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Chapter 2: Q52E (page 83)

A system consists of two identical pumps, \(\# 1\) and \(\# 2\). If one pump fails, the system will still operate. However, because of the added strain, the remaining pump is now more likely to fail than was originally the case. That is, \(r\; = P\left( {\# 2\;fails|\# 1 fails} \right) > P\left( {\# 2 fails} \right) = q\). If at least one pump fails by the end of the pump design life in \(7\% \) of all systems and both pumps fail during that period in only 1%, what is the probability that pump \(\# 1\)will fail during the pump design life?

Short Answer

Expert verified

The probability that the pump \(\# 1\) will fail during the pump design life is \(0.04\).

Step by step solution

01

Definition of Probability

Probability is a branch of mathematics concerned with numerical descriptions of the likelihood of an event occurring or a proposition being true. The probability of an event is a number between \(0{\rm{ }}and{\rm{ }}1\), with zero indicating impossibility and one indicating certainty.

02

Given Data

Denoted events

\(\begin{array}{l}A = \{ {\rm{ pump }}\# 1{\rm{ fails }}\} \\B = \{ {\rm{ pump }}\# 2{\rm{ fails }}\} \end{array}\)

We are given that

\(\begin{array}{l}P(B) &=& q\\P(B\mid A) &=& r > q\end{array}\)

The pumps are identical (given in the exercise), which indicates that the probabilities are the same

\(\begin{array}{l}P(A) &=& P(B) &=& q\\P(B\mid A) &=& P(A\mid B) &=& r\end{array}\)

03

Calculation for the determination of probability

We are given the probability that at least one pump fails by the end of pump design life. The probability is actually a union of events A and B (event A occurs or event B occurs or both occur)

\(P(A \cup B) = 0.07.\)

We are also given the probability of an event that all systems and both pumps fail during the period. The probability is the intersection of events A and B (both A and B fail)

\(P(A \cap B) = 0.01\)

The Multiplication Rule

\(P(A \cap B) = P(A\mid B) \cdot P(B)\)

04

Further calculation for the determination of probability

Using The Multiplication Rule, we have

\(0.1 = P(A \cap B) = P(A\mid B) \cdot P(B)\mathop = \limits^{(1)} rq\)

(1): we are given the probabilities.

The union can be tricky to find with the information we have. The following is true

\(\begin{array}{l}0.7 = P(A \cup B)\mathop = \limits^{(1)} P(A \cap B) + P\left( {{A^\prime } \cap B} \right) + P\left( {A \cap {B^\prime }} \right)\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\mathop = \limits^{(2)} rq + (1 - r)q + (1 - r)q = rq + 2(1 - r)q\end{array}\)

(1): the union of events A and B can be represented as a union of three disjoint events, from which we can conclude the equality,

(2): In the same way we calculated the intersection of events A and B, we can calculate the intersections of the other two events.

05

Further Calculation for the determination of probability

\(\begin{array}{l}P\left( {{A^\prime } \cap B} \right) &=& P\left( {{A^\prime }\mid B} \right) \cdot P(B) &=& (1 - r)q\\P\left( {A \cap {B^\prime }} \right) &=& P\left( {{B^\prime }\mid A} \right) \cdot P(A) &=& (1 - r)q\end{array}\)

where we used the facts that \(P\left( {{A^\prime }\mid B} \right) = 1 - P(A\mid B)\)and \(P\left( {{B^\prime }\mid A} \right) = 1 - P(B\mid A)\), which are true because of the rule \(P(C) + P\left( {{C^\prime }} \right) = 1\),for any event C.

Now we have system

\(\begin{array}{l}rq &=& 0.01\\rq + 2(1 - r)q &=& 0.07\end{array}\)

From which, by substituting \(rq = 0.1\) into the second equation, we have

\(\begin{array}{l}rq + 2q - 2rq &=& 0.07\\ \Leftrightarrow 0.1 + 2q - 2 \cdot 0.01 &=& 0.72q - 0.01 &=& 0.07\\ \Leftrightarrow 2q = 0.08q = 0.04\end{array}\)

When we substitute \(q = 0.04\)into the first equation we get

\(rq = 0.01\;\;\; \Leftrightarrow \;\;\;r \cdot 0.04 = 0.01\;\;\; \Leftrightarrow \;\;\;r = 0.25\)

However, the answer to the question, the probability that pump #l will fail during the pump design life is

\(P(A) = 0.04\)

\(\begin{array}{l}P\left( {{A^\prime } \cap B} \right) &=& P\left( {{A^\prime }\mid B} \right) \cdot P(B) &=& (1 - r)q\\P\left( {A \cap {B^\prime }} \right) &=& P\left( {{B^\prime }\mid A} \right) \cdot P(A) &=& (1 - r)q\end{array}\)

where we used the facts that \(P\left( {{A^\prime }\mid B} \right) = 1 - P(A\mid B)\)and \(P\left( {{B^\prime }\mid A} \right) = 1 - P(B\mid A)\), which are true because of the rule \(P(C) + P\left( {{C^\prime }} \right) = 1\),for any event C.

Now we have system

\(\begin{array}{l}rq = 0.01\\rq + 2(1 - r)q = 0.07\end{array}\)

From which, by substituting \(rq = 0.1\) into the second equation, we have

\(\begin{array}{l}rq + 2q - 2rq &=& 0.07\\ \Leftrightarrow 0.1 + 2q - 2 \cdot 0.01 &=& 0.72q - 0.01 &=& 0.07\\ \Leftrightarrow 2q &=& 0.08q &=& 0.04\end{array}\)

When we substitute \(q = 0.04\)into the first equation we get

\(rq = 0.01\;\;\; \Leftrightarrow \;\;\;r \cdot 0.04 = 0.01\;\;\; \Leftrightarrow \;\;\;r = 0.25\)

However, the answer to the question, the probability that pump #l will fail during the pump design life is

\(P(A) = 0.04\)

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