0
$\begingroup$

I have to find the sample space and a few probabilities here and I am wondering about if I am going down the right track for these. If I am incorrect, then please point me in the right direction, but not give me the full process.

There are 5 students that must finish 6 problems. Each problem is only worked on by one student. A teacher assigns each student to each problem.

Find the Sample Space: Total of $5^6$ ways to assign problems. Any 5 students can do each of the problems, hence $5^6$ ways.

Find probability that student A gets two problems: $6 \choose 2$$*4^4$. This is because 2 problems for A to choose from and then 4 problems left for 4 other students.

Find prob that one student has three problems, another has two, and a third has one problem (this is were I am the most unsure): So there are $5 \choose 3$ ways to arrange the students into groups. But I am stuck on how to section off the problems. My original idea was to have $6\choose2$$3 \choose2$$1 \choose 1$. But I know that is incorrect. I have 3 groups of problems that are sectioned off in 3 ways. How do I go about thinking about this?

$\endgroup$
1
$\begingroup$

Find the Sample Space: Total of $5^6$ ways to assign problems. Any $5$ students can do each of the problems, hence $5^6$ ways.

$\checkmark\;$ Yes. Though I'd express it as: any $1$ of $5$ students can do each of $6$ problems.

( Note: this assuming students are assigned to a problem independent of whether any student is assigned any other problems )

Find probability that student A gets two problems: $\binom{6}{2} \cdot 4^4$ . This is because $2$ problems for A to choose from and then $4$ problems left for $4$ other students.

$\checkmark\;$ Yes; if you divide that count by the size of the sample space.

Note: this gives the probability that A gets exactly 2 problems. The probability of getting at least 2 problems is somewhat different.

Find prob that one student has three problems, another has two, and a third has one problem (this is were I am the most unsure): So there are ${5\choose 3}$ ways to arrange the students into groups.

But I am stuck on how to section off the problems. My original idea was to have ${6\choose 2}{3\choose 2}{1\choose 1}$. But I know that is incorrect. I have 3 groups of problems that are sectioned off in 3 ways. How do I go about thinking about this?

Count the ways to: select 1 of 5 students and assign 3 of 6 problems, then 1 of 4 students and 2 of 3 problems, and finally 1 of 3 students for the last problem. (The last 2 students have no problems to do.)

Then divide this count by the size of the sample space.


I thought about this way, but isn't that assuming that you give out 3 problems first, then 2 problems, then 1 ...

No. It doesn't matter how the task of assigning problems to students is managed (whether you collate the problems together before assigning them to a student, or assign a student to each problem in sequence), we only need to count the arrangements that result in the favoured space.

... and also assuming you give the first 3 to student 1. Because problem 1 could go to student 1, then the next problems to student 2, then the next to student 1, and the last problem to student 3.

No, we are counting the ways any $3$ of the problems might be assigned to any $1$ of the students, and also any $2$ of the other problems to any $1$ of the other students, and the remaining problem to and $1$ of the remaining students.

Also we could have had $5 \choose 1$ students and assigned them 2 problems, $4 \choose 1$ assign them 3 problems and $3 \choose 1$ assign 1 problem right?

That's right. Order of assignment doesn't matter when we are counting the results. (When you expand the binomial coefficients you'll find a lot of cancellable terms.)


Another way to look at it:

First we count how many ways can we can arrange the $5$ students in this manner: 2 students to put in an imaginary 'no problem' box, 1 student in a 'one problem' box, 1 in a '2 problem' box, and 1 student in a '3 problem box'.

Next count how many we ways can the $6$ problems fill these 'boxes' with the appropriate number of problems, $0, 1, 2, \& 3$.

By the universal principle of counting, we multiply the counts of these subtasks and get the size of the favoured space.

Then we divide by the size of the sample space and obtain the required probability measure.

$$\dbinom{5}{1,1,1,2}\dbinom{6}{3,2,1,0} = \dfrac{5!}{1!1!1!2!}\dfrac{6!}{3!2!1!0!}$$

$\endgroup$
  • $\begingroup$ I thought about this way, but isn't that assuming that you give out 3 problems first, then 2 problems, then 1 and also assuming you give the first 3 to student 1. Because problem 1 could go to student 1, then the next problems to student 2, then the next to student 1, and the last problem to student 3. Also we could have had 5 choose 1 students and assigned them 2 problems, 3 choose 1 assign them 3 problems and 4 choose assign 1 problem right? $\endgroup$ – Jack Armstrong Mar 3 '15 at 0:02
  • $\begingroup$ @Jack, I've edited the solution to answer that. $\endgroup$ – Graham Kemp Mar 3 '15 at 1:26
  • $\begingroup$ so we take what you solve the 5 and 6 choose stuff over the sample space of $5^6$. I like the depth that you put into answering this. $\endgroup$ – Jack Armstrong Mar 3 '15 at 21:17
  • $\begingroup$ but why do you include 0 in figuring out 6 problems. I know 0!=1 so it doesnt really matter, but what is the thought behind it? $\endgroup$ – Jack Armstrong Mar 3 '15 at 21:28
  • 1
    $\begingroup$ @Jack, the intermediate answer was not simplified so that it reflects the explanation (which includes that there was a group of two students not assigned any problem). As you note, the next step, simplification, is a matter of reducing the $1!$ and $0!$ terms to unity. $\endgroup$ – Graham Kemp Mar 4 '15 at 0:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.