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Larimani
03-22-2005, 11:42 AM
Right. I got a probabilty question with very important practical implications. My university exams are coming up very soon, and an answer to this question would help me in my revisions.

The known facts:
- There are 10 lectures in each course.
- Our exams for each course will consist of 7 essay questions of which we have to answer 2.
- The 7 essay questions will cover a minimum of 6 lectures (ie 2 questions could be on the same lecture)

My question is:
How many lectures should I revise so that I am 95% sure that at least 2 questions out of 7 I will have covered in my revision?
What about if I take a bigger risk and make that 85%?

P.S.: I'm obviously not a math student /images/graemlins/grin.gif

Ta
Paul2432
03-22-2005, 02:03 PM
This could be way off, but I figure as follows:

Number lectures studied / Chance of success

1 / 9%
2 / 26%
3 / 46%
4 / 91%
5 / 98%
6 / 100%

Paul
Larimani
03-22-2005, 02:18 PM
Brilliant. I'm quite surpised that there is such a big gap in chances of success between 3 & 4 lectures... If your calculations are right, it means that studying 4 lectures has the best work : result ratio. Which means the best strategy would be to revise 4 lectures for each course & then revise the rest if time permits.

Thanks for your help.
Larimani
03-22-2005, 02:25 PM
Also, could someone confirm Paul's results... as it is of extreme importance to me. Thanks.
Paul2432
03-22-2005, 04:51 PM
I've revised the numbers. These numbers I am much more confident. Instead of a calculation, I simply listed all the possible calculations and then counted. (made easy using MS Excel, I can e-mail you the file if you like).

Number lectures studied / Chance of success

1 / 9%
2 / 43%
3 / 76%
4 / 91%
5 / 98%
6 / 100%

Paul
Paul2432
03-22-2005, 04:51 PM
I've revised the numbers. These numbers I am much more confident. Instead of a calculation, I simply listed all the possible calculations and then counted. (made easy using MS Excel, I can e-mail you the file if you like).

Number lectures studied / Chance of success

1 / 9%
2 / 43%
3 / 76%
4 / 91%
5 / 98%
6 / 100%

Paul
Larimani
03-22-2005, 05:46 PM
Ok. these numbers look more correct. thanks!