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Point) A multiple-choice test consists of 29 questions with possible answers of = guessing_ the number of correct answers is at least 9.d; e; Estimate the probabili...

Question

Point) A multiple-choice test consists of 29 questions with possible answers of = guessing_ the number of correct answers is at least 9.d; e; Estimate the probability that with random

point) A multiple-choice test consists of 29 questions with possible answers of = guessing_ the number of correct answers is at least 9. d; e; Estimate the probability that with random



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Tests A multiple-choice test has four choices for each answer.
a. What is the probability that a random guess on a question will yield the correct answer?
b. Suppose you need to make a random guess on three of the ten test questions. What is the probability that you will answer all three correctly?

In this problem, Let us define certain events 1st. Let even be the event that me nose the answer similarly A two is he does not no the answer. Yeah. Let e. The note that he gets correct answer. Now we are given that be even is 90% which is nine x 10, which makes 32 as one minus P. Even so this becomes one by 10 probability of giving the correct answer provided he knows the answer is one proversity of giving the correct answer provided he does not know the answer is one by four Because there are four choices, So he can choose any. That's why one x 4. What we need in the problem is probability that he guessed that is he does not know the answer. Given that mm gave the answer correct? My beast here. Um this has given us E. two and 2 B. He over a two divided by E and B. E. Can be recognized the even into the E over e. one plus Be a two into peace. He over here too, substituting this is one might turn into one x 4. He married by nine by 10 into one plus one by 10 into one by four. This comes us one day 37 correct options. See that's

Question says that you're gonna have 10 true and false questions on a past, and the student answers them all by guessing. And the first part of the question says, What's the probability that you get all correct? So if you think about that, there are 10 questions. Um, there are two answers per question. So if you think about that total number of possible answers, then our is 20 There. It's not 20. It's to to the 10 because you've got two possibilities each time. And then there's only one way to get them all right. There are two and is 1000 4 And so that would buy and E says find, probably getting athletes seven. So you'd have 1000 and 24 possibilities and and you have to have and seven of the. So if I just figure out what 10 June 7 is, I'm just gonna grab 120. Different possibilities are different ways to get seven right, so you have 120 divided. By that 10th 24 it was just 15 and a 128. Or if you want the approximate valuable 11 well, 12%. So that's my answer for pardon. Be

You know this problem? We are given that we have a multiple choice question multiple choice test with 25 questions. And so in this 25 and there are four options for each test, which means the probability of randomly guessing on one of them and getting it correct Will be a .25. And so we have our p value at this .25. Yeah. This means that the probability of X. Mhm. Following a binomial distribution is 25-6 Times .25 to the X Times .75 To the 25 again because it follows a binomial distribution. Now on a we want the probability that student answers more than 20 questions correct. So that probably the access greater than 20 Is equal to the sum as that goes from 21 2025 of our pr that's values, We know the p of access. And so this is the sum from access 21-25 of 25- six. 1.25 to the X. 1.75 25- X. So here we can just evaluate this on a calculator. We do this some of 25 choose x Times .25 to the x. .75 To the 25 -X. No matter what the sum to go from 21 to 25. And so this is going to give us 9.677 Times 10 to the -10. No B. We want the probability The student answers fewer than five questions correctly. So X is less than five. This is the sum as extras from 0-4 because we want to be less than fine enough paperbacks. She's the some as extras from 0 to 4 25- six. This .252 there 27 5 to the 25- X. So we can do with the exact same way we did the previous problem. We just change our bounds 20-4 in step, and that should give us uh huh 0.21 37.

So we're going to be looking at that independence problem again, just like the four boys, four girls being born doesn't affect the probability of the next one being born A boy. Um So what we're looking at is 10 multiple choice answers. So one we got a B C D R E. And this is going to continue on all the way to the 10th one where we have a B C D E N E. And what this means is what is the probability that we get the at least one correct when the first nine we thought were wrong. Yeah. And you know, there's a bit of a losing streak that is taking place and that might affect the Their their their capacity to answer correctly. But the probability of any event happening that you get it correctly is always going to be 20%. Since that is our Probability of getting one right, there's one right dancer and for wrong answers. So even though it might have been slow unlikely that they get nine rolling or wrong still won't affect the probability that they get the next one right are wrong. This is a concept known as independence. So the probability that they get their 10th question right, Regardless of whether they got their last nine right or wrong is still going to be 20


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