Question

in a certain four engine vintage aircraft, now quite unreliable, each engine has a 10% chance of failure on any flight, as long as it is carrying its one-fourth share of the load. but if one engine fails, then the chance of failure increases to 20% for each of the other three engines. and if a second engine fails, each of the remaining two has a 30% chance of failure. assuming that no two engines ever fail simultaneously, and that the aircraft can continue flying with as few as two operating engines, find each probability for a given flight of this aircraft.

Answer 1

To find the probability for a given flight of a vintage aircraft with unreliable engines, we need to consider each failure scenario separately. The probabilities of no engine failure, one engine failure, two engine failures, and three engine failures are calculated step-by-step. The probability for a given flight is the sum of the probabilities from each scenario, which in this case is 0.738.

Step-by-step explanation:

To find the probability for a given flight of this aircraft, we will consider each failure scenario separately and calculate the probability at each step.

Scenario 1: No engine failure

The probability of no engine failure is the complement of the probability of any engine failing, which is 1 - 0.1 = 0.9 for each engine.

The probability of all four engines not failing is the product of the individual probabilities, so 0.9 x 0.9 x 0.9 x 0.9 = 0.6561.

Scenario 2: One engine failure

The probability of one engine failing is 0.1 for each engine.

The probability of one engine failing and the other three engines not failing is 0.1 x 0.9 x 0.9 x 0.9 = 0.0729.

Scenario 3: Two engine failures

The probability of two engines failing is 0.1 x 0.1 = 0.01.

The probability of two engines failing and the other two engines not failing is 0.01 x 0.9 x 0.9 = 0.0081.

Scenario 4: Three engine failures

The probability of three engines failing is 0.1 x 0.1 x 0.1 = 0.001.

The probability of three engines failing and the other one engine not failing is 0.001 x 0.9 = 0.0009.

Adding up the probabilities from each scenario, the probability for a given flight of this aircraft is:

0.6561 (no engine failure) + 0.0729 (one engine failure) + 0.0081 (two engine failures) + 0.0009 (three engine failures) = 0.738.

Answer 2

The probabilities for the vintage aircraft engine failures are calculated using combinatorics and adjusted percentages based on the number of remaining engines: 0.6561 for all engines working, 0.2916 for one failure, and 0.03888 for two failures.

Step-by-step explanation:

To calculate the probabilities associated with the engines failing on a vintage aircraft, we model the scenario mathematically considering each engine failure event independently (assuming no two engines fail simultaneously). Initially, each engine has a 10% chance of failure. If one engine fails, the remaining engines then have a 20% chance of failure. Should a second engine fail, the remaining two would have a 30% chance of failure. Here are the probabilities for each scenario:

All engines work: (0.9
`)^4`0.6561

• One engine fails (but the aircraft can still fly): 4 * (0.1) * (0.9
  `)^3`0.2916
• Two engines fail (aircraft can still fly): 6 * (0.1
  `)^2`(0.9
  `)^2`(0.8) = 0.03888
• Three engines fail (the aircraft cannot fly with less than two engines): The probability for this scenario is not needed as the plane cannot continue with less than two engines.

The probabilities above are calculated by using combinations and accounting for the increased failure rates when an engine fails. This problem uses concepts from probability theory and combinatorics to account for different engine failure scenarios.