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 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.)
2 answers
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.
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.
9 answers