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.
