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This is an assignment question, but I am not interested in the answer.

This is the question:

A student is preparing for an upcoming exam. The professor for the course has given the class 30 questions to study from and plans to select 10 of the questions for use on the actual exam. Suppose that the student knows how to solve 25 of the 30 questions.

a) What is the probability that the student will get perfect on the test?

b) What is the probability that the student will get at least 8 questions correct on the test?

So let the event A be the event that a question will be on the test and B be the event that i know how to answer it. So P(A) = 10/30 and P(B) = 25/30 Now probability that a question will be on the test and i know how to solve it is P(AandB) = (10/30)*(25/30). Is this correct? If not why?

Also please make sure that you demonstrate how in the proper way, you can easily assert that the probability of getting less then 5 questions right is zero since if by chance 5 of the questions choosen by the professor where the five that i dont know how to solve, i still know how to solve the remaining 25 questions. Thanks for your help.

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Hint: There are $\binom{30}{10}$ different exams that the professor could set. How many of those will have all $10$ questions selected from the $25$ that the student knows how to solve? How many will have exactly $9$ such solvable questions? How many will have exactly $8$ such solvable questions? – Dilip Sarwate Feb 5 '12 at 0:55
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To iterate on your suggestion with smaller problem. So instead, let say there are 4 questions and the professor will choose two out of the 4 [(1,2) (1,3) (1,4) (2,3) (2,4) (3,4)]. The student knows how to solve for 3 out of the 4 questions [(1,2,3) (1,2,4) (1,3,4) (2,3,4)]. What i am confused about is why this is an exercise in counting and why i cannot apply what i mentioned in the question? – Adham Feb 5 '12 at 2:51
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Just to throw a kink in the works... That P(A & B) = (1/3)*(5/6) falls directly out of probability theory IF A and B are independent. Thus, that part would be correct if the probability of the professor choosing a question is independent of the probability that you know the answer. Consider this possibility: the questions vary in difficulty, and the 5 you don't know are among the hardest, but the professor wants some easy, some mid-range, and some toughies for the test; now p(A) and p(B) are no longer independent. – gung Feb 5 '12 at 3:05
So the probability that the professor picks a question out of the 30 is dependent on the probability that the i know how to solve it? Makes sense. It wouldn't make sense if the professor randomly selected any 10. But when we count the different combinations of questions the professor could choose, they all have equal weight even though it is more probable that the professor would choose questions with varying difficulties while the probability of choosing all easy questions is not as probable. – Adham Feb 5 '12 at 3:23
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But if they all have equal weight, doesnt that mean that they are independent? – Adham Feb 5 '12 at 3:24

1 Answer

up vote 3 down vote accepted

Your calculation helps for calculating the expected number of correct answers: you have found that the probability a particular sample question both is in the test and is answered correctly is $5/18$ (assuming independence). There are $30$ sample questions so the expected number of correct answers is $30 \times 5/18 = 25/3$, just over $8$.

This does not really help with the questions you actually posed about the probability of given number of correct answers, except as a check.

Instead you need to use something like counting or binomial methods. So the probability the first test question comes from the $25$ revised is $\frac{25}{30}$; if it does, then the probability the second test question comes from comes from the other $24$ revised is $\frac{24}{29}$, and so on down to the tenth test question with probability $\frac{16}{21}$. Multiply these together and you get ${25 \choose 10} /{30 \choose 10}$. The others are similar, but you also need to take into account that various orders of revised and unrevised questions are possible.

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Makes sense. What i did to understand your explanation was simplify the problem a bit. Prof. can choose out of 5 test questions, and i know how to solve for 4. Now i listed the ten possible combinations of test questions that the prof. can choose from. Then i listed all the possible combinations of questions i can answer (5 total) and then for each combination of questions i checked what the possibility of getting full marks for all possible combinations of tests and that gave the same results as suggested. Thanks – Adham Feb 5 '12 at 14:58
Now i will try to calculate what the probability is of getting less then 5 correct answers as a sanity check and see if i get zero. – Adham Feb 5 '12 at 15:00
One more thing, why does the order matter? So if i get one question wrong, why does it matter which question it is? – Adham Feb 5 '12 at 15:27
@Adham: In your original question, to get the probability of the first two questions being unrevised and the next eight revised would be ${10 \choose 2}{25 \choose 8}/{30 \choose 10}$. But in practice the unrevised questions can come in any position, so you have to multiply this by ${10 \choose 2}$ to get the probability exactly eight questions can be answered. – Henry Feb 5 '12 at 22:02

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