Tonight I have an exam with 10 short answer questions, of which 7 will be on the test, and I have only studied 7.
What are the odds that all 7 questions I studied will be on the test??
My classmates and I have been arguing but have no definitive answer so far 
So what exactly is the problem we're working with?
I'm interpreting it like this:
There's 10 questions you had to study, and you have a test of 7 questions. Then, you studied only 7 out of 10 questions. What are the odds that exactly those 7 questions will be on the test.
Am I interpreting that correctly?
yes TinoLaan that is it exactly. none of us are the best at math so no luck so far in figuring it out
Ah, in that case it's pretty simple really. The odds of the first question being one that you studied is simply 1/7. In fact, the odds of the first one being any of the 10 questions is 1/7, because there's 7 questions. Then the odds for the second question are 1/6. And so on.
Since there's 10 possible questions though, you have to account for that. The 7 questions you studied can be the questions on the test in (10 choose 7) = 120 possible combinations.
So the odds of exactly those 7 questions being on the test are
(10 choose 7) * (1/7*1/6*1/5*1/4*1/3*1/2*1/1) = ~21.4%
thanks Tino! my one classmate actually had the combinations correct but we were all scrambling on the %. Thanks again 
21% seems way to big...
Isn't it 1/(10 choose 7)=1/120=0.83% ?
Since you want to hit all 7 questions you studied you want your examiner to select the one specific set of 7 from all possible. I assume you don't care about the order of the selected questions on the test paper.
no i do not care about the order. and this is where my classmates and i had issues. i thought that it would actually be a minute chance that all 7 would make it, while they thought it would be a relatively high chance. still totally unsure what the actual answer is though
Originally posted by yougotfelted51
no i do not care about the order. and this is where my classmates and i had issues. i thought that it would actually be a minute chance that all 7 would make it, while they thought it would be a relatively high chance. still totally unsure what the actual answer is thoughIt's as datalog said.
10 questions, 7 will be chosen = 120 combination.
Only one combination will eventually be chosen for the test.
Therefore, the odds are (1 favorable case) divided by (120 possible cases) = 1/120 = 0.83%.-----------
If the order of the question matters, the total combinations are 604800 , which leads to odds of 1/604800
Seems like in the time you spent arguing about it you could have just studied the other 3 and had a 100% chance of 7 questions you know coming up 
This is the visual representation of what has already been said, where a-g are the ones you know and ######-######## are the ones you don't.
Combinations without repetition (n=10, r=7)
List has 120 entries.
{a,b,c,d,e,f,g} {a,b,c,d,e,f,######} {a,b,c,d,e,f,#######} {a,b,c,d,e,f,########} {a,b,c,d,e,g,######} {a,b,c,d,e,g,#######} {a,b,c,d,e,g,########} {a,b,c,d,e,######,#######} {a,b,c,d,e,######,########} {a,b,c,d,e,#######,########} {a,b,c,d,f,g,######} {a,b,c,d,f,g,#######} {a,b,c,d,f,g,########} {a,b,c,d,f,######,#######} {a,b,c,d,f,######,########} {a,b,c,d,f,#######,########} {a,b,c,d,g,######,#######} {a,b,c,d,g,######,########} {a,b,c,d,g,#######,########} {a,b,c,d,######,#######,########} {a,b,c,e,f,g,######} {a,b,c,e,f,g,#######} {a,b,c,e,f,g,########} {a,b,c,e,f,######,#######} {a,b,c,e,f,######,########} {a,b,c,e,f,#######,########} {a,b,c,e,g,######,#######} {a,b,c,e,g,######,########} {a,b,c,e,g,#######,########} {a,b,c,e,######,#######,########} {a,b,c,f,g,######,#######} {a,b,c,f,g,######,########} {a,b,c,f,g,#######,########} {a,b,c,f,######,#######,########} {a,b,c,g,######,#######,########} {a,b,d,e,f,g,######} {a,b,d,e,f,g,#######} {a,b,d,e,f,g,########} {a,b,d,e,f,######,#######} {a,b,d,e,f,######,########} {a,b,d,e,f,#######,########} {a,b,d,e,g,######,#######} {a,b,d,e,g,######,########} {a,b,d,e,g,#######,########} {a,b,d,e,######,#######,########} {a,b,d,f,g,######,#######} {a,b,d,f,g,######,########} {a,b,d,f,g,#######,########} {a,b,d,f,######,#######,########} {a,b,d,g,######,#######,########} {a,b,e,f,g,######,#######} {a,b,e,f,g,######,########} {a,b,e,f,g,#######,########} {a,b,e,f,######,#######,########} {a,b,e,g,######,#######,########} {a,b,f,g,######,#######,########} {a,c,d,e,f,g,######} {a,c,d,e,f,g,#######} {a,c,d,e,f,g,########} {a,c,d,e,f,######,#######} {a,c,d,e,f,######,########} {a,c,d,e,f,#######,########} {a,c,d,e,g,######,#######} {a,c,d,e,g,######,########} {a,c,d,e,g,#######,########} {a,c,d,e,######,#######,########} {a,c,d,f,g,######,#######} {a,c,d,f,g,######,########} {a,c,d,f,g,#######,########} {a,c,d,f,######,#######,########} {a,c,d,g,######,#######,########} {a,c,e,f,g,######,#######} {a,c,e,f,g,######,########} {a,c,e,f,g,#######,########} {a,c,e,f,######,#######,########} {a,c,e,g,######,#######,########} {a,c,f,g,######,#######,########} {a,d,e,f,g,######,#######} {a,d,e,f,g,######,########} {a,d,e,f,g,#######,########} {a,d,e,f,######,#######,########} {a,d,e,g,######,#######,########} {a,d,f,g,######,#######,########} {a,e,f,g,######,#######,########} {b,c,d,e,f,g,######} {b,c,d,e,f,g,#######} {b,c,d,e,f,g,########} {b,c,d,e,f,######,#######} {b,c,d,e,f,######,########} {b,c,d,e,f,#######,########} {b,c,d,e,g,######,#######} {b,c,d,e,g,######,########} {b,c,d,e,g,#######,########} {b,c,d,e,######,#######,########} {b,c,d,f,g,######,#######} {b,c,d,f,g,######,########} {b,c,d,f,g,#######,########} {b,c,d,f,######,#######,########} {b,c,d,g,######,#######,########} {b,c,e,f,g,######,#######} {b,c,e,f,g,######,########} {b,c,e,f,g,#######,########} {b,c,e,f,######,#######,########} {b,c,e,g,######,#######,########} {b,c,f,g,######,#######,########} {b,d,e,f,g,######,#######} {b,d,e,f,g,######,########} {b,d,e,f,g,#######,########} {b,d,e,f,######,#######,########} {b,d,e,g,######,#######,########} {b,d,f,g,######,#######,########} {b,e,f,g,######,#######,########} {c,d,e,f,g,######,#######} {c,d,e,f,g,######,########} {c,d,e,f,g,#######,########} {c,d,e,f,######,#######,########} {c,d,e,g,######,#######,########} {c,d,f,g,######,#######,########} {c,e,f,g,######,#######,########} {d,e,f,g,######,#######,########}
thanks metza!
we actually spent all off 2 minutes on the discussion, and i posted this thread in the exam room 8 minutes before it started 
anyways only 4 of the 7 we studied were on the exam, thank goodness it was so easy 
Originally posted by unshpe
Originally posted by yougotfelted51
no i do not care about the order. and this is where my classmates and i had issues. i thought that it would actually be a minute chance that all 7 would make it, while they thought it would be a relatively high chance. still totally unsure what the actual answer is thoughIt's as datalog said.
10 questions, 7 will be chosen = 120 combination.
Only one combination will eventually be chosen for the test.
Therefore, the odds are (1 favorable case) divided by (120 possible cases) = 1/120 = 0.83%.-----------
If the order of the question matters, the total combinations are 604800 , which leads to odds of 1/604800
If order doesn't matter, surely you couuld have more than one favourable case?
Originally posted by bennisboy
Originally posted by unshpe
Originally posted by yougotfelted51
no i do not care about the order. and this is where my classmates and i had issues. i thought that it would actually be a minute chance that all 7 would make it, while they thought it would be a relatively high chance. still totally unsure what the actual answer is thoughIt's as datalog said.
10 questions, 7 will be chosen = 120 combination.
Only one combination will eventually be chosen for the test.
Therefore, the odds are (1 favorable case) divided by (120 possible cases) = 1/120 = 0.83%.-----------
If the order of the question matters, the total combinations are 604800 , which leads to odds of 1/604800
If order doesn't matter, surely you couuld have more than one favourable case?
You are confusing things.
It's 1 favorable case out of way less possible combinations (120 instead of 604800). Or elsely put, its 7! (aka 7*6*5*4*3*2*1) favorable combos out of those 604800. Either way, the result is 0.83%.
GL!
Originally posted by yougotfelted51
thanks metza!we actually spent all off 2 minutes on the discussion, and i posted this thread in the exam room 8 minutes before it started
anyways only 4 of the 7 we studied were on the exam, thank goodness it was so easySo, what are the odds that all 3 of the questions you didn't learn, come on exam?
Originally posted by VladimirN
Originally posted by yougotfelted51
thanks metza!we actually spent all off 2 minutes on the discussion, and i posted this thread in the exam room 8 minutes before it started
anyways only 4 of the 7 we studied were on the exam, thank goodness it was so easySo, what are the odds that all 3 of the questions you didn't learn, come on exam?
Lets assume the questions he hasn't learned are 1, 2 and 3.
All the favorable (although unfortunate for him
) scenarios are the ones that look like 1 , 2 , 3 , _ , _ , _, _ . That translates into finding all combos of 4 out of the subset { 4,5,6,7,8,9,10 } , which results in 35 combos.
Divide it by all the combos possible (see above ,120) and it results in 35/120 = 29%
Hope I didn't screw up, it's late at night
GL
