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Marriage Anniversaries Suppose all the months of the year are equally likely as marriage anniversaries. Glen and Shahid are two randomly selected married males (unr...

Question

Marriage Anniversaries Suppose all the months of the year are equally likely as marriage anniversaries. Glen and Shahid are two randomly selected married males (unrelated).a. What is the probability that they were both married in August?b. What is the probability that Glen OR Shahid was married in August? Hint: The answer is not $2 / 12$ or $1 / 6$. Refer to Guided Exercise $5.25$ if you need help.

Marriage Anniversaries Suppose all the months of the year are equally likely as marriage anniversaries. Glen and Shahid are two randomly selected married males (unrelated). a. What is the probability that they were both married in August? b. What is the probability that Glen OR Shahid was married in August? Hint: The answer is not $2 / 12$ or $1 / 6$. Refer to Guided Exercise $5.25$ if you need help.



Answers

Three married couples have purchased theater tickets and are seated in a row consisting of just six seats. If they take their seats in a completely random fashion (random order), what is the probability that Jim and Paula (husband and wife) sit in the two seats on the far left? What is the probability that Jim and Paula end up sitting next to one another? What is the probability that at least one of the wives ends up sitting next to her husband?

The birthday problem, asks in a group of some number people say 30 unrelated people. What's the probability? At least to have the same birthday is to make this easier. We ignore February trying night. We don't care about that sweet soon. There's only 365 days. We assume that, um, each person is equally likely to be born eyes equally likely to become their birthday on any the 365 days. It's so if we assume that each of the three and 65 days is equally likely for any given a person's birthday. So how would we try to figure out this probability using random digits to do one repetition of the process? Well, Ah, okay. We could for each person, generate a random number from one, 2 365 Casey. There's online random number generators, or if you're using a random digit table, you could look at three digits at a time and ignore it if it didn't fall in the range. 12 365. Do you generate a random number for each person? So in total, you'll have 30 random numbers in this range from 1 to 36 365 and then see if any of the two numbers are the same. And it's so then, uh, that would correspond to two or more people of the group having the same birthday. So heart be asks us to do this for five times, so I'll leave you to do that. Um, so simulate the process five times. So each time generate 30 random variables, random numbers in this and then record. Whether or not you got me that were the same, we'll see what happened. In part, C says that the theoretical probability. So the true probability is actually 0.71 Are you surprised? This does seem quite high. Ah, it seems somewhat unlikely that there would be in a group of 30 people. But if you think about it, 30 people is quite a large group, and in particular there's a lot of sub different pairings. Um, if you look at the number of pairs of people in a group of 30 people, that number is quite large. So you know, depends on perhaps what you got from the simulations or this would be a fun game to do the next time you're in math class, have everybody go around and say they're probability are, say their birthday and see if any two people have the same birthday and it's likely that they will.

If two people are selected and random the populated that they do not have The sympathy day that this day and month is 365 over renders its end under 64 on the thread goes this far So okay, say Sarandos and survive over Remember, 65 times 364 65 suddenly for for the very festive isn't all the days are available. So for you For for him to choose ese allowed the truth any off the 365 days in a year. All right, so visit number I'm this Is this in one and in person, too? Um, he is, uh, no longer allowed to truth one off the days before. Which means the day that this one have chosen it's no longer if I live or to be chosen by their again. Pisser. Okay, so So basically, does the explanation. Like this statement is correct then. Secondly, you're supposed to explain why this is so okay. That's what we've done. So second least, if three people are selected at random, find the probability that they all have different bad days. So for them to have different birthdays, they should be the fifth person. 3 6500 to 5. Second person 360 for 3 65 Dead person. 3 63 over. But isn't it? And then Principalities is really lovely. I bought, uh uh, very bigness. Indigent dude. 99.2 person pounded off. Then see, um, the person stays if three people are selected. Uh, it's random. Find the probability that at least two off them have the same better t so at listed to off them have the same bed today. Means that ah, to off them all. Three off them and the sympathy. So what we'll do is to do to work out where No, no them is. What of the sympathetic? Then we work out the compliment off it. All right, so that way we have already done. Ah, very big a part ofit. Ah, we have done the that the But so is the compliment off. See, which we're doing Because what is happening is these people, all of this they do not share the same brevity. So if I subject to those that do not share the same birthday, I'm left with those that share at least to off them. Sharing a bed. So I'll just say one my mass 99.2% which is a little point little eight, which is roughly about eight a little bit, because it'll aid Thio piss and ears. No, no, is it opens at eight. On the market Descent. Ah, Then that's a sort of instead of my mama is set up in 008 Uh um, a part of me on that one, then Number D if, um, do you say's if 20 people are selected at random, find the probability that at least two off them have the same birthday. So in that kiss, we'll need toe find, at least too. Which means if we can find zero off them having the symbol, they weaken just subject to that and then get at least to all family. So how do you go about it? We were, uh we were, uh, on D. Hey, we you want at least two people will share the bed day, So, basically, if we know all the people wouldn't share, they didn't just, uh, work of the compliment. So, in gems off all the people, uh, let me put another page. We would say uh, those would wouldn't share that. There was a one and then my mass all those would own share. Okay, so this is one man us. So for those who don't share, uh, they're Toto. We all know that there are 365 and they are under 65 years. Is the denominator for each and every person off the 20 people. So this one will be powered 20 and ah, then, uh, the human later now is what made us so the new marriage, the way in which it is governed. Legent pigeons car planted those follows. So the 1st 1 will have all the days a violin bow, the 2nd 1 30 64 over presence defiles the dead one 3 63 All of us 65 And it's continuous like that. I threw a gator. Very left. 1/20 1 Um and I think the 20th 1 would be ahh 34 device off the race. It's the fact because it's up to bat now to summarize this venue, Merida would suit you say three, because prisons before it isn't wife visited for president to be going down. So it's some combination. So we can say 365 Victorio divided by 300 40 on 344. Victoria. So that's the humanity has become that you murdered. So basically we can put it. Is that so? It will be 3 65 Victory are divided by every number 44 Victoria, I do not see him step up the attic chocolate to work out this one. My mom will always be Paris. But basically, if you do this, you should be able to get in from our cochlear. Huh? I celebrated for one too. So it's a little point for 12 Then the last question say's Ah, how many large groups is needed to give? Is there a 0.5 chance off? At least two people have the same birthday out. We're good guidance e We do. We have syrup in 412 already so we can put trial ed. One more person to our group. We have got 20 with this figure with 20. So if we add an additional person, that should increase our probability. Um, so if you do that trial in era, then the last question would give you Ah, 23. So if you just end one more person, you'll be closer to little 0.5

The birthday problem. Question it says, What's the probability that out of a group six of six dudes, at least to have the same birth month So what? We're gonna start flipping around? I'm gonna say, what's the probability? But no one, minus the probability that nobody has the same months. Okay, so we're gonna go one minus if I'm born, listed and born in December rate than the Net first and has a month to, um, not be born in that month and then and then nine and then eight and then seven, because we don't want them to have the same birthday. So let's see what that gives me. So I'm gonna go one minus, and then we do it slightly differently. It's going to be 11 times, 10 times, nine times eight, prime seven invited by, um, 12 Frieze of the Fifth Power. It just makes it a little easier on the calculator to do it that way. Go and it gives me this. If I kick that and actually get an approximation, it's about 0.777 All right, so that's my answer.

So in this exercise we were given the information that there are 366 days in a year. All all birthdays are equally likely in all birthday months are equally likely. The party of the problem asks us. What is the probability that two grand and people are born on the same months? And the probability is one over 12? Because the problem tells us that ah person is born that all birthday months are equally likely in there are 12 months and a year, so it probably is one of 12 for B. We have a group of people. What and we're asked to show to know what is the probability that to our foreign during the same month, so to born on same month, that is the same as subtracted, so attracting one from the probability that everyone was born in a different month. There are 12 months total, so there are 12 to the end possible total outcomes since every person has told months so they can be born in if everyone is born in a different month, the first person has 12 options and the second person has 11 options. The third person has 10 all the way until the end person, which has 13 minus and options. Therefore, by the multiplication rule. The probability that everybody is born and on a different month is 12 times I'm another probability. But the that all the possible outcomes where everybody is run on a different month is 12 times 11 times 10 doctor, about times 13 miners end and the probability that everybody is born in a different month. ISS This number 12 times 11 times 10 all the way to times 13 months n divided by the total outcomes which is 12 to the end. And this is the probability that everybody is born in a different month. Therefore, the probability that two people are born in the same month is one minus 12 times, 11 times 10 times all the way to 13 hundreds and over 12 to the end. And this is the answer. Patsy asks us by what is the minimum end so that we can have but the probability that people are born on the same month be creator than 50% through that 0.5 by observation, we observed that the probability that so the probability that winning five people for the probably to be greater than 1/2. So the answer is that we need five people and I believe would be used. Your calculator. You get that one. If they're If N is equal to five, we get that this number is 2057 or 20570.50 to 1.


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