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Naddika [18.5K]
3 years ago
10

A product is made up of components A, B, C, D, E, F, G, H, I, and J. Components A, B, C, and F have a 1/10,000 chance of failure

during useful life. D, E, G, and H have a 3/10,000 chance of failure. Component I and J and a 5/10,000 chance of failure. What is the overall reliability of the product?
Mathematics
1 answer:
lesantik [10]3 years ago
4 0

Answer:

The overall reliability is 99.7402 %

Step-by-step explanation:

The overall reliability of the product is calculated as the product of the working probability of the components.

For components A,B,C and F we have :

P(failure)=\frac{1}{10000}

⇒

P(Work)=1-\frac{1}{10000}=\frac{9999}{10000}=0.9999

For components D,E,G and H we have :

P(failure)=\frac{3}{10000}

⇒

P(Work)=1-\frac{3}{10000}=\frac{9997}{10000}=0.9997

Finally, for components I and J :

P(failure)=\frac{5}{10000}

⇒

P(Work)=1-\frac{5}{10000}=\frac{9995}{10000}=0.9995

Now we multiply all the working probabilities. We mustn't forget that we have got ten components in this case :

Components A,B,C and F with a working probability of 0.9999

Components D,E,G and H with a working probability of 0.9997

Components I and J with a working probability of 0.9995

Overall reliability = (0.9999)^{4}(0.9997)^{4}(0.9995)^{2}=0.997402

0.997402 = 99.7402 %

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<em>Additional comment</em>

You are given a value in liters (14 liters) and asked for the equivalent in quarts. That means you want to change the units from liters to quarts. To do that, you can multiply the given value (14 liters) by a conversion factor that has quarts in the numerator and liters in the denominator. That is what the fraction 1/0.95 is in the above. You will note that units of liters cancel in this equation.

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You will notice that we treat units just like any variable. They can be multiplied, divided, cancelled, raised to a power. Only terms with like units can be added or subtracted.

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