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combinatorics/probability problem -- please help

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combinatorics/probability problem -- please help [#permalink] New post 03 Apr 2013, 16:53
I downloaded a file with some sample combinatorics and prob problems, and trying to solve this one:

There are three secretaries who work for four departments. If each of the four departments has one report to be typed out, and the reports are randomly assigned to a secretary, what is the probability that all three secretaries are assigned at least one report?

The answers sheet says it's 8/9. I get 1/5. Here's how I calculate this:

number of possible arrangements:

sec1 sec2 sec3
4 0 0 => 3 arrangements (3!/2!)
3 1 0 => 6 arrangements (3!)
2 2 0 => 3 arrangements
2 1 1 => 3 arrangements, and the only condition satisfying at least 1 report/secretary

arrangements total: 3+6+3+3=15
arrangements where each gets at least one report: 3

Probability of last arrangement: 3/15=1/5

So where am I going wrong? Or is the known answer for this problem wrong?
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Re: combinatorics/probability problem -- please help [#permalink] New post 07 Apr 2013, 10:21
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Dixon wrote:
I downloaded a file with some sample combinatorics and prob problems, and trying to solve this one:

There are three secretaries who work for four departments. If each of the four departments has one report to be typed out, and the reports are randomly assigned to a secretary, what is the probability that all three secretaries are assigned at least one report?

The answers sheet says it's 8/9. I get 1/5. Here's how I calculate this:

number of possible arrangements:

sec1 sec2 sec3
4 0 0 => 3 arrangements (3!/2!)
3 1 0 => 6 arrangements (3!)
2 2 0 => 3 arrangements
2 1 1 => 3 arrangements, and the only condition satisfying at least 1 report/secretary

arrangements total: 3+6+3+3=15
arrangements where each gets at least one report: 3

Probability of last arrangement: 3/15=1/5

So where am I going wrong? Or is the known answer for this problem wrong?




Probability = favorable outcomes / Total Outcomes.

Total Outcomes = Number of ways in which Reports can be assigned to the secretaries.
There are 4 reports and 3 secretaries and each report can be assigned to any one of the 3 secretaries
So First Report can be assigned in 3 ways, simiarly 2nd, 3rd, and 4th Report also can be assigned in 3 ways each
So total number of ways = 3 X 3 X 3 X 3 = 3^4 = 81

Favorable Outcomes = Each Secretary should get atleast one Report.
3 Secretaries can get 4 Reports in 4P3 ways. Now 1 Report left. That can be allotted to any one of 3 secretaries in 3 ways.
Hence Total permutations = 4P3 X 3------->(4!/ (4-3)!) X 3----------> 4! X 3------------>24 X 3 = 72

So the desired Probability is 72/81 = 8/9

Regards,

Narenn.

PS :- This concept will be thoroughly covered in my upcoming article on Permutations and Combination. You can find it in my signature - perhaps on tomorrow.
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Re: combinatorics/probability problem -- please help [#permalink] New post 09 Apr 2013, 13:25
Thank you. I kinda get the explanation, hope to find some more similar problems to get a bit more practice and really grasp these concepts.
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Re: combinatorics/probability problem -- please help [#permalink] New post 09 Apr 2013, 22:12
Expert's post
Dixon wrote:
I downloaded a file with some sample combinatorics and prob problems, and trying to solve this one:

There are three secretaries who work for four departments. If each of the four departments has one report to be typed out, and the reports are randomly assigned to a secretary, what is the probability that all three secretaries are assigned at least one report?

The answers sheet says it's 8/9. I get 1/5. Here's how I calculate this:

number of possible arrangements:

sec1 sec2 sec3
4 0 0 => 3 arrangements (3!/2!)
3 1 0 => 6 arrangements (3!)
2 2 0 => 3 arrangements
2 1 1 => 3 arrangements, and the only condition satisfying at least 1 report/secretary

arrangements total: 3+6+3+3=15
arrangements where each gets at least one report: 3

Probability of last arrangement: 3/15=1/5

So where am I going wrong? Or is the known answer for this problem wrong?

The reason you are not getting the correct answer is that you are assuming the letters are identical.
You say:
sec1 sec2 sec3
3 1 0 => 6 arrangements (3!) (Here, you get 6 arrangements becoz sec1 could have 3 or 1 or 0 letters and so on but which 3 letters does sec1 have? The 4 depts will have 4 different letters)

Instead, try to solve it like this:

Total no of ways of distributing 4 letters among 3 secretaries = 3*3*3*3 = 81 (such that some secretaries may get no letters)

To ensure that each secretary should get at least 1 letter, the distribution will be 2, 1, 1.
Select 2 letters out of the 4 and select one secretary out of the three to give the 2 letters => 4C2 * 3C1
Now, you have 2 letters to give to 2 secretaries => 2!
Number of favorable ways = (4*3/2) * 3 * 2

Required probability = 36/81

Try out my blog. I have discussed grouping identical/distinct items in various ways on it.
http://www.veritasprep.com/blog/categor ... om/page/2/

Scroll down to the last post. It is the first post on Combinatorics. The 10-12 posts above it are all on probability/combinatorics
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Last edited by VeritasPrepKarishma on 26 Mar 2014, 19:41, edited 1 time in total.
Removed the double counting
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Re: combinatorics/probability problem -- please help [#permalink] New post 24 Mar 2014, 23:00
I dont think the answers stated here are correct. Correct answer is 39/81.

Total number of outcomes = 81.
Probability that each gets atleast 1 = 1 - probability that each doesnt get atleast one report

=================================
Probability that each doesn't get atleast one report
=================================

Let reports be W,X,Y,Z and secretaries be A,B,C.

Now we have to find ways that atleast one of the secretary has 0 reports assigned to him.

We will do this for one secretary (find the number of outcomes that secretary has for 0 reports assigned) and then add up for all 3 secretaries.

==================================================
Number of outcomes for which "A" doesnt get assigned any report (W,X,Y,Z).
==================================================

A = 0, B = 1, C = 3
A = 0, B = 3, C = 1
A = 0, B = 2, C = 2

So above combination is the only way A is assigned 0 reports. We are not done here. For each of the above cases, there is more than one outcome associated with it.

e.g. A=0, B=1, C=3

B=W, C=XYZ
B=X, C = WYZ
B=Y, C=WXZ
B=Z, C=WXY

4 OUTCOMES. Multiply this by 2 since we have two cases of (1,3). = 8

e.g. A=0, B=2, C=2

B=WX
B=WY
B=WZ
B=XY
B=XZ
B=YZ

So there are 6 outcomes for (2,2) case.


So total number of outcomes for where secretary A does not get any report assigned = 8 + 6 = 14

Number of outcomes for which secretaries A, B & C all don't get any reports assigned = 14 * 3 = 42

Probability that atleast one secretary is assigned 0 reports = 42 / 81 = 1/2.

WE ARE NOT DONE YET.

We have not considered the case where all 4 reports are assigned to a single secretary. There would be 3 such outcomes.

So the probability that atleast 1 report is assigned to each secretary is 39/81.
Re: combinatorics/probability problem -- please help   [#permalink] 24 Mar 2014, 23:00
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