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Toss a dice twice: Let E be the event that the sum of the resulting tosses is odd, let F be the event that at least one of the tosses lands on 1,and let G be the ev...

Question

Toss a dice twice: Let E be the event that the sum of the resulting tosses is odd, let F be the event that at least one of the tosses lands on 1,and let G be the event that the sum is 5. Assume that (1,2) and (2,1) are considered different outcomes, that is, we care about the order of the tosses Which of the following describes (E nF) - G. (Suggested approach: rule out options that do not describe the set rather than list out all of the set's elements and find match)0 ((1, 2). (1,4). (1, 6)

Toss a dice twice: Let E be the event that the sum of the resulting tosses is odd, let F be the event that at least one of the tosses lands on 1,and let G be the event that the sum is 5. Assume that (1,2) and (2,1) are considered different outcomes, that is, we care about the order of the tosses Which of the following describes (E nF) - G. (Suggested approach: rule out options that do not describe the set rather than list out all of the set's elements and find match) 0 ((1, 2). (1,4). (1, 6)} 0 ((1,4), (4, 1). (2, 3). (3.2)} 0 ((1,2).(1,61,(2, 1),(6,1)} 0 ((1, 1), (1,.3), (1, 5), (3,1), (5,1)}



Answers

Find the probability for the experiment of tossing two dice. The sum is odd.

Bird 24. Um, A They ask if e n f are mutually exclusive and the answer will be no. And the reason for this is that there is the number of four which satisfies both ese conditions, and they asked to find probability of the event of either E or F. So to do this, we're gonna have to use the question What? And the probability of Evangelion for us. Probability of f minus provocative P and f. Okay, so to fill out this, um, go find probability of P you first. And for that it has to be a number greater than three. And there is 445 and six. So that is three over his tricks. Does P of e so defined probability of ass. It has to be a number less than far. And for a number to be lessened, there are 2/3 because there is 1231234 possibilities out of six. And finally, to have both e and f, there's only one possibility, and that would be four. So this only one possibility out of six. All right, so now we can simplify this and we got three over six plus four over six minus 16 So the probability is one. And if we really think about it, that makes a lot of sense because in any case, we're always going to get a number. If we take one between one and six that meets either the condition of E or F. So this was like a really long way to explain such a simple concept. If you really think about it. So now we're gonna turn to be And this thing about this, um so are Ian f mutually exclusive. And if I think about it, any number that is less than three will not be divisible by three. So I think the NF are mutually exclusive, so the answer is yes, and because they're mutually exclusive, we have pretty easy formula compared to the last one. And to find the probability of P and that would be p clean class. And I mean to say it so usually we'll have this part of the equation right here, but because they are mutually exclusive, this is basically zero because say, Okay, that's better. So this was the equation, and I want to say, was that p the probability of P and F. We're basically be zero because they are mutually exclusive and they cannot happen at the same time, so that occurrence will never happen. So now we just need to think about what the probability of event is and if the number is divisible by three. So there are two possibilities of that. There's number three and number six. So to over six would be probability of Event E. And the probability of F is if the number is less than three, which is true over six because the only numbers less than three are one into, and those are two occurrences, so the answer would be 446 or 2/3.

So in this question we are basically given the sample space of rolling a two sided um to dies basically. And for us to first identify certain events. So the event end is basically the event that the sum is at least nine. So we're going to look At all of the events in the entire sample space of Rolling two Dice. So for the sum to be at least nine, nothing in our first role, in our second role nothing as well. Third role, We have one event 3, six, 4th row, we have 4,5, 5th rule, we have 54 and six through. We have 63 So that constitutes all of the events where the sum is at least nine. Now T at least one of the dies is a truth at least one is it too? So we have 1 to at least one of the devices are too and then we have 21 two, two or three 4 to 5 26 We have 32 again and then we have 42 52 Nothing else over there. And then six two. And then we have the third event at For at least one of the dies is up five at least one is five. So we have 15 to fly 35 four or 5. And then we have my warning 525354. Fine fine. 56 and six. Bye. So we have our three events, N. T. And F. Now first we are asked to find the probability of and so the probability of end is basically 1234 over. There's 36 events Which is 1/9 and that's equal to point 11 The probability that the sum is at least part B. The probability of N. Even X. Is basically the probability according to conditional events. The quality of the intersect with death divided by the probability of. So we are given the event F. If it's basically this event, and if we have 123456789 10, 11. So we're just going to Use events. So we're going to take our denominators 11. And the probability of any given f. So we're going to see where the intersect. So the intersect at 45 and five, so that's true out of 11 and that's equal to point one heat. Now, the third we asked the probability of and given T. So we're given the event T. And we have 1234567 8, 9, 10, 11 Events in T. And the probability of any given T. So that let's see how many intersect here. So we have Let's see if 36 intersex. So it doesn't for 55 for 60, none of them, because at least one of the ties has to be true and that's never going to happen with this event. So We basically have zero. Our third Cancer here is zero. And were asked to determine from the previous answers whether or not the events N and F are independent and whether or not, and and t are independent. So we're going to compute here. So to see an N. F are independent. We're basically going to take the probability of 10 times the probability of f. So a probability of N. We have 1/9 probability of event F. So we have 123456789 10 11 11/36. So Times 11/36. So 1/9, times 11/36 is approximately point for three or now the probability of and intersect with that and intersect with F. We basically had two events out of 36. So that's two out of 36 Which is find 0.56. And so we say that they are not independent because the probabilities are not this is not equal to this. So then an F. Or not independent. Now R. N. And T. Independent. Let's take a look. So we have event. And here we have event. See here now the probability of event and intersecting with event T. was basically zero. So the probability of an intersect with teen was zero. Now the probability of end we know is 1/9. The probability of T. We know is 234567 You know it's 11 over 36. And we know that these are not equal. So again they are not independent. So and until they are not independent but they appear because their intersection is zero, they appear to be mutually exclusive. So we have our answer to part mm B. C. Andy

For problem 4.20 to dissect road, and we're supposed to find the probabilities of certain events occurring. But first, in part, they were asked, why is the set 234 through 12? Not a useful sample space. So it recall that for most of these problems were looking at the probability of certain events occurring on the role of two dice. So one type of outcome may be the some of the dice. For example, the dice made some to three if we roll a one in a to or if we roll a two and a one. So there are two ways for this event to occur. So it might be tempting to define the samples basis simply by the different types of events that can occur. For example, 2343 12. However, the problem with this is that some events are more probable than others. There are only two ways to roll a sum of three, but there are four ways to roll a sum of four, for example, so recall that a sample space, by definition must contain elements that are equally likely. So a set of outcomes with different probabilities, does not qualify as a sample space in the first place. Moreover, we may be interested in events that are defined differently than the sum, as we will see in part B of this problem. So to summarize that set of numbers 23 12 doesn't meet the definition of a sample space because the elements are not equally likely. But on a practical level, we might also be interested in different types of events that can occur that are not part of the numbers. Two through 12 for example, we might be interested in the event that the white guy is a one moving on to party. Whereas what is the probability that the White diet is odd so we can define that event? Is white guys on? And one way to look at this problem is we're willing to dice, and we're only interested in the outcome of the white die. So we're just going to ignore the black dye for a part B. So we might be interested to simplify the question in redefining the sample space as what the white dykan be. So we know that it could be any number from one through six, and if that's our sample space than the number of events in which the white dye is odd is equal to three. That's 13 and five. So the probability of a is equal to three over six. My formally, we might just note that the size of the sample space of six. So the probability of a is one over half. That's the probability of the white diving odd and then moving on to Part C. What is the probability that some is six so you can look up the chart in the chapter for that shows all of the outcomes when rolling two dice, and you can look at all count of all the number, all the outcomes in which the 76 and what you'll find is the number of outcomes is equal to you five. And this is a sample of space that is the size 36. There are 36 different outcomes in that sample space. Different elements. So the probability of ruling some of six is equal to 5/36 Party were asked What is the probability that both dice turn up odd? So let's just define that So going back to the chart. If you count up all of the outcomes in which both guys, they're odd, you will find that there are nine. In this case, the probability of bulls dice being odd is 9/36 or 1/4. And then finally for party were asked, What is the probability that the number on the black dye is larger than the number on the white die again going back to the chart and Chapter four? With the outcomes, we can see that there are 15 outcomes where the black guy has a larger number than the white die. So the probability of this event is there for 15/36 which is equal to 5/12.

In this question, we're asked to consider the set of interviews. 1234 and fun. When we're told that one integer is selected at random. What is the probability that it is up? So all numbers 13 and five says three of the numbers out of five. So if we put this into a decimal, we're gonna get no 50.6. So we've redefined, aren't oh, old makes it poverty. No point six. So, Paul So why Puppy says it turns just selected at random. One a time with replacement. So this means we have independence. Find the probability that knee that is odd. Exactly. One of them is old on both ends all. So let's start with the 1st 1 and that Nida Well, so we have the probability of O is no point. Sex is the probability that it's not. Ott is one minus that, but should know quite four. So for both of them to be not barred, we need no point for times. No 0.4 gives us no point 16 and then let's look at one of the result, which is one is off on. Then that means we need the other one to not be on. So this time we're doing no 10.6 times no point for which gives us no point 24 on the for the hospital. It says go filled. This is just like the neither ought. But no 0.6 this time. No 0.6 times, Not quite six. Then we're gonna equal s are no point 36 to those of the three ounces about pop.


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