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# Probability Difficulties

• Silver
Joined: 04.02.2010
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
• 14 replies
• Bronze
Joined: 12.10.2011
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?
• Silver
Joined: 04.02.2010
yes TinoLaan that is it exactly. none of us are the best at math so no luck so far in figuring it out
• Bronze
Joined: 12.10.2011
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%
• Silver
Joined: 04.02.2010
thanks Tino! my one classmate actually had the combinations correct but we were all scrambling on the %. Thanks again
• Bronze
Joined: 13.05.2011
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.
• Silver
Joined: 04.02.2010
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
• Basic
Joined: 20.05.2013
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 though
It'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

• Bronze
Joined: 18.06.2010
1 devided by a combination of 10 elements class 7 => 1/120 => around 1%
• Bronze
Joined: 28.01.2012
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,######,#######,########}
• Silver
Joined: 04.02.2010
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
• Bronze
Joined: 11.04.2011
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 though
It'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?
• Basic
Joined: 20.05.2013
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 though
It'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!
• Bronze
Joined: 22.02.2008
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 easy
So, what are the odds that all 3 of the questions you didn't learn, come on exam?
• Basic
Joined: 20.05.2013
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 easy
So, 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