VeritasKarishma wrote:
Lapetiteflo wrote:
Hi,
I have Big troubles with this question.
We havé total possibilities N =3^4
Then we can say that the probability to give at least one report to each secretary is the opposite of giving no report to at least one secretary.
So evaluating this proba :
-ways to choose 1 secretary among 3 (she will not be assigned any report)=3
-ways to affect 4 reports to 2 secretaries (it can be just one if ever) : examples (1122) assigns 2 first reports to secretary 1 and two last one to secretary 2. There are 2^4 possibilities (1111,1112,1122,....)
So result is 1-(3*2^4)/3^4=11/27.
The true result is 12/26.... Could you explain where is my mistake ?
Thanks a lot
Marie
"We havé total possibilities N =3^4"
What are these total possibilities? They are the ways in which the 4 reports can be distributed among 3 secretaries so that all reports may go to one secretary, the reports may be distributed among 2 secretaries or they may be distributed among all 3. So such cases are included (4, 0, 0), (0, 4, 0), (2, 2, 0), (1, 3, 0), (1, 1, 2), etc
In how many ways can you give 4 reports to only 1 secretary? You choose the secretary who will get the reports in 3C1 = 3 ways
In how many ways can you give 4 reports to only 2 secretaries? Choose the 2 secretaries in 3C2 ways. Each secretary must get at least one report so you can distribute them in 2 ways:
'1 and 3' - Choose one report in 4C1 ways and give it to one secretary in 2C1 ways. The other secretary gets the other 3 reports.
or
'2 and 2' - Choose 2 reports in 4C2 ways and give to the first secretary. The other secretary gets the other two reports.
Total number of ways is 3C2 * (4C1*2C1 + 4C2) = 42 ways
Note that in your method 2^4 is incorrect. It includes the ways in which all 4 reports go to one secretary.
So this means that in 3 + 42 = 45 ways, at least one secretary gets no report. In the rest of 81 - 45 = 36 ways, each secretary gets at least one report.
Required Probability = 36/81 = 12/27
Quote:
I was wondering in the '1 and 3' choice - why isn't it like this - 4C1*3C1*3C3*2C1 (One out of 4 reports, One out of 3 Supervisors, and 3 out of 3 reports and One out of 2 supervisors)
and in the '2 and 2' choice - 4C2 * 3C1 * 2C2 * 2C1 (Two out of 4 reports, one out of 3 supervisors and 2 out of 2 reports, one out of 2 supervisors).
It would be 60, which is wrong. I'd just like to understand why.
The only error here is highlighted. You do not pick one of the two secretaries.
Say you have 3 Secys - S1, S2, S3
When you pick 3C1 and then 2C1, you are arranging them as first and second. But you do not need to do that.
Just 3C1 (or 3C2 which is the same) is enough. It gives you 2 secys.
Say you get S1 and S2. Each will get 2 reports. When you choose 2 of the 4 reports to give to S1, say you choose R1 and R2 to give to S1, it gives all cases in which the reports can given to the 2 secys. This is so because there will be a case when R3 and R4 are given to S1 and R1 and R2 are given to S2.
_________________
Karishma
Veritas Prep GMAT Instructor
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