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Unformatted text preview: 5.62 A certain airplane has two independent alternators to provide electrical power. The probability that a given alternator will fail on a 1-hour flight is .02. What is the probability that (a) both will fail? (b) Neither will fail? (c) One or the other will fail? Show all steps carefully. P(An alternator will fail) = 0.02 ⇒ P(An alternator will not fail) = 1 – 0.02 = 0.98 (a) P(Both will fail) = P(The first will fail and the second will fail) = 0.02 * 0.02 = 0.0004 (b) P(Neither will fail) = P(The first will not fail and the second will not fail) = 0.98 * 0.98 = 0.9604 (c) P(One or the other will fail) = P(The first will fail) + P(The second will fail) – P(Both will fail) = 0.02 + 0.02 – 0.0004 = 0.0396 In solving the above problems, we have used multiplication law and addition law of probability to calculate the probabilities for the different scenarios of the alternators failing. This computation could be crucial for the probabilities for the different scenarios of the alternators failing....
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This note was uploaded on 04/16/2011 for the course BUS 415 taught by Professor Bobsmith during the Spring '10 term at University of Phoenix.
- Spring '10