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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.

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

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

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

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

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.

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

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

\(P(A) = 0.04\)