# Probability - Combinatorial

• August 15th 2009, 04:04 AM
Robb
Probability - Combinatorial
Im not sure if to post it in here, or not. Seems kinda basic...

A student prepares for an exam by studying a list of ten problems. She can solve six of them. For the exam, the instructor selects five problems at random from the ten on the list given to the students. What is the probability that the student can solve all 5 problems on the exam?
• August 15th 2009, 06:14 AM
garymarkhov
Quote:

Originally Posted by Robb
Im not sure if to post it in here, or not. Seems kinda basic...

A student prepares for an exam by studying a list of ten problems. She can solve six of them. For the exam, the instructor selects five problems at random from the ten on the list given to the students. What is the probability that the student can solve all 5 problems on the exam?

Interesting problem. I'll take a crack at it, but I can't promise that I'll deliver the right answer.

There are 10 ways the professor could choose the first question, 9 ways to choose the second, ..., 6 ways to choose the 5th question. So he has 10*9*8*7*6 = 30,200 ways to choose the five questions (using factorial notation, that would be 10 factorial divided by 5 factorial).

Of those 30,200 PERMUTATIONS, there are many redundancies (choosing questions 5, 8, 4, 1, and 9 is the same as choosing 9, 5, 1, 8, 4). How many redundancies? Well, there are 5 ways to choose the first of 5 questions, 4 ways to choose the second of 5 questions, ..., 1 way to choose the fifth of 5 questions. That's 5*4*3*2*1, or 5 factorial. So we divide 30,200 by all the redundancies: 30,200/(5*4*3*2*1) = 252. So there are 252 COMBINATIONS, or genuinely different sets of questions the teacher could choose.

The student knows 6 questions, which makes for 6 factorial PERMUTATIONS, but only 6*5*4*3*2*1/(5*4*3*2*1) = 6 genuinely different COMBINATIONS.

So it seems to me that the probability of the student knowing all 5 questions is the number of combinations she knows divided by the number of combinations the teacher might spring on her, or 6/252 = 0.0238.

That's my guess. I'm open to being proven wrong.
• August 15th 2009, 06:26 AM
Plato
Quote:

Originally Posted by garymarkhov
So it seems to me that the probability of the student knowing all 5 questions is the number of permutations she knows divided by the number of permutations the teacher might spring on her, or 6/252 = 0.0238.

That answer is correct by there is a much simplier way to get it.
$\frac{\binom{6}{5}}{\binom{10}{5}}=0.02381$
From the six she knows choose five, divide by 10 choose five.
• August 15th 2009, 06:35 AM
garymarkhov
Quote:

Originally Posted by Plato
That answer is correct by there is a much simplier way to get it.
$\frac{\binom{6}{5}}{\binom{10}{5}}=0.02381$
From the six she knows choose five, divide by 10 choose five.

Yes, formulas make almost everything easier... except that they don't help you understand the intuition behind why you're using them! (Happy)
• August 15th 2009, 07:02 AM
Plato
Quote:

Originally Posted by garymarkhov
except that they don't help you understand the intuition behind why you're using them!

But you mixed up the concepts of permutation and combination.
This problem is strictly about the content of the test: combinations.
It has nothing to do with the order in which the questions are presented: permutations.
Therefore, the student may not benefit for using a mix of the two even if it gives the correct answer.
• August 15th 2009, 07:26 AM
garymarkhov
Quote:

Originally Posted by Plato
But you mixed up the concepts of permutation and combination.
This problem is strictly about the content of the test: combinations.
It has nothing to do with the order in which the questions are presented: permutations.
Therefore, the student may not benefit for using a mix of the two even if it gives the correct answer.

Sorry about the mixup - I did fix it shortly after I made it and before it could do any damage.

When I learned combinations and permutations in high school, I would have benefited greatly from knowing that combinations were just permutations with the redundancies removed. Don't you think it's useful (and cool) to know that finding the number of permutations is the first step toward finding the number of combinations? That's not a quirky way of doing things - that's exactly what the formulas prescribe.

Anyway, I sure don't want to cause any contention here. This website is great, you're a great contributor, and I hope to learn plenty from you and the site going forward!

Cheers,

Gary