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Probability

by shankar.ashwin » Thu Sep 22, 2011 11:15 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
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by GMATGuruNY » Thu Sep 22, 2011 12:04 pm
shankar.ashwin wrote: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 indi » Thu Sep 22, 2011 6:22 pm
Hi,
Can we do the question with repetitive permutation method, as we have do distribute 4 files to 3 people. n+r-1 C r-1..Not able to get answer but the concept is not clear.
Thanks
GMATGuruNY wrote:
shankar.ashwin wrote: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 knight247 » Mon Oct 10, 2011 4:32 am
Hey Mitch,
Hope you can verify my logic.
Since we have four departments A B C D and three typists T1 T2 T3. Each department could be serviced by any of the three so the total number of possibilities are 3*3*3*3=81

Now, since the four files have to be be assigned to the 3 typists there will always be one secretary who gets two files. So one of the options is AB C D and calculation has to be done on those lines.

Out of the four files we can make groups of 2 in 4C2 ways. And this group of 2 can be assigned to any one of the typists in 3C1 ways. While the remaining two files can be assigned to the remaining two typists in 2P2 ways.

So the total number of ways is 4C2*3C1*2P2=36
So the probability of our desired outcome is 36/81=4/9

Is my reasoning correct so is my answer purely coincidental?? Hope u can verify!! Thanks
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by knight247 » Mon Oct 10, 2011 4:34 am
P.S. Mitch, is there any way you could show how to solve this problem using the complementary i.e.
P(Each one gets assigned atleast one file)=1-P(One person does not get any file). Thanks!!
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by GMATGuruNY » Tue Oct 11, 2011 9:55 am
knight247 wrote:P.S. Mitch, is there any way you could show how to solve this problem using the complementary i.e.
P(Each one gets assigned atleast one file)=1-P(One person does not get any file). Thanks!!
Since there are 4 files, and each file could be assigned to any of the 3 typists, the number of ways to assign the files = 3*3*3*3 = 81.

One of the 3 typists is assigned all 4 files:
Number of typists who could receive all 4 files = 3.

One of the 3 typists is assigned 3 files, another is assigned 1 file:
Number of typists who could receive the 3 files = 3.
Of the 4 files, number of combinations of 3 that could be assigned to this typist = 4C3 = 4.
Number of remaining typists who could receive the one remaining file = 2.
To combine these options, we multiply:
3*4*2 = 24.

2 typists are assigned 2 files each:
Number of pairs of typists who could be assigned the files = 3C2 = 3.
Number of pairs of files that could be assigned to the first typist = 4C2 = 6.
The second typist must receive the remaining pair of files, so the number of options for the second typist = 1.
To combine these options, we multiply:
3*6*1 = 18.

Thus:
The total number of ways in which one or more of the typists could NOT receive a file = 3+24+18 = 45.

Thus:
The total number of ways in which each typist is assigned at least 1 file = 81-45 = 36.

P(each typist is assigned at least 1 file) = 36/81 = 4/9.

P.S. The reasoning in your post above looks good.
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by knight247 » Mon Oct 17, 2011 11:42 pm
GMATGuruNY wrote: One of the 3 typists is assigned all 4 files:
Number of typists who could receive all 4 files = 3.
Hey Mitch,
I'm not quite sure of this part. Coz we know that
P(All of them recv atleast 1 file)+P(EXACTLY 1 of them does not get a file)=1
If this is the case then can we consider 1 typist receiving all 4 files? Wouldn't that be the number of ways that two typists receive no files which we are not supposed to include? Hope u can comment.
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by GMATGuruNY » Tue Oct 18, 2011 2:37 am
knight247 wrote:
GMATGuruNY wrote: One of the 3 typists is assigned all 4 files:
Number of typists who could receive all 4 files = 3.
Hey Mitch,
I'm not quite sure of this part. Coz we know that
P(All of them recv atleast 1 file)+P(EXACTLY 1 of them does not get a file)=1
If this is the case then can we consider 1 typist receiving all 4 files? Wouldn't that be the number of ways that two typists receive no files which we are not supposed to include? Hope u can comment.
EXACTLY ONE is not the opposite of ALL.
The opposite of ALL is NOT ALL:

P(ALL of the typists get at least 1 file) + P(NOT ALL of the typists get at least 1 file) = 1.

My solution above accounts for the different ways in which NOT ALL of the typists get at least 1 file.
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by gmat6087 » Wed Oct 10, 2012 10:52 am
GMATGuruNY wrote:
shankar.ashwin wrote: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.
Mitch,
Kindly correct me where I am wrong.
81 is the total ways.

Now 4 dept(d1,d2,d3,d4) 3 typists(t1,t2,t3)

d1 has 3 options 3(3 typists).
after d1 , d2 has 2 options 2(2 typists left).
After d1 and d2, d3 has 1 option 1(1 typist left)
After d1,d2 and d3, d4 has 3 options 3(can give the letter to any one of the 3 typist)
multiplying we get 3*2*1*3=18

Please help me here

Regards,
Satya
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by GMATGuruNY » Thu Oct 11, 2012 2:58 am
gmat6087 wrote:
shankar.ashwin wrote: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
Mitch,
Kindly correct me where I am wrong.
81 is the total ways.

Now 4 dept(d1,d2,d3,d4) 3 typists(t1,t2,t3)

d1 has 3 options 3(3 typists).
after d1 , d2 has 2 options 2(2 typists left).
After d1 and d2, d3 has 1 option 1(1 typist left)
After d1,d2 and d3, d4 has 3 options 3(can give the letter to any one of the 3 typist)
multiplying we get 3*2*1*3=18

Please help me here

Regards,
Satya
In assigning d1, d2 and d3 each to a DIFFERENT typist and then assigning d4 LAST, the solution above overly restricts the number of ways the files can be PAIRED.
Since d4 is assigned last, the only pairings included in the solution above are those that include d4:
d1d4, d2d4, d3d4.

We still need to count the arrangements in which one of the 3 typists is assigned d1d2, d1d3, or d2d3:
Number of ways to assign these pairings = 3. (Any of the 3 typists.)
Number of pairings that could be assigned to this typist = 3. (d1d2, d1d3, or d2d3.)
Number of ways to arrange the remaining 2 files = 2! = 2.
To combine these options, we multiply:
3*3*2 = 18.

Adding these 18 arrangements to the 18 arrangements counted in your post, we get:
Total number of ways to assign the files = 18+18 = 36.

P(each typist is assigned at least 1 file) = 36/81 = 4/9.
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