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03 Apr 2013, 17:53
Kudos
Bookmarks
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?
OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/there-are-th ... 31926.html
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?
OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/there-are-th ... 31926.html
07 Apr 2013, 11:21
Kudos
Bookmarks
Dixon
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?
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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