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Probability random selection Question

  • Thread starter Lukedawgus
  • Start date
  • #1
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?
 

Answers and Replies

  • #2
jamesrc
Science Advisor
Gold Member
476
1
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.
 
  • #3
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?
 
  • #4
jamesrc
Science Advisor
Gold Member
476
1
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)
 
  • #5
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.
 

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