Airplane Engine Problem: One reason
commercial airplanes have more than one
engine is to reduce the consequences should
an engine fail during flight (Figure 9-7c). Under
certain circumstances, some counterintuitive
things happen when the number of engines is
increased. Assume that the probability that
any one engine will fail on a given flight is 0.1
(this is high, but assume it anyway).
Figure 9-7c
a. For a plane that has four engines, calculate
the probabilities that zero, one, two, three,
and all four engines fail during the given
flight. Show that the probabilities sum to 1,
and explain the significance of this fact.
b. If the plane will keep flying as long as no
more than one engine fails, what is the
probability that the four-engine plane keeps
flying?
c. Suppose a different kind of plane has three engines of the same reliability and that it,
too, will keep flying if no more than one
engine fails. What is the probability that the
three-engine plane keeps flying?
d. Based on your computations in this
problem, which is safer, the four-engine
plane or the three-engine plane?
a. To calculate the probabilities, we need to use the binomial probability formula. The formula is:
P(X = k) = (nCk) * p^k * (1-p)^(n-k)
Where:
P(X = k) is the probability of exactly k successes,
n is the number of trials or engines,
k is the number of successes or engines failing,
p is the probability of success or one engine failing,
(1-p) is the probability of failure or one engine not failing, and
(nCk) represents the number of combinations of n items taken k at a time.
In this case, we have n = 4 engines and p = 0.1 probability that an engine fails.
For zero engines failing (k = 0):
P(X = 0) = (4C0) * (0.1^0) * (0.9^4) = 1 * 1 * 0.9^4 = 0.9^4
For one engine failing (k = 1):
P(X = 1) = (4C1) * (0.1^1) * (0.9^3) = 4 * 0.1 * 0.9^3
For two engines failing (k = 2):
P(X = 2) = (4C2) * (0.1^2) * (0.9^2) = 6 * 0.1^2 * 0.9^2
For three engines failing (k = 3):
P(X = 3) = (4C3) * (0.1^3) * (0.9^1) = 4 * 0.1^3 * 0.9
For all four engines failing (k = 4):
P(X = 4) = (4C4) * (0.1^4) * (0.9^0) = 1 * 0.1^4 * 1
Now we can calculate these probabilities:
P(X = 0) = (0.9)^4 = 0.6561
P(X = 1) = 4 * 0.1 * (0.9)^3 = 0.2916
P(X = 2) = 6 * 0.1^2 * (0.9)^2 = 0.0486
P(X = 3) = 4 * 0.1^3 * 0.9 = 0.0036
P(X = 4) = 0.1^4 = 0.0001
To show that the probabilities sum to 1, we add them up:
P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.6561 + 0.2916 + 0.0486 + 0.0036 + 0.0001 = 1.0
The significance of this fact is that the sum of the probabilities represents the total probability of all possible outcomes. In this case, it means that the probability of either zero, one, two, three, or all four engines failing during the flight is 1 or 100%. This represents the exhaustive set of possibilities.
b. Since the plane will keep flying as long as no more than one engine fails, we need to calculate the probability of having either zero or one engine failing.
P(X <= 1) = P(X = 0) + P(X = 1) = 0.6561 + 0.2916 = 0.9477
Therefore, the probability that the four-engine plane keeps flying is 0.9477 or 94.77%.
c. For the three-engine plane, we use the same approach as before, assuming the probability of an engine failing is 0.1.
P(X = 0) = (0.9)^3 = 0.729
P(X = 1) = 3 * 0.1 * (0.9)^2 = 0.243
P(X = 2) = 3 * 0.1^2 * (0.9) = 0.027
P(X = 3) = 0.1^3 = 0.001
P(X <= 1) = P(X = 0) + P(X = 1) = 0.729 + 0.243 = 0.972
Therefore, the probability that the three-engine plane keeps flying is 0.972 or 97.2%.
d. Comparing the probabilities, we can see that the three-engine plane has a higher probability (97.2%) of keeping flying compared to the four-engine plane (94.77%). Therefore, based on the computations in this problem, the three-engine plane is safer.