(b) probability extension: jerry did only 6 problems of one assignment. what is the probability that the problems he did comprised the group that was selected to be graded? (round your answer to four decimal places.)

You are asking for the second part (part b) of a problem and some needed  information is missed.

The relevant information missed is the amound of problems that the professor assigned and the number of problems that were going to be selected to grade.

When you have that information you can use combinatory theory to find the amount of different groups of problems that can be formed under the given conditions, and then the probability that the group done by the student is exactly the group selected by the professor.

 Suppose that the missing information is that the professor grades 6 problems out of 15 to grade.

Then the number of possible groups of 6 out of 15 problems is:

C(6,15) = 15! / [6! (15 - 6)! ] = 15! / [6! 9!] = 15*14*13*12*11*10 / (6*5*4*3*2) =

= 5005

Since the student did only five problems that is only one of those 5005 groups and the probablity is 5 / 5005 = 1 / 1001 = 0.0009999 ≈ 0.0009.

Answer 0.0009 (for the hypotethical case of grading 6 problems out of 15).