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Chapter 2: Q80E (page 90)

Consider the system of components connected as in the accompanying picture. Components \({\rm{1}}\) and \({\rm{2}}\) are connected in parallel, so that subsystem works iff either 1 or 2 works; since \({\rm{3}}\)and\({\rm{4}}\) are connected in series, that subsystem works iff both\({\rm{3}}\)and\({\rm{4}}\)work. If components work independently of one another and P(component i works) \({\rm{ = }}{\rm{.9}}\)for \({\rm{i = }}{\rm{.1,2}}\)and \({\rm{ = }}{\rm{.8}}\)for \({\rm{i = 3,4}}\),calculate P(system works).

Short Answer

Expert verified

The system works is \(\begin{array}{c}{\rm{P( system works ) = 0}}{\rm{.9964}}\\{\rm{ = 99}}{\rm{.64\% }}\end{array}\)

Step by step solution

01

Definition of Independence

Independence If the probability of one event is unaffected by the occurrence or non-occurrence of the other, two events are said to be independent in probability. Consider the following example for a better understanding of this definition.

02

Given parameters

Given: Five valves

\(\begin{array}{l}{\rm{P( component 1 works ) = 0}}{\rm{.9}}\\{\rm{P( component 2 works ) = 0}}{\rm{.9}}\\{\rm{P( component 3 works ) = 0}}{\rm{.8}}\\{\rm{P( component 4 works ) = 0}}{\rm{.8}}\end{array}\)

The components are self-contained.

03

Finding the system works

For independent events, use the following multiplication rule:

\({\rm{P(A and B) = P(A)*P(B)}}\)

For any two events, the following is the general addition rule:

\({\rm{P(A or B) = P(A) + P(B) - P(A and B)}}\)

Use the multiplication rule for independent events:

\({\rm{P( component 1 and 2 work ) = P( component 1 works )*P( component 2 works )}}\)\({\rm{ = 0}}{\rm{.9*0}}{\rm{.9 = 0}}{\rm{.81}}\)

If component \({\rm{1}}\)or component \({\rm{2}}\)work, the top subsystem will also work. For any two occurrences, apply the general addition rule:

\({\rm{P}}\left( {{\rm{top sub system works }}} \right){\rm{ = P( component\;works ) + P( component\;works ) - P(component\;andwork )}}\)\({\rm{ = 0}}{\rm{.9 + 0}}{\rm{.9 - 0}}{\rm{.81 = 0}}{\rm{.99}}\)

If both components \({\rm{3}}\) and \({\rm{4}}\)are operational, the bottom subsystem will function. For independent events, use the multiplication rule:

04

Explanation of the solution

\({\rm{P}}\left( {{\rm{bottom subsystem works }}} \right){\rm{ = P( component\;works )*P( component\;works)}}\)\({\rm{ = 0}}{\rm{.8*0}}{\rm{.8 = 0}}{\rm{.64}}\)

The two subsystems are independent, because each component is independent. Use the multiplication rule for independent events:

\({\rm{P}}\left( {{\rm{both subsystems work}}} \right){\rm{ = P}}\left( {{\rm{top subsystem works}}} \right){\rm{* P}}\left( {{\rm{bottom subsystem works}}} \right)\)\({\rm{ = 0}}{\rm{.99*0}}{\rm{.64 = 0}}{\rm{.6336}}\)

The system will then work if either the top subsystem works or the bottom subsystem works. Use the general addition rule for any two events:

\({\rm{P}}\left( {{\rm{system works}}} \right){\rm{ = P}}\left( {{\rm{top subsystem works}}} \right){\rm{ + P}}\left( {{\rm{bottom subsystem works}}} \right)\)

\(\begin{array}{c}{\rm{ = 0}}{\rm{.99 + 0}}{\rm{.64 - 0}}{\rm{.6336}}\\{\rm{ = 0}}{\rm{.9964 = 99}}{\rm{.64\% }}\end{array}\)

Therefore,The system works is \(\begin{array}{c}{\rm{P( system works ) = 0}}{\rm{.9964}}\\{\rm{ = 99}}{\rm{.64\% }}\end{array}\)

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Most popular questions from this chapter

An insurance company offers four different deductible levels鈥攏one, low, medium, and high鈥攆or its home owner鈥檚 policy holders and three different levels鈥攍ow, medium, and high鈥攆or its automobile policyholders. The accompanying table gives proportions for the various categories of policyholders who have both types of insurance. For example, the proportion of individuals with both low homeowner鈥檚 deductible and low auto deductible is .06(6% of all such individuals).

贬辞尘别辞飞苍别谤鈥檚

Auto N L M H

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d. Based on your answer in part (c), what is the probability that the two categories are different?

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