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2 [4 marks] In a given semester; a large university will processes many grades In the past; 0.2% ofall grades have been erroneously reported. Assume you are taking ...

Question

2 [4 marks] In a given semester; a large university will processes many grades In the past; 0.2% ofall grades have been erroneously reported. Assume you are taking four courses at this university in one semester.a) What is the probability that all your grades are correctly reported?b) What is the probability that exactly three of your grades are reported incorrectly?

2 [4 marks] In a given semester; a large university will processes many grades In the past; 0.2% ofall grades have been erroneously reported. Assume you are taking four courses at this university in one semester. a) What is the probability that all your grades are correctly reported? b) What is the probability that exactly three of your grades are reported incorrectly?



Answers

In a history class, a set of exams earned the following grades: $4 \mathrm{A's}, 8 \mathrm{B's}$ 22 C's, 10 D's, and 3 F's. If a single student's exam is chosen from this class what is the probability that it received a B grade?

So there are six different options for grades that you can get. And if you take four courses, well, then for each one there are six different options you can get in a B, C, D E f or W Between did he multiply these together number of a different possible record cards, or 1296?

In question 4.37. We're told there are four times as many students who pass the statistics course than those who fail and that a student is selected at random from the class. So what is the probability that that student passes the course? So let's first take a look at her sample stays. It's the one outcome is fail and the other outcome is passed. But since we're four more times as likely to pass as fail, we can account for this inequality in the sample space by listing past four times. So that is a proper sample space. And now the probability of passing is equal to the number of passes in the sample space. So that's four divided by the total size of the sample states, which is five

So in the given question, we are told that A and B are appearing for an exam, they appears forward. They appeared for an exam and we are told that the chance that they will succeed, let's say we can take this sp of A has given us 0.3 and the probability that be succeeds has given us 0.4. And now we are told to find the probability that is or B fails. Right? So what we need to find over here is we have to find the probability of a compliment. Union. Be compliment right. A compliment Is the event where A is failing and be compliment is the event where B is failing the exam. So we have uh property by which we can find the probability of a union be asked P of a capacity of B minus P of a intersection. B. Where B F. A. In the section B is the probability that the event A and B happens. So using this expansion or the property we can right be off a compliment, Union, be complement, which is the probability that a R B happens is equal to P F. A compliment must be a big compliment. That is the probability of that defense minus the probability of a compliment intersection B compliment, which means both A and B fails. So since we have P of A over here, we can find p of a compliment as one minus 0.3, which would be equal to 0.7 and we of be complement would be equal to one minus 0.4, which is equal to 0.6. And in the next step, what we need to know is that the case where we should know that these events A and B are independent events, Right? Are independent events, which means if a fails or succeeds. It has nothing to do with the fact that be would fail or success succeed. Right? So the event that a fails or succeeds and the event that be fails or succeeds are entirely independent events. And if they are independent events, what we can write is that for independent independent events. P of a intersection, we can be found us B. Of A plus P R. A. Times they are me. So let's keep this in mind before substituting the values in the abo uh formula. So what we will have over here is B of a complemented 0.7 pair of be complemented. What is being complemented? It is 0.6 minus. We have P. L. A. Times P F. B. Which would be equal to P of 0.3 times 0.4. Here we have instead of P. Of A and B. Of B. We are taking pf a intersection. A compliment, intersection B compliment. Right? So they will have to take B of a compliment which is 0.7 times B f p f B complement, which is 0.6. So this is what we have in our formula. Right? So now what we can do is we can take 0.7 plus 0.6 which would be equal to 1.3 minus 0.7 times. 0.6 is equal to 0.42 and 1.3 minus 0.42 is equal to for two is equal to 0.88 So this is the required probability of our idea that either A or B fails the sequel to 0.88 And in the question we can say that there are four options out of which options see is giving us the answer 0.88 So this is the correct answer. So I hope you understood the method. Thank you.

All right, this question asks us about a binomial experiment with 15 trials and a probability of success of 150.28 So part eh asks for the probability that there exactly four successes and that just plugging into the binomial distribution formula you get 15. Choose for times 0.28 to the number of successes times probability of failure to the remaining number of trials, which is the same thing as saying, If you have a T 84 by a no meal, pdf pdf because we're only interested in one probability 15 trials 0.28 success rate and we're interested in X equals four. And both of those answers work out 2.2261 All right, And then Part B asks for the probability the X is greater than or equal to three, which equals one minus the probabilities. We don't want soapy zero plus p one plus p too, which can be written as one minus the key mood of probability. So Bynum CDF 15 trials 150.28 probably of success, and we're adding all the probabilities upto X equals two, and that works out to 0.835 five


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