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03 34 A hiker leaves point A shown in Figure 2.13 below. choosing at random one path from AB; AC, AD and AE. At each subsequent junction she chooses another path at...

Question

03 34 A hiker leaves point A shown in Figure 2.13 below. choosing at random one path from AB; AC, AD and AE. At each subsequent junction she chooses another path at random_ but she does not immediately return on the path she has just taken: You meel the hiker at point X-Figure 2.13: Paths for HikerWhat is the probability that the hiker came via point B? What is the probability that the hiker came via point C? What is the probability that the hiker came via point D? What is the probability that t

03 34 A hiker leaves point A shown in Figure 2.13 below. choosing at random one path from AB; AC, AD and AE. At each subsequent junction she chooses another path at random_ but she does not immediately return on the path she has just taken: You meel the hiker at point X- Figure 2.13: Paths for Hiker What is the probability that the hiker came via point B? What is the probability that the hiker came via point C? What is the probability that the hiker came via point D? What is the probability that the hiker came via point E? What is the probabilily that the hiker came via point B and C? Consider both paths B and € mutually exclusive What is the probability that the hiker came via point B and C? Consider both paths B and € independent. What is the probability that the hiker came via point B or D?



Answers

Probability The point $P(3,2)$ is translated along one of the following four vectors chosen at random: $\langle- 3,0\rangle,\langle- 1,-4\rangle,\langle 3,-2\rangle,$ and $\langle 2,3\rangle .$ Find the probability of each of the following. a. The image of $P$ is in the fourth quadrant. b. The image of $P$ is on an axis. c. The image of $P$ is at the origin.

In this problem, we have a scenario where somebody is going home and there are two possible routes. The first route has to railroad crossings and the second has four crossings. The person will be laid if on route at least half of the railroad crossings on the route. Um, he gets stopped at due to a train crossing and the probability of getting stopped at any individual crossing is 0.1. So for party were asked which route should be taken either route one or two to minimize the probability of being late? Let's calculate the probabilities of being laid for each of the roots. So the probability of being late given that we take Route one, So Route one has four crossing. So we have to be stopped at at least two of them in order to be late to is at least half of them. Now this is equal to the probability that were stopped sequel to one, minus the probability that were stopped at either zero or one crossings, and this is equal to one minus. So the probability of being stopped at zero crossings is 0.1. Sorry, that's should be 0.9 to the exponents for 0.9 is the probability of not being stopped at any individual crossing. So to not be stopped at four, the probability is 0.92 exponents for plus the probability of being stopped at one. So we have to have three successes. So that's 30.19 exponents three and one where were stopped. So that 0.1 to explain it one. And then there are four times to arrange these four ways to arrange these. So we get a probability of 0.5 to 3. And now for the second route, the probability of being late given that we take this route is the probability of being stopped at one or both crossings. And then we have to subtract from this intersection. And that's because being stopped at one intersection and being stopped at two intersections are not mutually exclusive events. This equals 0.1 plus 0.1 minus 0.1 squared. It comes out to 0.19 And so the conclusion is to take the first route because the probability of being laid is lower. Now, for part B, we're told that he takes either route with equal probability. And if he is late, what is the probability that he took the first route? So what is the probability that the first rate was taken, given that he is late and here we can apply Bayes rule? So this is equal to the probability of the first rate and being laid divided by the probability of being laid. So the numerator is theirs. Probability of taking the first Rufus half times the probability of being late 0.5 to 3. That's the probability of being late, given that the first rate has taken and in the denominator, we use the the rule of the law of total probability to find the probability of being late and thats half times zero point 05 to 3, plus half times 0.19 and this comes out to 0.216 So, given that he is late, the probability that he took the first route is 0.216

Let's start to look at this example, find the probabilities and indicate whether the 5% guard guideline for cumbersome calculations is used. Okay. Medical helicopters in a study. Okay, in a study off medical helicopter usage and patient survival, results were obtained from 47,637 patients. 47,000 637 patients transported by helicopter on 111,000 and 34. Patient transported by ground 111874 patients 111 874 patients were transported by ground. Okay, if one off the 159,511 patients in the studies randomly selected, what is the probability that this subject was transported by a helicopter? Okay, so this is nothing but the division of these two. The division of these 2. 47,637 637 divided by 111,874. And this turns out to be 0.4 to 5 years. This turns out to be 0.4258 It's This is my probability. If five of the subjects in the study are randomly selected without replacement. What is the probability that all of them were transported by helicopter without replacement? What is the probability that all of them were transported by helicopter? This is part now we're looking at part D at fat B. Barbie is going to be against the probability for one patient is the division of these 247637 divided by 111874 But now we have five such patients and this is without replacement. This is without replacement tree. Yeah. So what is the probability? In this case, I can simply divide this and multiplied five times. What I can see that this is nothing but zero point 4 to 58 Raise to five. Or this is 0.13 990 point 01399 Or I could write a zero point 014 So there is only 1.4% chance that this happens

In this study, it was given that 47,637 patients were transported by helicopter on 11 or 111,874 patients were transported by ground. The total is 15951 When the total is 159511 the total is 159 511 The first question is, if one patient is selected randomly. Second, what is the probability that the subject was transported by a helicopter? So this is nothing but 47,000 47,637 637 The These are more favorable outcomes upon total number of outcomes. And what does this turn out to be? This turns out to be 0.299 This is the probability 0.29 nine. This is my party. The second one is Barbie is a five of the subjects in the study a randomly selected without replacement. This is important without replacement. What is the probability that all of them were transported by helicopter? Okay, because this is without replacement. We can say that all of these are independent wait with or without replacement. Okay, without replacement. All right. So all of these are dependent. If they have without replacement, it is. All of these are dependent. So part B is going to be Let's take first patient. What is the probability off him or her being transferred by helicopter transported by helicopter? It is going to be this 159511 Now looking for the second patient. Second patient is 176 three 17636 Why? Because we have said the first patient aside, we're considering that he was transported by the helicopter and now we're setting him aside. So this reduces by one and this reduces by one also. 159510 This carries on still five terms. 76543 So this is going to be 47633 upon 159 five zero, seventh. This is the probability that I want. And what is just turning out to be This will be 0.239 This will be 0.239 This is the probability that I want

So we have the traffic lights and the probability of being stopped at a traffic light we're saying is .4 and there are three different traffic lights that uh we're dealing with and in part day we want to know what's the probability that you get stopped on exactly two of these lights. And so the probability of getting stopped just point to and what To excuse me, .4 we want two of those. .6 is not getting stopped and we want to end up having one of those and that's going to be times a combination of three, choose to and that value is three. So if we take three Times .4 Squared times .6, that gives us a probability of .288. So about a 30% chance of that happening Now, we want to know which one is bigger. The probability of being stopped at more than one Or the probability of the number of stops being at no more than one less than or equal to one. And this value is going to be two and three. So let's take that .28. And then we're going to find out the probability of three. And the probability of three is going to be that three, choose three Being stopped at all. three times .6-0 at so this is that's one, this is one. So it's going to be just adding on to that answer the .4 to the third power and that comes out to be .352. So this probability is its complement is going to be 1- that .352. And we can see that this is going to be more likely. This comes out to the point .648. So this probability is more likely than the other of being stopped at more Now on part C. They said you make four trips, you have a total of 12. So basically our probability distribution would go from if I let us stand for the number of stops and this is the probability of S and you'd have it going from zero all the way up to 12. We would be using that combination of there are 12, choose as the number of stops. We would have that 120.4 to the S. Power and then we would have the 0.6 to the 12 minus s power. And I'm going to put in just a few of these. I'm not going to do right down the entire distribution but this is very very unlikely for this one to happen. This is the only .002. And this next one of being stopped at exactly one is only .017. Um Being stopped at 2.06. I would round that to four and then I'll just get these last couple And then you could get all the rest. This one is basically 1.00004 Zeros. And to this one is um point that can read it right three times 10 to the negative fourth powers of 123 and a three. So these are teeny tiny down here, and you can find the rest with that same distribution formula.


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