Probability that a random sample will have certain elements 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.
 A: 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.  
