The probability that an airplane engine will fail in a transcontinental flight is 0.004. Assuming that engine failures are independent of each other, what is the probability that on a certain transcontinental flight, a four-engine plane will experience:
(c) More than two engine failures? (Note: In this event, the airplane will crash. Round your answer to nine decimal places.)
First I am assuming that you know the AND and OR theorem if you dont know it you can ask me.
Now we need more than two engine failures so three engines can fail OR all engines can fail.
the probability of three engine failing is (0.004)*(0.004)*(0.004) which is equal to 0.000000064.
the probability of all engine failing is (0.004)*(0.004)*(0.004)*(0.004) which is equal to 0.000000000256.
Now we will add these probabilities to get the final answer which is 0.000000064256.
The situation calls for binomial distribution because all criteria are satisfied:
1. Bernoulli trials (i.e. either success or failure, one of exactly two outcomes).
2. Probability of success (i.e. failure of engine) is known (p=0.004) and remains constant throughout the trials.
3. The number of trials is known (n=4) and is a constant.
4. All trials are random and independent of each other.
The probability of x successes out of n trials each with probability p is given by:
B(x,n,p)=C(n,x)p^x(1-p)^(n-x)
where
C(n,x)=n!/(x!(n-x)!) is combination of x objects taken from n.
Probability of more than two engines failures out of four,
Probability=B(3,4,.004)+B(4,4,.004)
=C(4,3).004^3(.996)^1+C(4,4).004^4(.996)^0
=4(.004^3)(.996)+3(.004^4)(1)
=2.54976*10^(-7)+2.56*10^(-10)
=2.55232*10^(-7)
Note that this is slightly more probable than the sum of 3 failures out of 3 plus 4 out of 4.
To find the probability of more than two engine failures on a certain transcontinental flight:
Step 1: Identify the probability of an engine failure.
The probability of an engine failure in one transcontinental flight is given as 0.004.
Step 2: Determine the number of engine failures that would result in a crash.
In this case, a crash occurs if more than two engine failures happen on the flight.
Step 3: Calculate the probability of more than two engine failures.
To find this probability, we need to calculate the probability of exactly three engine failures, the probability of exactly four engine failures, and sum them up.
Let's calculate the probability of exactly three engine failures:
P(Three engine failures) = (probability of failure) * (probability of failure) * (probability of failure) * (probability of no failure)
P(Three engine failures) = 0.004 * 0.004 * 0.004 * (1 - 0.004)
Now, let's calculate the probability of exactly four engine failures:
P(Four engine failures) = (probability of failure) * (probability of failure) * (probability of failure) * (probability of failure)
P(Four engine failures) = 0.004 * 0.004 * 0.004 * 0.004
Finally, to find the probability of more than two engine failures (and hence a crash):
P(More than two engine failures) = P(Three engine failures) + P(Four engine failures)
Calculate the values and round the answer to nine decimal places.
To find the probability of more than two engine failures on a certain transcontinental flight, we need to calculate the probability of exactly 0, 1, and 2 engine failures and subtract that from 1.
The probability of a single engine failure is 0.004. Since engine failures are assumed to be independent, we can use the binomial probability formula:
P(x=k) = C(n, k) * p^k * (1-p)^(n-k)
where:
- P(x=k) is the probability of getting exactly k engine failures
- n is the number of trials (in this case, the number of engines, which is 4)
- k is the number of successes (engine failures) we are interested in
- p is the probability of success (engine failure) on a single trial
For exactly 0 engine failures, we have:
P(x=0) = C(4, 0) * (0.004)^0 * (1-0.004)^(4-0)
C(4, 0) is the number of ways to choose 0 out of 4, which is 1.
P(x=0) = 1 * 1 * (0.996)^4
For exactly 1 engine failure, we have:
P(x=1) = C(4, 1) * (0.004)^1 * (1-0.004)^(4-1)
C(4, 1) is the number of ways to choose 1 out of 4, which is 4.
P(x=1) = 4 * 0.004 * 0.996^3
For exactly 2 engine failures, we have:
P(x=2) = C(4, 2) * (0.004)^2 * (1-0.004)^(4-2)
C(4, 2) is the number of ways to choose 2 out of 4, which is 6.
P(x=2) = 6 * (0.004)^2 * (0.996)^2
Now we can calculate the probability of more than two engine failures:
P(more than two engine failures) = 1 - P(x=0) - P(x=1) - P(x=2)
P(more than two engine failures) = 1 - [1 * 1 * (0.996)^4] - [4 * 0.004 * 0.996^3] - [6 * (0.004)^2 * (0.996)^2]
Calculating this expression will give us the probability of experiencing more than two engine failures on a certain transcontinental flight.