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SmoKing2012
United States385 Posts
For the final exam in one of my classes, the teacher gave us a list of 5 potential essay questions. She's going to pick 3 of them to put on the test, and we have to answer one of them. If I pick 2 essay questions to study, what are the chances that at least one of the questions I studied shows up on the test?   | ||
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EGMachine
United States1643 Posts
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DeathSpank
United States1029 Posts
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Jonoman92
United States9103 Posts
On May 04 2009 08:44 DeathSpank wrote: study four of them. /thread Well the most he would need to study in order to definitely get one he studied for would be three... | ||
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SmoKing2012
United States385 Posts
On May 04 2009 08:39 Machine[USA] wrote: figure it out yourself, its not our test. Thank you for your completely useless contribution. I thought ppl here liked math problems.. | ||
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Myrmidon
United States9452 Posts
The probability you get 1 or 2 questions that you studied is 1 minus the probability that you get 0 questions you studied. The probability you get unlucky and get 0 questions is the probability that... Imagine she is choosing the 3 questions to put on your exam in order: First question has 3/5 chance of being bad for you. Given that the first question you didn't study, the second question has 2/4 chance of being bad. Given that first two questions you didn't study, the third question has 1/3 chance of being bad. So the probability you get no questions you studied is (3/5)(2/4)(1/3) = 0.1 Probability you get at least one question you studied is 1 - 0.1 = 0.9 = 90%. | ||
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SmoKing2012
United States385 Posts
On May 04 2009 08:50 Myrmidon wrote: I think it's 90%. The probability you get 1 or 2 questions that you studied is 1 minus the probability that you get 0 questions you studied. The probability you get unlucky and get 0 questions is the probability that... Imagine she is choosing the 3 questions to put on your exam in order: First question has 3/5 chance of being bad for you. Given that the first question you didn't study, the second question has 2/4 chance of being bad. Given that first two questions you didn't study, the third question has 1/3 chance of being bad. So the probability you get no questions you studied is (3/5)(2/4)(1/3) = 0.1 Probability you get at least one question you studied is 1 - 0.1 = 0.9 = 90%. Looks good at a glance, thanks. | ||
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n.DieJokes
United States3443 Posts
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Myrmidon
United States9452 Posts
Yeah I hope I didn't make a fool of myself, too. I think an alternative way to think of it is there being (5 choose 3) combinations of possible exam questions. (5 choose 3) = 5! / [(3)!*(5-3)!] = 120 / [6 * 2] = 10. Only 1 of the 10 combinations is bad for you: the one that contains all 3 test questions you didn't study. | ||
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Lemonwalrus
United States5465 Posts
Possible Essay Questions: A,B,C,D,E. Possible combinations: (Made really simple since order doesn't matter) ABC BCD CDE ABD BCE ABE BDE ACD ACE ADE Therefore, assuming you studied A and B, all of the outcomes except for CDE would have one essay that you were prepared for, so you have 90% chance to encounter one. I'd suggest at least familiarizing yourself with a third one though just in case. | ||
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Crunchums
United States11144 Posts
On May 04 2009 08:34 SmoKing2012 wrote: For the savvy.. For the final exam in one of my classes, the teacher gave us a list of 5 potential essay questions. She's going to pick 3 of them to put on the test, and we have to answer one of them. If I pick 2 essay questions to study, what are the chances that at least one of the questions I studied shows up on the test? 1 - 1/(5 choose 2) = 9/10 = 90% | ||
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Racenilatr
United States2756 Posts
On May 04 2009 08:34 SmoKing2012 wrote: For the savvy.. For the final exam in one of my classes, the teacher gave us a list of 5 potential essay questions. She's going to pick 3 of them to put on the test, and we have to answer one of them. If I pick 2 essay questions to study, what are the chances that at least one of the questions I studied shows up on the test? isn't it 40%? coz it's the chance of ONE question and your studying 2/5 of the potential questions right? | ||
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Pawsom
United States928 Posts
On May 04 2009 10:11 Racenilatr wrote: isn't it 40%? coz it's the chance of ONE question and your studying 2/5 of the potential questions right? no | ||
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n.DieJokes
United States3443 Posts
On May 04 2009 08:55 n.DieJokes wrote: cool, binomial probability question. So we do, 1-binomialcdf (n,p,k-1) or (5,.4,0) so .92224. If I screwed up anywhere someone call me on it, it's been a while. I checked it, this is the answer. | ||
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foppa
Canada451 Posts
3/5 chance that you have not prepared the essays if the teacher does a random draw: on the first draw there will be a 3/5 chance that you have not prepared that question. on the second draw there will be a 2/4 chance you have not prepared. thus the probability of getting 2 draws of "no" will be 3/5 * 2/4 = 6/20 or 3/10 or 30% thus knowing that the probability of knowing both questions + the probability of knowing one of the questions will be equal to 70%. I would advice you to not be a lazy piece of crap and study for 4 as there will be a 100% chance of you knowing at least 1 of the questions on the exam. if you study for 3 the probability of getting a question you can answer will be of 90%. make your choice wisely. /good luck ps: i have a math exam next week and probability is one of the units i gotta know so you can believe my post ^-^ | ||
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DeathSpank
United States1029 Posts
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Lemonwalrus
United States5465 Posts
On May 04 2009 11:36 foppa wrote: there is a 70% chance that you will have prepared for one of the essays. 3/5 chance that you have not prepared the essays if the teacher does a random draw: on the first draw there will be a 3/5 chance that you have not prepared that question. on the second draw there will be a 2/4 chance you have not prepared. thus the probability of getting 2 draws of "no" will be 3/5 * 2/4 = 6/20 or 3/10 or 30% thus knowing that the probability of knowing both questions + the probability of knowing one of the questions will be equal to 70%. I would advice you to not be a lazy piece of crap and study for 4 as there will be a 100% chance of you knowing at least 1 of the questions on the exam. if you study for 3 the probability of getting a question you can answer will be of 90%. make your choice wisely. /good luck ps: i have a math exam next week and probability is one of the units i gotta know so you can believe my post ^-^ There are THREE draws, not two. | ||
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foppa
Canada451 Posts
On May 04 2009 08:55 n.DieJokes wrote: cool, binomial probability question. So we do, 1-binomialcdf (n,p,k-1) or (5,.4,0) so .92224. If I screwed up anywhere someone call me on it, it's been a while. wrong ^-^ you can only use binomialcdf when success or failure is determined randomly with the same probability of success each time the event occurs in this case if you pick one essay whether its a yes or a no the probability will be out of 5 (2/5 success, 3/5 failure), the second draw will be out of 4 since one essay been eliminated by the first draw | ||
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Jonoman92
United States9103 Posts
On May 04 2009 09:48 Lemonwalrus wrote: Since there are so few possible combinations, I just solved it this way. Possible Essay Questions: A,B,C,D,E. Possible combinations: (Made really simple since order doesn't matter) ABC BCD CDE ABD BCE ABE BDE ACD ACE ADE Therefore, assuming you studied A and B, all of the outcomes except for CDE would have one essay that you were prepared for, so you have 90% chance to encounter one. I'd suggest at least familiarizing yourself with a third one though just in case. Once again, simplicity and common sense wins. | ||
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foppa
Canada451 Posts
im gonna fail xD yes you are right I calculated the outcome of the first 2 draws. why am I doing math at 11 at night is beyond me the 3rd draw indeed gives you a probability of a 90% of having studied for at least 1 of the essays | ||
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SerpentFlame
415 Posts
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Ivs
Australia139 Posts
On May 04 2009 08:55 n.DieJokes wrote: cool, binomial probability question. So we do, 1-binomialcdf (n,p,k-1) or (5,.4,0) so .92224. If I screwed up anywhere someone call me on it, it's been a while. It is not a binomial, the trials aren't independent. | ||
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