Probability student exam problem

AI Thread Summary
The discussion revolves around calculating probabilities for a student who knows how to solve 15 out of 20 exam problems. The main queries include finding the probability of answering all 10 questions, exactly 8 questions, and at least 9 questions correctly. Participants clarify that the total combinations of questions can be calculated using binomial coefficients, specifically 20C10 for the exam questions and 15C10 for the problems the student can solve. The conversation emphasizes the need to determine how many combinations of good and bad problems can be selected for each scenario. Ultimately, the focus is on applying combinatorial methods to derive the required probabilities.
themadhatter1
Messages
139
Reaction score
0

Homework Statement



A class is given a list of 20 study problems from which 10 will be on an upcoming exam. If a student knows how to solve 15 of the problems, find the probability that the student will be able to answer (a) all 10 of the questions on the exam. (b) exactly 8 questions on the exam, and (c) at least 9 questions on the exam.

Homework Equations




The Attempt at a Solution



Well for a I found 20C10 which is the total possible combinations of questions that could be on the test then I found 20C15 which is the total number of combinations of the problems the student solved. Would you divide 20C15 by 20C10? I'm not sure how to solve this problem.
 
Physics news on Phys.org


The student can solve 15 problems, but these problems are given. The student knows, which ones he can solve.
From the viewpoint of the student, there are 15 good problems and 5 wrong ones. What is the probability that all the 10 problems in the test are selected from the good ones? What is the probability, that 8 good and 2 wrong problems all selected? What is the probability that at least 9 good problems are selected?

ehild
 


ehild said:
What is the probability that all the 10 problems in the test are selected from the good ones?
ehild

Would that be 10/15 or 1/3? because there are 10 problems being chosen and 15 good ones.
 


The elementary event is which problem is chosen. You said correctly already that 20C10 is the total number of the possible set of questions: you can select 10 out of 20 in (20!/10!10!) ways.
How can you select 10 question out of 15?

ehild
 


ehild said:
The elementary event is which problem is chosen. You said correctly already that 20C10 is the total number of the possible set of questions: you can select 10 out of 20 in (20!/10!10!) ways.
How can you select 10 question out of 15?

ehild

15C10 ways?
 


Yes.

ehild
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Back
Top