitute in the equation (1)

$$\begin{gathered} {\left(0.90\right)}^{\mathrm{n}-1}\left\{0.90+\mathrm{n}\left(0.10\right)\right\}={\left(0.90\right)}^{46-1}\left\{0.90+46(0.10\right\} \\ =0.{90}^{45}\left(0.90+4.6\right) \\ =0.{90}^{45}\left(5.5\right) \\ =0.0480 \end{gathered}$$

Thus, the value of n when$K=1$ is 46 in order to get the level of confidence at least 0.95.