# Probability random selection Question

Ok, I am sorta stuck on this, and was hoping someone could nudge me in the right direction. The question is:

An instuctor gives her class a set of 10 problems with the information that the final exam will consist of a random selection of 5 of them. If a student has figured out how to do 7, what is the probability that he or she will answer correctly
a) all 5 problems.
b)at least 4 of the problems.

This seemed so simple. To start, I took 10 choose 5 as the possible test question combinations, since they were to be randomly picked. Then I took 7 choose 5, for the 5 answers the student would know. Then I divided 7 choose 5 by 10 choose 5. This worked out right, and answered question a) correctly.

I tried to apply this same line of thinking to b), taking 7 choose 4 and dividing it by 10 choose 5, but my answer is way off according to the book.

So, did I just get lucky on the first one, or am I on the right line of thought?

Related Introductory Physics Homework Help News on Phys.org
jamesrc
Gold Member
You're fine with the first one. For the second one, it asks for the probability of getting at least 4 questions right, meaning 4 or 5 right. The way you answered it, you found the probability of getting exactly 4 questions right. For your numerator, you have to count the number of ways to get 4 right and add to that the number of ways to get 5 right.

Yeah, I tried that, it seemed logical. I get 7 choose 5 = 21, and 7 choose 4 = 35. Since 10 choose 5 = 252, this gives about 20%. My book says 50%, so am I just working the numbers the wrong way here?

jamesrc
Gold Member
You book has the right answer; I just didn't help enough. When you do C(7,4) and get 35, that says that there are 35 combinations of the 7 known questions taken 4 at a time, but what about the 5th question that the student gets wrong? There are three questions that the student doesn't know, so there are 3 times 35 ways for the student to get exactly 4 right. (3*35 + 21)/252 = 50%

So a more complete way to write the number of ways the student will get exactly 4 right would be C(7,4)*C(3,1)

(when I write C(n,r), I mean the number of combinations of n items taken r at a time)

I never would have thought of that, as evidenced by the fact that I turned in my homework with 20% as my answer. Oh well, it's still good to know the right answer, so I can see where i messed up.

Thanks Alot.