My daughter is taking a college history course and has an exam next week. The professor has given the students 11 essay prompts ahead of time. Of those 11, five will be on the test, and the students have to write two of the essays (1-2 pages long each).
She and I were talking about it and wondering - what are the odds that any given prompt will be on the exam? What’s the best way to devote most of her time studying to the least number of prompts? Is the professor using some magical formula, or is this just an interesting way to make the students study all of the material and hope that their favorite topic comes up?
If we assume that the the odds are completely random (or otherwise unknowable to the student), then the odds of any one particular prompt being one of the five topics for the exam are (1/11) + (1/11) + (1/11) + (1/11) + (1/11) = .091 + .091 + .091 + .091 + .091 = 45.5%. (This assumes that my somewhat rusty stats skills are still working properly. )
My guess is that the professor is using the eleven prompts to make sure that the students study and prepare for a range of topics from the class content for the final, while narrowing things down a bit (i.e., not expecting the students to study every single thing covered over the course of the class).
I suspect what you and your daughter are looking for is a formula that suggests the fewest number of topics that she would have to prep, while ensuring that she’ll have prepped at least two of the five topics that wind up on the final.
I don’t have the time to do the math, but I’m pretty sure that the answer – if she wants to absolutely assure that she has two topics covered – would be eight. If she preps eight, then she has left three topics unstudied, and the worst-case scenario would be that, of the five topics that wind up on the exam, she can only write on two of them (because the remaining three happen to be the three that she ignored).
Preparing only seven topics would yield a small – but non-zero – chance that she will wind up seeing only one topic on the final that she’s ready to answer. And, obviously, prepping fewer than seven would increase that chance (as well as having a chance that she winds up zero-for-five on the final).
I’m pretty sure that’s the right answer, but I can’t tell if you got there by luck, or if you just didn’t show all your work.
From what I see, there are two ways to reach the answer.
Method #1:
Five slots on the exam, numbered 1-5.
For slot #1, pick one of the questions from the pool of 11. Odds of being selected for slot #1, 1/11.
For a question to be chosen for slot #2, it has to have been passed over for slot #1. Odds of that are 10/11, and now the candidate pool has been reduced to 10. So after passing that trial, the odds of being chosen for slot #2 are 1/10. Altogether, the odds of being chosen for slot #2 are 10/11 * 1/10 = 1/11.
To be chosen for slot #3, a question has to have been passed over for slots #1 and #2. Odds of surviving those trials are 10/11 and 9/10. After that, the odds of being picked for slot #3 are 1/9. So altogether, the odds of being chosen for slot #3 are 10/11 * 9/10 * 1/9 = 1/11.
You can see the progression here. For slot #4, the odds are 10/11 * 9/10 * 8/9 * 1/8 = 1/11. For slot #5, it’s 10/11 * 9/10 * 8/9 * 7/8 * 1/7 = 1/11.
For a question to be picked for the exam, it just needs to get picked in any one of those trials, so these probabilities are additive. Add them up, and you get 5/11, or 45.5%.
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Method #2:
What are the odds of a question not getting picked for slot #1, and then not getting picked for slot #2, and then…not getting picked for any of the exam slots?
Odds of not getting picked for slot #1: 10/11
Odds of not getting picked for slot #2: 9/10
…#3: 8/9
…#4: 7/8
…#5: 6/7
For a question to not get picked, it has to survive all of these trials, so these probabilities are multiplicative: 10/11 * 9/10 * 8/9 * 7/8 * 6/7 = 54.5%.
If that’s the odds of not getting picked for any exam slot, then the odds of getting picked for any exam slot must be the complement: 100% - 54.5% = 45.5%.
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