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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

Given, \({A_i} = \) {awarded project i}, for \(i = 1, 2, 3\). Use the probabilities given there to compute the following probabilities, and explain in words the meaning of each one.

a. \(P\left( {{A_2}|{A_1}} \right)\)

b. \(P\left( {{A_2} \cap {A_3}|{A_1}} \right)\)

c. \(P\left( {{A_2} \cup {A_3}|{A_1}} \right)\)

d. \(P\left( {{A_1} \cap {A_2} \cap {A_3}|{A_1} \cup {A_2} \cup {A_3}} \right)\)

One of the assumptions underlying the theory of control charting is that successive plotted points are independent of one another. Each plotted point can signal either that a manufacturing process is operating correctly or that there is some sort of malfunction.

Human visual inspection of solder joints on printed circuit boards can be very subjective. Part of the problem stems from the numerous types of solder defects (e.g., pad non wetting, knee visibility, voids) and even the degree to which a joint possesses one or more of these defects. Consequently, even highly trained inspectors can disagree on the disposition of a particular joint. In one batch of 10,000 joints, inspector A found 724 that were judged defective, inspector B found 751 such joints, and 1159 of the joints were judged defective by at least one of the inspectors. Suppose that one of the 10,000 joints is randomly selected.

a. What is the probability that the selected joint was judged to be defective by neither of the two inspectors?

b. What is the probability that the selected joint was judged to be defective by inspector B but not by inspector A?

a. A lumber company has just taken delivery on a shipment of \({\rm{10,000 2 \times 4}}\)boards. Suppose that 20% of these boards (\({\rm{2000}}\)) are actually too green to be used in first-quality construction. Two boards are selected at random, one after the other. Let A 5 {the first board is green} and B 5 {the second board is green}. Compute \({\rm{P(A),P(B)}}\), and \({\rm{P(A}} \cap {\rm{B)}}\) (a tree diagram might help). Are \({\rm{A}}\) and \({\rm{B}}\) independent?

b. With \({\rm{A}}\) and \({\rm{B}}\) independent and \({\rm{P(A) = P(B) = }}{\rm{.2,}}\) what is \({\rm{P(A}} \cap {\rm{B)}}\)? How much difference is there between this answer and \({\rm{P(A}} \cap {\rm{B)}}\)part (a)? For purposes of calculating \({\rm{P(A}} \cap {\rm{B)}}\), can we assume that \({\rm{A}}\)and \({\rm{B}}\) of part (a) are independent to obtain essentially the correct probability?

c. Suppose the shipment consists of ten boards, of which two are green. Does the assumption of independence now yield approximately the correct answer for \({\rm{P(A}} \cap {\rm{B)}}\)? What is the critical difference between the situation here and that of part (a)? When do you think an independence assumption would be valid in obtaining an approximately correct answer to \({\rm{P(A}} \cap {\rm{B)}}\)?

The computers of six faculty members in a certain department are to be replaced. Two of the faculty members have selected laptop machines and the other four have chosen desktop machines. Suppose that only two of the setups can be done on a particular day, and the two computers to be set up are randomly selected from the six (implying 15 equally likely outcomes; if the computers are numbered1, 2,鈥�, 6, then one outcome consists of computers 1 and2, another consists of computers 1 and 3, and so on).

a. What is the probability that both selected setups are for laptop computers?

b. What is the probability that both selected setups are desktop machines?

c. What is the probability that at least one selected setup is for a desktop computer?

d. What is the probability that at least one computer of each type is chosen for setup?
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