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Conect0.75incomect781250.78125InconectAt least one of the answers above NOT comec:point)fair coin lossed three times and the events A, B. and are delined a5 follows...

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Conect0.75incomect781250.78125InconectAt least one of the answers above NOT comec:point)fair coin lossed three times and the events A, B. and are delined a5 follows:At least one head observed B : At least two heads are obsenved The number heads observed oddFind the following probabililies by summing Ine probabilities 0f the approprate sample points (note that 'even numbery{a) P(C) = P(B or C) = P((pot A) or B or C)78125

conect 0.75 incomect 78125 0.78125 Inconect At least one of the answers above NOT comec: point) fair coin lossed three times and the events A, B. and are delined a5 follows: At least one head observed B : At least two heads are obsenved The number heads observed odd Find the following probabililies by summing Ine probabilities 0f the approprate sample points (note that ' even numbery {a) P(C) = P(B or C) = P((pot A) or B or C) 78125



Answers

When a balanced dime is tossed three times, eight equally likely outcomes are possible: $$\begin{array}{llll} \hline \text { HHH } & \text { HTH } & \text { THH } & \text { TTH } \\ \text { HHT } & \text { HTT } & \text { THT } & \text { TTT } \\ \hline \end{array}$$ Let $A=$ event the first toss is heads, $B=$ event the third toss is tails, and $C=$ event the total number of heads is 1 a. Compute $P(A), P(B),$ and $P(C)$ b. Compute $P(B | A)$ c. Are $A$ and $B$ independent events? Explain your answer. d. Compute $P(C | A)$ e. Are $A$ and $C$ independent events? Explain your answer.

This is a binomial probability question. So I have gone ahead and written out the binomial distribution up here. We're being asked multiple questions on this problem. Personally, I think easiest way to handle this is simply for us to go through. And let's just find the probability of each of the different scenarios. Meaning, you know, if we're tossing five coins, we could have one head to heads three EDS. Foreheads five heads. Right? So let's just find the probability of those scenarios happening. And then after we've got the probability of each of those individual scenarios, we can look at A, B, C and D, and we can add in whatever probabilities we need to for each of those answers. Okay, so I want to go ahead and I want to solve out for the probability of getting one head. I want to solve out for the probability of getting two heads. I want to solve that for the probability of getting three heads do foreheads. And I also want to do five heads, okay. And what we're gonna do, we've got this binomial distribution up here, so I'm just going to go ahead and fill that out for each of these and then plugging in a calculator to figure out what our probability, what the what the chances are of each of these scenarios happening? Do you remember what I'm saying? Plugging in the calculator? We can plug in this formula as it is, um, to the calculator. Don't forget you also have the ah binomial pdf function. Possibly, if you have a calculator with that that you may be able to just plug in, um, an r p and Q and just solve it out that way. You guys talked about that was talked about in the last section of the book. So if you're not sure how to use your calculator for it doesn't matter. We have this formula right here. You don't have to have a binomial the probability function on your character, but if you do, you can use as well. So if we want the probability of one head, remember, for the binomial distribution, there's four important things here and is the number of trials that we're doing well in this case, we're flipping a coin five times. So N is going to be five throughout this entire problem. Okay, are is the number of successes that's going to change based on which of these probabilities were trying to find. So if I'm starting with finding the probability of one head, then that means we're doing five trials. But we only plan on getting one success I have. The problem is, getting one hit ahead is a success, right? Based on these questions that were being asked ahead is what we're considering a success. So we only want one success in this case, OK, but are will be different each time. P and Q P is the probability of success on any given trial. And Q is the probability of failure on any given trial. Well, for a coin flip, that's pretty straightforward. Either you get heads and it's a success or you get tales and it's a failure, right? It's a 50 50 shot, so R P and R Q will both be 0.5 because there is a 50% chance a 50 hopes a route peon siddiq you. There's a 50% chance a 50 50 shot that we will get either a head or a tail. Okay, so that's very straightforward for that meeting. The rest of my formula for finding the probability of one head would be taking 10.5 to the first power because that's Oregon and then 0.5 as our Q meaning our chances of, um, failure. And that is going to be taken to the n minus R, which would be five minus one. Or in this, his four right. So again, you can either plugged that into a calculator. Or you can just use the binomial pdf function on your calculator. Either way, what you should get if you sellable this out is zero point. Hopes one 56 to 5, which we will around to simply 0.156 I believe said to round actually didn't see it around. You know what? That does not sit around, so we will just keep it where it's at. Doesn't look like the book specifically, says Durant's. We'll just leave it that So we want to find the probability of two heads. All right, well, it's gonna be kind of the same idea. We still have five trials, but this time we want to head. So all that really changes is that my are is now, too, because our is the number of successes. So we wanted one head before. Now we want to. It's still going to be a probability of 0.5, but now it's 0.5 squared because again, we changed our. It's still gonna be 0.5 for my Q as well. But now it's five minus two, meaning we're actually going to be taking it to the third. So our is the only thing that changes, but it does change multiple parts of our formula. Okay, so again, you can your stick all that into your calculator, whether you're using the p the binomial functional, whether you're just plugging this exact math. And either way, you should get 0.31 to 5 0.31 to 5. Okay, so then, if we've got a probability of three, it's same idea. We have a total number of trials of five, but now we want three successes. Hopefully, by now, you guys are kind of getting the process that we're going through here. So now it's 0.5 to the third power and be 0.5 to the five minus three. All right, so now we put all that in two R calculator and feel free to be pausing this video to plug in your calculator if you want. I just don't want to make you sit through me plugging it. And every time we get 0.31 to fuck. Okay, so then probability of foreheads. So we'd have told number of trials of five. We're looking for four successes. That means we'd have 0.5 to the fourth power, and then we'd have 0.5 to the five minus four, Which would just be what? Okay, if we plugged that into our calculator, then we should get point 31 No, we get 0.31 to 5. Yeah. Yep. No, no. I would my captain. I just wasn't keeping up. We should get 0.156 to 5. That makes more sense. I noticed that there is some similarities here because our probability of success and our probability of failure are exactly the same. So here I took 0.5 to the fourth and 0.5 to the first, which is exactly what I did here. It was just in different order. Just what? I was confused there. Michael was gonna be 0.3125 but it was just being a little slows off. Lastly, finally, we've got probability of five heads, so that is five choose five. Meaning we have five trials and we want five successes. So that would also mean we're taking 0.5 to the fifth power. But here we're taking 0.5 to the five minus five, which means it's really to the zero power. So it's really not even doing anything for us if we're being honest. Finally, we calculate that, and that should give us 0.0 3125 Okay, so we have finally now found the probability of all our different scenarios. Okay, which is great. Now we can finally go about finding our actual answers. Part A wants us to find the probability of at least four heads. At least four heads, meaning for is the lowest number that were supposed to have for is the lowest amount of heads. That is okay, Meaning if we want to find the probability of that, four is acceptable. Five is also acceptable, cause that's higher, right for is the lowest amount of heads we can have. We are allowed to have higher amounts than that four and five are the only possible ones, so are answered part A then would be taking our probability of getting foreheads, which is 50.156 to 5, and adding that to our probability of getting five, which is 50.31 to 5. So we take 0.15 65 and add that 2.31 to 5 and you get a probability of 0.1875 That would be our probability of at least four heads. So then we take a look at part B. Part B now is asking us to find the probability of at least three heads, meaning now three is the lowest we can have. So three is acceptable for is acceptable. Five is acceptable. Well, we already know that foreheads plus five heads gave us 0.1875 right from the last problem, so we might as well use that then we need to add that to the new part, which is that three heads is also acceptable. So we need to add in that zero point 3125 So add in point 3125 and the chances of us getting three or more would be 0.5. Meaning there is a 50% chance that if you flip a coin five times, you will get 34 or five heads. Three or more heads. Which makes sense. Because, remember, we there is one scenario that we didn't solve for here, which is the probability of getting zero heads right. That is technically possible. You could flip a coin five times. They get zero heads. We didn't solve that up above because it doesn't apply to any of the questions being asked here. Part C asked what the probability of at least two heads is. All right, well, we know that three heads plus four heads plus five heads was 0.5. So now we just need to add in the chances of us getting two heads. All right, well, looking up here, the probability of us getting to heads out of five flips is again 50.31 to 5. So we're gonna add 0.31 to 5. Looks like that gives us a probability of 0.8125 81% chance. Roughly right. Finally, Part D were asked what the probability is of us getting at least one head. What we already know that two heads plus three heads plus four heads like that that handrail there plus four heads plus five heads 0.8125 So now we just need to add in the probability of us getting one head, which is right here 0.156 to 5. If we add in 0.156 to 5, we get 0.96875 So around 97% chance that if we flip a coin five times, we at least get one head, Which makes sense, Right? 50 50 shot, five chances. You would think you would get at least one head, not guaranteed, but pretty close.

Are a If I list all the possibilities for three coins, I am missing. One already tenses. And hence sales head steals and head still steals. Then, you know, sales have been going to the ham's and ah fans. Ah, hail. Yeah. I don't feel deal to, um, you right now we're removing kill, kill tail. So we have all of these doing part A. We're gonna have the probability three heads. There's only this one. At least one of the queen shows head they're seven. And there that's my answer for a and then or be, he had at least one of the coin. So you weigh for part B 18 over one minus 18 which equals 18 over 7/8 which is the same as 18 Divided by 7/8 which is the same as 18 times 87 eight and eight. Cancel out and I'm with 1/7

So we have 500 coins that were flipped and we've been looking at the number of heads and we have 01 two three. When we have three, Let's see was it being thrown, TOS three coins are being tossed not one repeatedly. And here was the actual frequency. And so for zero there were 54 for one, there were 190 198 and there were 58 for three. And we want to find for part a the empirical uh Probability and the empirical probability is going to end up being for this first event is going to be 54 out of 500 Second event. 190 Out of 500 198 Out of 500. And then this last 158 Out of 500. And so that's approximately, let's see if we take the basically we multiply the top and bottom by by two. We can get it out of 1000. But 54 divided by 500 comes out to be 10.8%. 4.108, 1 90, divided by 500. Comes out to be .38, 1 98. About it by 500 comes out to be .396 And 58 out of 500. Mhm comes out to be 0.116 Now, part B, we want to find what the theoretical probability is and we know that the sample space is. We can have heads. Heads, heads we can have heads. Heads tails we can have heads, tails, heads, we can have tails, heads, heads we can have a tails, tails. Heads we can have tails. Heads tails we can have tail. Excuse me, This should be heads, Tails tails and we can have tails tails tails. So the likelihood for the first event, the probability of having no heads, zero heads is right there and I have one out of eight which is .125. The probability of having one hat is we can see that there are three ways out of eight and three divided by eight, is that .375. And the probability of two is also 3/8 Which is .375. And the probability of three Is equal to uh 1/8 again, which is that .125. Uh huh. So we can see that these are relatively close. Now, Part C asked us what was the expected frequency? And so if I take this 18 and multiply it by 500. I'm just going to put this right here. We would expect there to be 62.5 rules for each of these for part C. Now obviously it looks obviously we can't have 62.5 but that would be the expected value. And if I take this 3/8 of 0.375 times 500 we would expect there to be 100 7.5 of each of these. Uh huh. Sorry about that. So again we're not getting what we expect here. Um It looks like I Multiplied that incorrectly because that should be higher than that so I'm not reading it right that's 1 87. I'm sorry I can't see it. That was 187. Can't read the calculator. And then if you add those rolls up they would come up so we can see that they're pretty close you're off by a little bit. And then how would we simulate it with a calculator? Um If we're going to use a calculator we could have rand aunt and we could get numbers from um one and 2 and have it generate one would end up meaning it was ahead to could end up meaning it was a tail and we would generate three numbers and we could do this 500 times. And so for instance, if it if it spit out um 112 that would mean that we had a had a head and a tail. And so we would say we have two heads And we can continue to do that with 500 trials would be quite cumbersome, but we can do it with a random miniature function.

Question number 15 Christian E Given and equal three. The financial binomial probability p off X Equal key equal in c k dot People came that one minus B Power in mine escape equal factorial in over factorial key boy in minus key with Korean The People K that one minus p or in minus k. When flipping a coin the probability off flipping Higgs end off flapping tales is equal and thus probability equal one over to 4.5 Evaluate every week at K equal three. So probability off X equal three equal pictorial three over. Factorial three multiplied by three minus three factorial multiplied by 0.543 multiplied by one point one minus 10.5 for three minus three equals 1/8 equal 4.125 equal 12.5% question number be given an equal three We would use Ah, the above the above definition. Okay, When flipping a coin, the probability off flapping heads and off flapping tales is equal and thus probability equal one over to 4.5 very with that key equal to so property off X equal to equal pictorial three over factory in tool not deployed by three minus two factorial multiplied by 5.5 for two, multiplied by one minus 0.5 for three, both minus two equal 3/8. Equal 37.5%. Yeah, question number See result at Port E and be probability of X equal three. Equal one of her eight equal 12.5% probability off X X equal to equal 3/8 equal 37.5%. Add the two probabilities since it's not possible to obtain more than three heads when flipping a coin three times. So probability off X is bigger than or equal to equal probability of X equal to plus probability of X equal. Three. Equal three overeat plus 1/8 equal fall over eat equal one of her too equal 50%


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