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Airplane Engine Problem 1: One reason airplanes are designed with more than one engine is to increase the planes' reliability. Usually a twin-engine plane can make it to an airport on just one engine should the other engine fail during flight. Suppose that for a twin-engine plane, the probability that any one engine will fail during a given flight is $3 \%$ a. If the engines operate independently, what is the probability that both engines fail during a flight? b. Suppose flight records indicate that the probability that both engines will fail during a given flight is actually $0.6 \% .$ What is the probability that the second engine fails after the first has already failed? c. Based on your answer to part $b$, do the engines actually operate independently? Explain.

So we have a situation that we're going to assume that having a flat tire Has a probability of .1 and the probability of having engine problem is equal 2.05. We're going to assume that these two events are independent. And so that the likelihood of a flat does not have any impact on the engine or the engine does not have an impact on the flat. And you may want to draw a Venn diagram of this and look at here is the probability of a flat, here's the probability of an engine. Um and we have the intersection then would be the product of these two, the product of These and this happening is going to be .005. So this, I'll do that in red is .005 in here, which means this would be .045 out here and this would be .095 in here. And so the if we sum up all these or take the .1 Plus The .5 and then subtract away the .005, We find that the red regions add up 2.145, which means this region outside here, One minus that answer is going to end up being this .855. So now let's look at the questions and I've drawn that information down. So we want to find what's the probability that we don't have a flat, no flat. And so that's just the complement of this .9 Or it would be everything outside of this circle in the circles .1 so outside the circle of the .9 we want to find what's the probability that there's no engine problem and the no engine problem is going to be again the complement of this is going to be 0.95 and that's the region outside of this e circle. Now we want to find the probability that there is neither of this up a little bit that there is neither neither a problem with a flat nor A an engine problem, which means we're outside in this region outside here and that probability is .855. See how, if we draw a graph about it, makes it much easier to see. Now we want to find what's the problem that we have, both of them have a problem and that's going to be this little region right here, which is the .005 and then finally we want to know what's the probability that you have at least one problem. So at least one. Well, at least one problem means you have to be either in this circle this circle or both and we know that when we have these three red regions together, we get this. So that probability Is going to be a .145. So if you can draw a diagram, I do recommend that. And this is assuming that we have independence between those two problems. Yes

So we're going to do the for the first problem, we're going to the probability that a new oil filter is needed. So we're going to do the probability of oil filter. And so for our numerator, we're going to put in filter intersects with oil in a world where you need an oil filter. So that's .2 divided by 0.25 Yeah. Mhm. This is forfeits. And the second problem, we're going to find the probability that the oil had to be changed. So for this problem this is going to be no, denoted as. Oh, so this is the probability that oil intersects with the filter. This is gonna be point to over 0.4 and this probability is going to be one hand, which can be good 0.4 0.5 And this can be .8.

The probability of exactly three engines functioning properly echoes the term data can change .95 Cubed and hands. The coefficient is for Times points 95 Cube Times .05, and the answer is .171.

All right. So a poem is we have Owen, we're doing a simple random sample, which means we're taking 350 randomly. A simple kind of sample of 300 the engines and probably one of each engine. In a population running is 10.999 and exit poll number number running. And we're supposed to show a couple things. First is supposed to show that X is ah binomial, Let's get first. So for extra by, no meal needs to satisfy five properties. First is that we need to have in identical trials. This is true because we're doing the same experiment every time describing engine, turning it on, seeing what happens then. Second is that they're only two possible outcomes. This is again true because every engine will either run or not run. Then we need to have that the n Well, I do. Yes, you do. In in trials are independent. This is true, because if I turned in Injun on whether it runs or not, will have no impact on whether or not the next one runs or not, we need to have a probability of success equal a pee and then probability of failure he's equal to. And I wrote that, sloppy. I'm sorry, right? That better is equal one minus B, which is the case because we have other successes in this case, 0.99 then failure every 0.1. And in. The last thing is that X is equal to the total number of success is. And that is also true because we are letting ex count the total number of interests that are running after an hour so that we have shown that it is by no meal. So ex is by no meal. All right now for B here's was to find mu of X and signal Vex all supposed to interpret our results. All right, well, this is actually pretty straightforward. Mu of X is equal to, and by normal distribution is in times P, which is 350 times 0.999 which is 350 3 49.65 And that just means that if I do some I do for any simple random sample, the average or expected well, I will use expected value. The average number of running engines will be 349.65 In other words, this is Well, this is This is what is trying to say earlier. This is the expected value. So if I were to say, how many do you expect to run? Given a simple experiment, you would say 349.65 because that is the average average is tthe e expected value. And so we've interpreted. Let's see, it's that one. And then we need to do staring deviation of X that is equal to the square root of the variance of X, which is equal to the square root of N. P. Q is equal to this word of 350. His 0.999 has 0.1 is equal to 0.591 And what does that mean? That means that for any supporting a sample, the number of running engines will be within 0.591 of you have X, which was equal Thio. What do we say? It was equal to 3 49.65 In other words, if we think of the binomial CDF, I think the binomial CDF um, the average is going to be here. This is gonna and then right here. But it's too full, Stephen. So if I draw this kind of draw an example here, if I do the experience over and over again, I would expect that because I know the standard deviations 0.591 I respect that 60. That's 68% of the experiments that I worked hard to keep doing this simple random sample over and over again. I would expect that about 60% them would result in the average being well that the number of running engines being let's see of one sir deviation away from the mean is 10.591 So if I do this experiment over and over again, or take 350 random ones and I check how many ran, I would expect that the averages 349.65 and at every experiment that 60% of experiments would lie within 349.65 minus one standard deviation, which was or 2.59 2 343 149.65 plus 0.59 which would be 349 point. Those six 2 300 50 point 24. So, yes, if I did, like conceited 10 experiments. If I did 10 experiments, I would expect that of those 10 like roughly like 6 to 7 of them, or 60% of those 10 so 6 to 7 of them would result in the number of engines running, being somewhere between 304 9.6 to 350.24 In other words, I would expect 60% of time for all of them to work, since the whole number between these two numbers is 3 50 But that is like going way. The odd or trying to answer the simple answer is this furnace to standard, simple random sample. The number of running engines will be within 0.591 of New Vex. All right, and now the last one and was see fine. Probably the X is less cynical. 3 48 It's the probability and X is less than equal to 348. Well, most, um, a little note here. Most calculators, like you're using a T I 84. We'll give you the probability that X is less than or equal something So if we're using, uh, most car caters to answer this problem, we would just do the binomial CDF of. We have 350 trials, probably of successes 0.999 and we want to go up to 348. And that will get me. I used my online calculator 0.0 49 ish. So pretty unlikely another way to get the same answer. But if you were to do it by not using the sea of calculator and calculate the terms individually, you could do the probability of X. A single to 348 is equal to the probability earlobes one minus the probability that X is greater than 348 which is equal to one minus probability. That X is equal to 349 minus probability that X that's equal to 3 50 and I would be a simple way to do it as well. But if you calculate those two things and subtracting from one, you would get this answer. So the question was that if we have two of them fail out of this simple random sample experiment, um, and we have to fail What does that say about the advertised success rate of point 999? Well, 0.49 implies for the probability, all right, that little friendly probably that X less than or equal to 348. Being equal 2.49 implies that is very unlikely. Two engines fail, so the advertised of 99.9% is most. That's a bad most, isn't it most likely inaccurate? Because if we have to fail, that's incredibly unlikely to happen. So it would suggest that this advertised rate is false and we are done.