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distribution problem

by mehaksal » Thu Aug 30, 2012 7:19 am
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?


ans : [spoiler]8/9[/spoiler]
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by neelgandham » Thu Aug 30, 2012 8:12 am
Solution here - https://www.beatthegmat.com/tough-probab ... 47818.html
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by GMATGuruNY » Fri Aug 31, 2012 12:24 am
In a certain company, three typists service for four departments. If the departments each send a file to the typists at random, what is the probability that every typist will receive at least one file?

A. 8/9
B. 64/81
C. 4/9
D. 16/81
E. 5/9
Let's call the 4 files A, B, C and D.

Number of ways the files can be assigned:
Since each file can be assigned to any one of the 3 typists:
Number of options for A = 3.
Number of options for B = 3.
Number of options for C = 3.
Number of options for D = 3.
To combine the options above, we multiply:
Total possible assignments = 3*3*3*3 = 81.

Now we need to determine how many ways the files can be assigned so that every typist is assigned at least 1 file.
For every typist to be assigned at least 1 file, one typist must be assigned a pair of files, while the other 2 typists are each assigned 1 file.

Case 1:
One typist is assigned AB, one typist is assigned C, and one typist is assigned D.
Number of ways to arrange the 3 elements AB, C and D = 3! = 6.

Remaining cases:
In the arrangements above, AB can be replaced with any pair of files.
Thus, the result above needs to be multiplied by the number of pairs that can replace AB.
Number of pairs that can be made from the 4 files = 4C2 = 6.

Multiplying the two results, we get:
Number of ways to assign the files so that every typist is assigned at least 1 file = 6*6 = 36.

P(every typist gets at least 1 file) = 36/81 = 4/9.

The correct answer is C.
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by mehaksal » Fri Aug 31, 2012 11:39 am
Lets say secretaries are called a,b,c. no. of ways of distributing the letters are:

a b c
------
4 0 0
0 4 4 set 1
4 0 4
-------
1 3 0
0 1 3 set 2
0 3 1
1 0 3
3 1 0
3 0 1
-------
1 2 1
1 1 2 set 3
2 1 1
------
2 2 0
2 0 2 set 4
0 2 2

no. of ways for set 1=3
no. of ways for set 2= 6* 4C3=24
no. of ways for set 3 = 3*4C2*2=36( because there are two ways of distributing the single 2 letter(s) after 2 are picked)
no. of ways for set 4=3*4C2=18

Probability = set3/(set1+set2..+set4)=36/81=4/9



CAN SOMEBODY PLEASE EXPLAIN THE ABOVE SOLUTION??
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by mehaksal » Wed Sep 05, 2012 2:13 am
I expect a reply!
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by GaneshMalkar » Wed Sep 05, 2012 6:52 am
mehaksal wrote:I expect a reply!

Let me try to simplify ....

There will be broadly 4 Ways in which 4 reports distributed among 3 seces

1. Each sec get at least 1 report = 211,121,112
2. Exactly 1 sec gets 0 report = 310,301,013,130,103,031
3. Exactly 2 sec gets 0 report = 400, 004, 040
4. Exactly 2 sec gets 2 reports = 220,022,202


1. Each sec get at least 1 report = 211,121,112

211 can be selected in 4C2 (2 reports out of 4 reports) * 2C1 (1 report out of 2) * 1C1 (1 report out of 1) = 6 * 2 * 1 = 12 ways
So above 3 in 12 * 3 = 36 ways

2. Exactly 1 sec gets 0 report = 310,301,013,130,103,031

310 can be selected in 4C3 (3 reports out of 4 reports) * 1C1 (1 report out of 2) = 4 * 1 = 4 ways
So above 6 in 6 * 4 = 24 ways


3. Exactly 2 sec gets 0 report = 400, 004, 040

400 can be selected in 4C4 (4 reports out of 4 reports) = 1 way
So above 3 in 3 * 1 = 3 ways

4. Exactly 2 sec gets 2 reports = 220,022,202

220 can be selected in 4C2 (2 reports out of 4 reports) * 2C2 (2 reports out of 2 reports)= 6 ways
So above 3 in 3 * 6 = 18 ways



total number of ways = 36 + 24 + 3 + 18 = 81 ways

So the P(Atleast one report/sec ) = condition 1 / total number of ways

= 36 / 81 = 4 / 9
If you cant explain it simply you dont understand it well enough!!!
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by GaneshMalkar » Wed Sep 05, 2012 7:02 am
mehaksal wrote:I expect a reply!

One more way to solve this :)

P(atleast 1 report / sec ) = 1 - P(0 report for 1 sec) - P(0 reports for 2 sec)

P(0 report for 1 sec) = 3 * (2 ^ 4 - 2) / 3 ^ 4 = 42 / 81
where 2 ^ 4 = 4 reports 2 sec can be assigned in this way
- 2 = Since there will be two combination which will assign all reports to sec1 or sec2
3 = total no of secs
3 ^ 4 = total number of ways; 4 reports 3 secs


P(0 reports for 2 sec) = 3 / 3 ^ 4 = 3 / 81



P(atleast 1 report / sec ) = 1 - P(0 report for 1 sec) - P(0 reports for 2 sec)
= 1 - 42/81 - 3/81 = 1 - 45/81 = 36/81 = 4/9
If you cant explain it simply you dont understand it well enough!!!
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by mehaksal » Wed Oct 03, 2012 4:26 am
I wonder if wud b able to solve such ques in 2 mins!
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