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Chapter 2: Q78E (page 89)

A boiler has five identical relief valves. The probability that any particular valve will open on demand is .96. Assuming independent operation of the valves, calculate P(at least one valve opens) and P(at least one valve fails to open).

Short Answer

Expert verified

The probability is \({\rm{P}}\left( {{\rm{At least one valve fails to open}}} \right){\rm{ = 18}}{\rm{.463\% }}\)

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

Finding the probability that any particular valve

Given: Five valves

\({\rm{P( One valve opens ) = 0}}{\rm{.96}}\)

Complement rule:

\({\rm{P(notA) = 1 - P(A)}}\)

For independent events, use the following multiplication rule:

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

A MINIMUM OF ONE VALVE OPEN

Use the complement rule to help you:

\({\rm{P}}\left( {{\rm{One valve does not open}}} \right){\rm{ = 1 - P}}\left( {{\rm{One valve opens}}} \right)\)

\({\rm{ = 1 - 0}}{\rm{.96 = 0}}{\rm{.04}}\)

We can use the multiplication rule for independent events because the rivets are independent.

\(\begin{array}{*{20}{l}}{{\rm{P}}\left( {{\rm{Five valves do not open}}} \right)}\\{{\rm{ = P}}\left( {{\rm{One valve does not open}}} \right){\rm{*}}...{\rm{* P}}\left( {{\rm{One valve does not open}}} \right)}\end{array}\)

\(\begin{array}{c}{\rm{ = P( One valve does not open }}{{\rm{)}}^{\rm{5}}}\\{\rm{ = 0}}{\rm{.0}}{{\rm{4}}^{\rm{5}}} \approx {\rm{0}}{\rm{.0000001024}}\\{\rm{ = 1}}{\rm{.024*1}}{{\rm{0}}^{{\rm{ - 7}}}}\end{array}\)

Use the complement rule again:

\({\rm{P}}\left( {{\rm{At least one valve opens}}} \right){\rm{ = 1 - P}}\left( {{\rm{Five valves do not open}}} \right)\)

\(\begin{array}{l}{\rm{ = 1 - 0}}{\rm{.0000001024}}\\{\rm{ = 0}}{\rm{.9999998976}}\\{\rm{ = 99}}{\rm{.99998976\% }}\end{array}\)

03

Finding the probability

AT LEAST ONE VALVE FAILS TO OPEN

Since the rivets are independent, we can use the multiplication rule for independent events.

\(\begin{array}{*{20}{l}}{{\rm{P}}\left( {{\rm{Five valves open}}} \right)}\\{{\rm{ = P}}\left( {{\rm{One valve opens}}} \right){\rm{ *}}\left( {{\rm{One valve opens}}} \right){\rm{\;*}}....{\rm{*P}}\left( {{\rm{One valve opens}}} \right)}\end{array}\)

\({\rm{ = (P( One valve opens )}}{{\rm{)}}^{\rm{5}}}{\rm{ = 0}}{\rm{.9}}{{\rm{6}}^{\rm{5}}}{\rm{\gg 0}}{\rm{.81537}}\)

Use the complement rule again:

\({\rm{P}}\left( {{\rm{At least one valve fails to open}}} \right){\rm{ = 1 - P}}\left( {{\rm{Five valves open}}} \right)\)

Use the complement rule again:

\({\rm{P}}\left( {{\rm{At least one valve fails to open}}} \right){\rm{ = 1 - P}}\left( {{\rm{Five valves open}}} \right)\)

\(\begin{array}{c}{\rm{ = 1 - 0}}{\rm{.81537}}\\{\rm{ = 0}}{\rm{.18463}}\\{\rm{ = 18}}{\rm{.463\% }}\end{array}\)

\(\begin{array}{c}{\rm{P( At least one valve opens ) = 0}}{\rm{.9999998976}}\\{\rm{ = 99}}{\rm{.99998976\% }}\end{array}\)

Therefore, the probability is \({\rm{P}}\left( {{\rm{At least one valve fails to open}}} \right){\rm{ = 18}}{\rm{.463\% }}\)

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