Calculating Probabilities for Studying Hours in Multiple Subjects

[libragirl79]

Suppose a student has 15 hours to study for exam in 5 subjects. She puts 15 plain and 4 chocolate mini candy bars into a paper bag and starts studying the first subject. At the end of every hour, she pulls a candy bar at random from the bag. If it is plain, she eats it and continues to study the same subject. If it is chocolate, she throws it away and moves along to the next subject.
a) What is the probability that she devotes precisely three hours to each subject?
b) What is the probability that she devotes three hours to one subject, then four hours to two subjects and two hours to two remaining subjects?


So, I mapped it out and for a) got the sequence of PPC PPC PPC PPC PP (P for plain, C for choc candy) and then used the hypergeom form to get (15 choose 2) times (4 choose 1) divided by (19 choose 3) and get 43% but it's only for the first subject, the first PPC. But if I multiply by the rest of the subjects, the numbers are too big.
For the second one, PC PPPC PPPC PC P, and for this one I am not sure if it would also be hypergeom? I also though I could use neg bin by letting X equal the success=chocolate, but the numbers are not working out.

Any help is appreciated! Thanks!

 

[Ray Vickson]

Suppose a student has 15 hours to study for exam in 5 subjects. She puts 15 plain and 4 chocolate mini candy bars into a paper bag and starts studying the first subject. At the end of every hour, she pulls a candy bar at random from the bag. If it is plain, she eats it and continues to study the same subject. If it is chocolate, she throws it away and moves along to the next subject.
a) What is the probability that she devotes precisely three hours to each subject?
b) What is the probability that she devotes three hours to one subject, then four hours to two subjects and two hours to two remaining subjects?


So, I mapped it out and for a) got the sequence of PPC PPC PPC PPC PP (P for plain, C for choc candy) and then used the hypergeom form to get (15 choose 2) times (4 choose 1) divided by (19 choose 3) and get 43% but it's only for the first subject, the first PPC. But if I multiply by the rest of the subjects, the numbers are too big.
For the second one, PC PPPC PPPC PC P, and for this one I am not sure if it would also be hypergeom? I also though I could use neg bin by letting X equal the success=chocolate, but the numbers are not working out.

Any help is appreciated! Thanks!

Let's assume she studies the subjects in specified order 1,2,3,4,5.

In your solution for (a) you should NOT use the hypergeometric distribution, because the actual order of candy events is important here. The first PPC (in that order) has probability = (15/19)(14/18)(4/17). The second PPC in that order (given that the first one has occurred) has probability = (13/16)(12/15)(3/14), etc.

Part (b) is similar.

RGV

 

[libragirl79]

Thank you!