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Suppose the probability that the control system used in a spaceship will malfunction on a given flight is 0.001. Suppose further that a duplicate but completely independent control system is also installed in the spaceship to take control in case the first system malfunctions. Determine the probability that the spaceship will be under the control of either the original system or the duplicate system on a given flight.

Short Answer

Expert verified

On each given flight, the probability that the spacecraft will be controlled by either the unique or duplicating systems is 0.999999.

Step by step solution

01

Given information

The probability of a spaceship's control scheme malfunctioning on a particular flight is 0.001.

A backup, though entirely separate, the control system is also placed in the spaceship in case the primary system fails.

02

Defining events

Let:

\(O = \)The event that the spaceship will be under the control of the original system on a given flight.

\(D = \) On a particular flight, the spaceship will be under the direction of the duplicated systems.

03

Computing the required probability

According to the Additional rule of probability, the probability of occurrence of event A or B is given by:

\({\bf{P}}\left( {{\bf{A}} \cup {\bf{B}}} \right){\bf{ = P}}\left( {\bf{A}} \right){\bf{ + P}}\left( {\bf{B}} \right) - {\bf{P}}\left( {{\bf{A}} \cap {\bf{B}}} \right)\)

The probability of event O is obtained as:

\(\begin{aligned}{}P\left( O \right) &= P\left( {{\rm{Original system does not malfunction}}} \right)\\ &= 1 - P\left( {{\rm{Original system malfunctions}}} \right)\\ &= 1 - 0.001\\ &= 0.999\end{aligned}\)

Now, the spaceship will be under the control of the same system only if the original system malfunctions.

So, the probability of event D is obtained as:

\(P\left( D \right) = P\left( {{\rm{Original system malfunctions but duplicate system functions}}} \right)\)

Since both the systems are independent; therefore,

\(\begin{aligned}{}P\left( D \right) &= P\left( {{\rm{Original system malfunctions}}} \right) \times P\left( {{\rm{Duplicate system functions}}} \right)\\ &= 0.001\left( {1 - 0.001} \right)\\ &= 0.000999\end{aligned}\)

Also, events O and D are mutually exclusive events because these events can't occur simultaneously.

Thus, \(P\left( {O \cap D} \right) = 0\)

Using the Additional Rule of Probabilistic, calculate the probability that the spaceship will be controlled by either the unique as well as duplicated systems on a specific flight as:

\(\begin{aligned}{}P\left( {O \cup D} \right) &= P\left( O \right) + P\left( D \right) - P\left( {O \cap D} \right)\\ &= 0.999 + 0.000999 - 0\\ &= 0.999999\end{aligned}\)

Therefore, the required probability is0.999999.

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