One professor grades homework by randomly choosing 5 out of 14 homework problems to grade.
(a) How many different groups of 5 problems can be chosen from the 14 problems?
(b) Probability extension: Jerry did only 5 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.)
(c) Silvia did 7 problems. How many different groups of 5 did she complete? What is the probability that one of the groups of 5 she completed comprised the group selected to be graded? (Round your answer to four decimal places.)

In order to solve this problem we must ask ourselves: Does the order matter? We can see that in this case the order in which the problems are chosen doesn't really matter, so we are talking about a combination:

[tex]_{n}C_{r}=\frac{n!}{r!(n-r)!}[/tex]

in this case, n=14 while r=5, so our combination will be:

[tex]_{14}C_{5}=\frac{14!}{5!(14-5)!}[/tex]

which yields:

[tex]_{14}C_{5}=2002[/tex] which will be the answer for part a.

b)

In order to find the probability we can use the probability formula:

[tex]Probability=\frac{desired \hspace{5}outcome}{number\hspace{5} of \hspace{5} possible \hspace{5} outcomes}[/tex]

so in this case there is only 1 desired outcome out of the 2002 possible outcomes so our probability is:

[tex]P=\frac{1}{2002}[/tex]

which is the same as:

P=0.04995%

c) The same applies to part c. In order to find the number of groups Silvia did, we can make us of the combination formula:

[tex]_{7}C_{5}=\frac{7!}{5!(7-5)!}[/tex]

which yields:

[tex]_{14}C_{5}=21[/tex]

so there are 21 different groups of 5 she did by solving 7 problems.

So we can use the same formula to find the probability that one of the groups she completed comprised the group selected to be graded so we get:

[tex]P=\frac{21}{2002}=\frac{3}{286}[/tex]

which is the same as:

P=1.04895%

(remember that in order to turn a decimal number into a percentage you need to multiply the decimal number by 100)