Hi kevincanspain,
essentially my approach comes down to the same result than yours. Although I wonder how you come up with the conclusion: "4 x 3 ways of assigning the reports so that a certain secretary gets two reports and the other two get one each"..?
If you did not calculate it by 4C(2, 1, 1) = 4!/(2!*1!*1!) = 4 * 3 = 12, how else did you come up with that result? Trial?

Cheers,
Nicolas

I too got 4/9. However, I used what I hope is a slightly more intuitive solution.

There is a one-third chance that any of the three secretaries gets the first document. Assume, without loss of generality, that the first document goes to a secretary A.

There is a 1/3 chance the second document also goes to A. Once A has two things to type, there is only one way to succeed. We need the next document to go to either B or C, which represents a 2/3 chance; then, the last document must go to the one remaining textless secretary, a 1/3 chance.

1/3 x 2/3 x 1/3 = 2/27

On the other hand, there is a 2/3 chance that either B or C (not A) gets the second document. Let's say B gets it (again without loss of generality). This means that in order to have all three secretaries with at least one document, at least one of the remaining two must go to C. To calculate that odds that C gets a document, we need to remember that the odds of all possibilities total to 1. Since the odds of either A or B getting a document is 2/3, there is a 4/9 chance that A and B will get both of the next documents--or, a 1 - 4/9 = 5/9 chance that C gets at least one of the two remaining documents.

2/3 x 5/9 = 10/27.

So, we have a 2/27 chance that the Doc 2 goes to the same secretary as Doc 1, and the remaining two documents are distributed among the remaining two secretaries; and, we have a 10/27 chance that the first two documents are spread among two different secretaries, and at least one of the remaining two documents goes two a third. Mutually exclusive possibilities add, so we end up with:

2/27 + 10/27 = 12/27 = 4/9

As the odds that each secretary gets at least one document to work with.

kevincanspain wrote:I would say there are 3^4 ways to assign the reports.

1 secretary gets 2 reports, the others get 1 each.

3 ways to decide which secretary gets two reports

4 x 3 ways of assigning the reports so that a certain secretary gets two reports and the other two get one each.


answer: 3 x 3 x 4/ 3^4= 4/9


alternatively:
undesirable assignments
4,0,0 3 ways
3,1,0, 4 x 3 x 2 ways = 24 ways
2,2,0 4c2 x 3 ways= 18 ways

45 undesirable assignments , thus 36 desirable

36/81= 4/9

4 x 3 ways of assigning the reports so that a certain secretary gets two reports and the other two get one each.

The way I got this was to think of the two secretaries that got one report each. The first of these secretaries could get any of the 4 reports, and the second could get one of the 3 remaining. The third secretary gets the two remaining reports

KapTeacherEli wrote:I too got 4/9. However, I used what I hope is a slightly more intuitive solution.

There is a one-third chance that any of the three secretaries gets the first document. Assume, without loss of generality, that the first document goes to a secretary A.

There is a 1/3 chance the second document also goes to A. Once A has two things to type, there is only one way to succeed. We need the next document to go to either B or C, which represents a 2/3 chance; then, the last document must go to the one remaining textless secretary, a 1/3 chance.

1/3 x 2/3 x 1/3 = 2/27

On the other hand, there is a 2/3 chance that either B or C (not A) gets the second document. Let's say B gets it (again without loss of generality). This means that in order to have all three secretaries with at least one document, at least one of the remaining two must go to C. To calculate that odds that C gets a document, we need to remember that the odds of all possibilities total to 1. Since the odds of either A or B getting a document is 2/3, there is a 4/9 chance that A and B will get both of the next documents--or, a 1 - 4/9 = 5/9 chance that C gets at least one of the two remaining documents.

2/3 x 5/9 = 10/27.

So, we have a 2/27 chance that the Doc 2 goes to the same secretary as Doc 1, and the remaining two documents are distributed among the remaining two secretaries; and, we have a 10/27 chance that the first two documents are spread among two different secretaries, and at least one of the remaining two documents goes two a third. Mutually exclusive possibilities add, so we end up with:

2/27 + 10/27 = 12/27 = 4/9

As the odds that each secretary gets at least one document to work with.

This is great: now we have 3 methods that point to 4/9 as the answer!

received a pm.

Here's how I solved it.

Probability = desired/total

Total = 3^4 = 81

Let's call the secretaries A, B, and C. Let's call the reports w, x, y, and z.

In order for each of the secretaries to type at least 1 report, we need to give 2 reports to one secretary (with each of the other two secretaries typing 1 report each). It doesn't matter which of the three secretaries takes which of the 2 reports. So there are 3C1 ways we can select any of the three secretaries to type 2 reports. But that secretary can type any 2 of the 4 reports. So, there are 4C2 ways to pull out any 2 reports from the 4 available.

So far, that's 3C1*4C2. Once we've assigned any 2 reports to any 1 of the secretaries, there are 2 reports left and two secretaries left. Suppose that secretary A was assigned reports w and x. Then either secretary B types y (with C typing z), or else the other way around. In other words, after we've assigned 2 reports to a secretary, order now matters, and we have 2 ways of assigning the 2 remaining reports to the two remaining secretaries.

Final answer:

Probability = (3C1 * 4C2 * 2)/81 = 4/9

Hi there. Here is how I get to 8/9.

It is tempting to go the (1-x) route here but it would be too studious in this case. Let's go straigth to P (Probability that all 3 secretaries are assigned at least one report).

P = Total Desired Outcomes / Total Possible Outcomes

Total Possible Outcomes = 3^4 (we all agree here). Let's now determine the Total Desired Outcomes (each secretary receives at least one report) step by step:

1 - Calculate the number of ways to choose 3 reports out of 4. This is combination as order doesn't matter = C(4,3) = 4. Althernatively, we can see that there are 4 ways to choose the 4th report to be allocated at the end, leaving 4 groups of 3 reports.

2 - Calculate the number of ways to allocate the 3 chosen reports to the 3 secretaries. This is a permutation: 3! = 3x2x1.

3 - Calculate the number of ways to allocate the 4th report to any of the 3 secretaries (yes, there are many ways) = 3 (more formally a combination C(3,1) = 3).

4 - Total Desired Outcomes = Ways to choose the 3 reports x Ways to allocate the 3 chosen reports x Ways to allocate the 4th report = C(4,3) x 3! x C(3,1) = 4 x 3 x 2 x 3

CONCLUSION:

P = Total Desired Outcomes / Total Possible Outcomes = (4 x 3 x 2 x 3) / (3 x 3 x 3 x 3) = 8/9.

I hope there's no flaw. Yet, I must admit that such a proba being 8/9 (close to 1) is not intuitive; 4/9 is more appealing to me.

PUNITKAUR: it may be time to tell us where this problem is coming from and whether their proposed solution (8/9) could be wrong.

THANKS//

You are double-counting:

Let's call the three secretaries A,B and C, and the four reports 1, 2, 3, 4

A-1 B-2 C-3 A- 4

is no different from A-4 B-2 C-3 A-1

Although I solved it differently the first time (see my post above), here's a couple of other approaches:

Write out four slots for the reports:
_ _ _ _
w x y z

The question about desired outcomes essentially asks how many orders we can create from 4 items (the secretaries) in which 2 are duplicated (one secretary types both reports). If there were 4 distinct items, the order is simply 4!. If there are four items with one duplicate, the number of orders is 4!/2!. Since any one of the three secretaries can be duplicated, there are 3 cases, with each case having 4!/2! orders (desired outcomes = 3x4!/2!).

Here is one more way (a method I advocate my students use), which is probably the simplest and most consistent with other probability problems--the slot method. First, you start by asking how many cases there are - two of the 4 reports will be written by the same secretary, so that is 4C2 cases. Then, select any one case (they are all numerically identical) and calculate the possibilities for each slot. Suppose that the first two slots (w and x) are by the same secretary; then we would have the following options for each slot:

3 1 2 1
_ _ _ _
w x y z

Any one of the three secretaries can type report w, but then there is only one option for slot x. Then any of the remaining 2 secretaries can type report y and the remaining 1 secretary types report z.

Thus, the desired outcomes is 4C2 cases times 3x1x2x1 for each case.

If all 3 secretaries do at least 1 report one secretary will do 2 reports.

There are 6 ways in which this can happen: 1st and 2nd department's reports are done by same secretary and the rest are done by the other two secretaries, in the same way: 1st & 3rd, 1st & 4th, 2nd & 3rd, 2nd & 4th, 3rd & 4th.

The probability of each of those hapening is: 1/3 * 2/3 * 1/3

Multiply that by 6(the number of outcomes) and you get 4/9.

A little late on this one, but here's how I did it. Could anyone tell me why this is wrong?

1. 4 0 0

The first possibility is that one secretary does all 4 reports. Number of ways this can happen = 3 (because either of the 3 secretaries may be the one doing all 4 reports). Therefore, total 3 ways.

2. 3 1 0

2nd possibility is one secretary does 3 reports and another secy does 1, with 1 secy not doing any report. Number of ways the secy doing 3 can be chosen = 3 (either of the 3). Number of ways the secy doing 1 report can be chosen = 2 (any of the remaining 2). Therefore, total 6 ways.

3. 2 1 1
The third possibility is that one secy does 2 reports and both the others do one each.
Number of ways this can happen = 3 (just choosing the secy that does 2 reports)
Therefore, total 3 ways.

Thus, there are total (3+6+3) ways that reports can be distributed. Of which applicable to us (at least 1 report being done by each secretary) are only 3 (third case). Therefore, the required probability = 3/12=1/4

I saw no reason to go in permutations here (3^4 etc.). Could anyone explain to me why my answer is wrong?

Sorry, there is one more possibility - 2 secys doing 2 reports each

4. 2 2 0

Number of ways this can be done is 3 (choosing one secy out of 3 who will do no report).

So total number of ways = (3+6+3+3) = 15
Relevant number (case with each secy doing at least 1 report) = 3
Therefore probability = 3/15 = 20%

is this right?

Total ways - (3C1)^4 assigning each report to 1 secretary

Favorable ways - 4C1 (assign 1 to S1) X 3C1(assign another to S2) X2C1 (assigning another to S3)

This can be done in 3 ways

Probability = 3! X4C1 X3C1X2C1 / (3C1)^4

=3X4X3X2/3X3X3X3 = 8/9

The correct answer is not 8/9 but [spoiler]4/9[/spoiler]. Check my explanation here:

https://www.beatthegmat.com/probability-t91579.html

We should recognize that the following outcomes are possible:

-- one secretary gets all 4 reports while the other 2 get none
-- 2 secretaries each get a pair of reports while the third gets none.

Thus, P(each secretary gets at least 1 report) = 8/9 is WAY TOO HIGH.

Be sure the result of your calculations makes sense before you select an answer choice.