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 recieve at least one file?
A. 8/9
B. 64/81
C. 4/9
D. 16/81
E. 5/9
There are two convincing answers for this question and I'm finding it difficult to select one. Can you please explain.
Soution - I
Total outcomes = 3 * 3 * 3 * 3
Each file can be sent to any of the three typists. So file 1 can be sent to typist 1,2 or 3. Similarly for the rest of the files.
Favourable Outcomes = 4C3 * 3! *3
First, we need to select 3 files out of 4 and distribute them to each of the three typists so that each one of them has one file. So we select the files in 4C3 ways and multiply it by 3! because 3 typists are involved and each of the three could get the 3 files in 3! ways. (So as typist 1, I could get file no 1, 2 or 3; As typist 2 I can get any of the files, excepting the one that typist 1 has already got; As typist 3 I can get only 1 file - the file that neither typist 1 or 2 have got. Hence 3 * 2 *1 or 3!)
Now, we have one file remaining of the four. This file has to be given to one of the three typists. Since it could be any of the three typists, we can say that that one file can be distributed in 3 ways.
Probability = Favourable outcomes/ Total outcomes
= (4 * 3 * 2 *3)/ (3 * 3 * 3 * 3)
= 8/9
Solution - II
Three typists A, B, and C.
In how many ways can one or two typists receive no file?
(1). A or B or C received all four files, 3 ways;
(2). A received no file, then 4 files will be distributed among B amd C:
2*2*2*2=16, just same as 3^4=81
The 16 ways include C=4, B=0 or C=0, B=4, which were counted in step (1). so, 16-2=14 favorable ways;
Same way, when B received no file: 14
C received no file: 14
Totally, in 3+14*3=45 ways, one or more typists will be leisure.
Therefore, 81-45=36 ways can satisfy the condtions.
36/81=4/9
GMAT Forum
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- tali62
- christiancryan
- Course Students
- Posts: 79
- Joined: Thu Jul 31, 2003 10:44 am
Solution 2 is correct. Let me show it another way, using a sequential argument. It's fine to break up a problem like this into a sequence of steps, as long as you make sure that you're *naming* the departments and typists AS you go, not beforehand. What I mean will become evident. Also, this will SEEM very long, but that's because I can't draw it out, and I also want to make sure every step is crystal-clear.
Put the 4 department heads on one side of the room and the 3 typists on the other. Ask a department head to step forward -- it doesn't matter which -- and ask him or her to choose any typist. NOW call that department head "#1" and call that typist "A" (stick signs on them). Now #1 steps back into the shadows, and typist A steps back too. We've done nothing so far that requires probability -- SOME department HAD to go first, and that department HAD to pick some typist.
So we are now at
1=A
2
3
4
with 100% probability.
Now ask ANOTHER department head to step forward. Call that department head "#2." Ask him or her to pick a typist.
I would draw this as a "probability tree" branching to the right if I could, but since I can't, I'll call the two outcomes "Top Path" and "Bottom Path."
Top Path: 1/3 chance that #2 picks A also.
1=A
2=A
3
4
Bottom Path: 2/3 chance that #2 picks SOMEONE ELSE. We'll NOW call that person "B."
1=A
2=B
3
4
Now, let's keep going down the top path. #3 is going to pick.
Top-Top Path: 1/3 chance that #3 picks A also -- BAD! No way to distribute one to everyone now.
1=A
2=A
3=A
4
Top-Bottom Path: 2/3 chance that #3 picks someone else (call that person "B," which also means that the final person must be called "C" now):
1=A
2=A
3=B
4
Just to finish this out, let's look at #4 on this "Top-Bottom" path. Now there's a 1/3 chance that #4 picks C:
1=A
2=A
3=B
4=C
The total probability of that GOOD outcome is therefore (1/3)*(2/3)*(1/3) = 2/27.
Now let's go back to the original Bottom Path:
1=A
2=B
3
4
Now, #3 picks. If #3 picks C, we're done -- that's another GOOD outcome:
1=A
2=B
3=C
4
The chance of picking C at THAT point is 1/3. So the total chance of this second GOOD outcome is (2/3)*(1/3) = 2/9.
Now, if #3 picked A or B (a 2/3 chance at this point), we're not done yet. Notice the chance of this outcome at this point is (2/3)*(2/3) = 4/9.
1=A
2=B
3=A or B
4
Now, for this to wind up as still another GOOD outcome, we need for #4 to pick C -- which is a 1/3 chance *at this point.* So the total chance of this third GOOD outcome is (2/3)*(2/3)*(1/3) = 4/27.
Since there are 3 separate ways to achieve a good outcome, we add their probabilities:
2/27 + 2/9 + 4/27 = 12/27 = 4/9.
In Solution 1, there's an overcounting. (By the way, I recommend that you never use numbers to label BOTH files and typists -- a great way to confuse yourself!!!!) According to this method, what precisely you do is DOUBLE-COUNT an ultimate outcome.
For example, take ultimate outcome AABC -- that is, typist A gets files 1 and 2, typist B gets file 3 and typist C gets file 4. By your method, you would count this outcome twice: once, as "initial group of files" 2-3-4 distributed to A, B, and C, and then you give the one left over (= file 1) to A -- AND secondly, as "initial group of files" 1-3-4 distributed to A, B, and C, and then you give the one left over (= file 2) to A. You have to be sure that there's NO OVERLAP in your outcomes. The ultimate outcomes to count & to keep straight can be modeled by a sequence of 4 letters, each of which is an A, a B or a C. Each of these outcomes has a 1/81 chance of occurring, since each letter can pop up in its slot with a 1/3 chance.
By the way -- this problem, while a good "challenge" problem to exercise your probability chops, doesn't seem very GMAT-like: it requires a few too many scenarios.
Hope this is helpful!
Put the 4 department heads on one side of the room and the 3 typists on the other. Ask a department head to step forward -- it doesn't matter which -- and ask him or her to choose any typist. NOW call that department head "#1" and call that typist "A" (stick signs on them). Now #1 steps back into the shadows, and typist A steps back too. We've done nothing so far that requires probability -- SOME department HAD to go first, and that department HAD to pick some typist.
So we are now at
1=A
2
3
4
with 100% probability.
Now ask ANOTHER department head to step forward. Call that department head "#2." Ask him or her to pick a typist.
I would draw this as a "probability tree" branching to the right if I could, but since I can't, I'll call the two outcomes "Top Path" and "Bottom Path."
Top Path: 1/3 chance that #2 picks A also.
1=A
2=A
3
4
Bottom Path: 2/3 chance that #2 picks SOMEONE ELSE. We'll NOW call that person "B."
1=A
2=B
3
4
Now, let's keep going down the top path. #3 is going to pick.
Top-Top Path: 1/3 chance that #3 picks A also -- BAD! No way to distribute one to everyone now.
1=A
2=A
3=A
4
Top-Bottom Path: 2/3 chance that #3 picks someone else (call that person "B," which also means that the final person must be called "C" now):
1=A
2=A
3=B
4
Just to finish this out, let's look at #4 on this "Top-Bottom" path. Now there's a 1/3 chance that #4 picks C:
1=A
2=A
3=B
4=C
The total probability of that GOOD outcome is therefore (1/3)*(2/3)*(1/3) = 2/27.
Now let's go back to the original Bottom Path:
1=A
2=B
3
4
Now, #3 picks. If #3 picks C, we're done -- that's another GOOD outcome:
1=A
2=B
3=C
4
The chance of picking C at THAT point is 1/3. So the total chance of this second GOOD outcome is (2/3)*(1/3) = 2/9.
Now, if #3 picked A or B (a 2/3 chance at this point), we're not done yet. Notice the chance of this outcome at this point is (2/3)*(2/3) = 4/9.
1=A
2=B
3=A or B
4
Now, for this to wind up as still another GOOD outcome, we need for #4 to pick C -- which is a 1/3 chance *at this point.* So the total chance of this third GOOD outcome is (2/3)*(2/3)*(1/3) = 4/27.
Since there are 3 separate ways to achieve a good outcome, we add their probabilities:
2/27 + 2/9 + 4/27 = 12/27 = 4/9.
In Solution 1, there's an overcounting. (By the way, I recommend that you never use numbers to label BOTH files and typists -- a great way to confuse yourself!!!!) According to this method, what precisely you do is DOUBLE-COUNT an ultimate outcome.
For example, take ultimate outcome AABC -- that is, typist A gets files 1 and 2, typist B gets file 3 and typist C gets file 4. By your method, you would count this outcome twice: once, as "initial group of files" 2-3-4 distributed to A, B, and C, and then you give the one left over (= file 1) to A -- AND secondly, as "initial group of files" 1-3-4 distributed to A, B, and C, and then you give the one left over (= file 2) to A. You have to be sure that there's NO OVERLAP in your outcomes. The ultimate outcomes to count & to keep straight can be modeled by a sequence of 4 letters, each of which is an A, a B or a C. Each of these outcomes has a 1/81 chance of occurring, since each letter can pop up in its slot with a 1/3 chance.
By the way -- this problem, while a good "challenge" problem to exercise your probability chops, doesn't seem very GMAT-like: it requires a few too many scenarios.
Hope this is helpful!
- Guest
I thought about this one for awhile and came up with a decent solution.
First, the total possibilities, for the bottom of the probability fraction:
3 (possible typists for the first file) x 3 (for the second...) x 3 x 3 = 81 possibilities.
Now, there are three distributions of files that will satisfy the initial condition of "no typists empty-handed."
Typist A B C
2 1 1
1 2 1
1 1 2
These arrangements are all equally likely to happen, so we only need to find the possible arrangements for one grouping and then multiply this result by 3.
If the first typist gets two files, then we must calculate how many ways four files can be paired.
We can calculate this with 4C2 = 6.
If the first typist has two files, the other two typists have 2! or two possible arrangements of files (AB and BA).
6x2 = 12 possible arrangements of the form "211"
12x3 = 36 possible arrangments in which each typist gets a file
36/81 = successful/total = 4/9.
First, the total possibilities, for the bottom of the probability fraction:
3 (possible typists for the first file) x 3 (for the second...) x 3 x 3 = 81 possibilities.
Now, there are three distributions of files that will satisfy the initial condition of "no typists empty-handed."
Typist A B C
2 1 1
1 2 1
1 1 2
These arrangements are all equally likely to happen, so we only need to find the possible arrangements for one grouping and then multiply this result by 3.
If the first typist gets two files, then we must calculate how many ways four files can be paired.
We can calculate this with 4C2 = 6.
If the first typist has two files, the other two typists have 2! or two possible arrangements of files (AB and BA).
6x2 = 12 possible arrangements of the form "211"
12x3 = 36 possible arrangments in which each typist gets a file
36/81 = successful/total = 4/9.
6 posts Page 1 of 1