There are three secretaries who work for four departments.

Question

There are three secretaries who work for four departments. If each of the four departments have one report to be typed out, and the reports are randomly assigned to a secretary, what is the probability that all three secretary are assigned at least one report?

A. 8/9
B. 64/81
C. 4/9
D. 16/81
E. 5/9

Show Answer

Official Answer

C

A. 8/9 
B. 64/81 
C. 4/9 
D. 16/81 
E. 5/9

ps_dahiya · Accepted Answer

Total outcomes = 3^4 = 81

When each secretary is assigned atleast 1 report then two report will be assigned to 1 and 1 report each to remaining two.

Favourable outcomes = (Ways to select two report) * (Ways to select 1 seceratry) * (Ways to assign two report to two secretaries)
= (4C2) * (3C1) * (2!) = 36

Prob = 36/81

KarishmaB · Answer

"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

mahesh004 · Answer

Posting answer ----

For each report there are three possible outcomes: Secretary A, Secretary B, Secretary C 
Total outcome: 3^3=27 
Total outcome that three secretary are assigned at least one report: P(3,3)=3!=6 
Probability = 6/27=2/9

jlgdr · Answer

I got the same answer: Here's my reasoning First Report you have 3 choices Second Report you have 2 choices Third Report you have 1 choice Fourth report 3 choices again Then total number of ways is: 3*2*1*3=3^2*2 Total number of possibiliities = 3^4 So probability = 2/9 Please advice if this is the correct answer Cheers! J

KarishmaB · Answer

Here you are assuming there are 3 reports and 3 secretaries. But you are given that there are 4 reports (from 4 departments). Hence this is not correct.

KarishmaB · Answer

This is not correct. You have assumed that the 4th report must go to someone who already has a report. There is no such constraint. You can easily give the 1st and 2nd reports to secretary 1, 3rd report to secretary 2 and 4th report to secretary 3. But you have ignored all such cases.

The number of ways of ensuring at least one report goes to each secretary is 4C2 (select 2 reports out of 4 which go to the same person)*3C1 (select the person who must type 2 reports)*2! (since you have 2 reports left which you must distribute to the 2 remaining people such that each person gets one) = 36

Required probability = 36/81

JusTLucK04 · Answer

What is wrong with.. Choose 3 out of 4 letters...4C3..assign them to each secretary...4C3 * 3! ways Assign the remaining letter to any secretary: 3 ways = 72/81 :roll:

KarishmaB · Answer

Using your logic,

"Reports R1, R2 and R3 are selected and given to secretaries A, B and C respectively and then report R4 is given to A", is a different case from "reports R2, R3 and R4 are selected and given to secretaries C, B and A respectively and then report R1 is given to A".

But actually both cases are same because A gets R1 and R4, B gets R2 and C gets R3. So there is double counting.

pretzel · Answer

How do we get the total outcomes as  3^4 = 81 ?

Bunuel · Answer

Each of the four reports can be assigned to any of the three secretaries, so each report has 3 options, total = 3*3*3*3.

simba19 · Answer

Can someone explain why we must select the person who must type two reports (3C1)? That feels to me like double-counting, because in my mind, I feel like we already selected that person with 4C2. I'd like to know to understand completely so I don't continue to make mistakes on these kinds of problems. Thanks!

KarishmaB · Answer

So here is the sequence of events you must follow:
Pick two reports out of 4 (R1, R2, R3, R4). You can do this in 4C2 ways such as (R1, R2) or (R1, R3) or (R2, R4) etc
Now go to the secretaries and select one secy from the 3 and hand over the two reports to her. You can do this in 3C1 ways such as A or B or C.
Now you have 2 reports and 2 secys. You can give them the reports in 2! ways.

Lapetiteflo · Answer

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

rajatmehrotra20 · Answer

Departments : A,B,C,D
Secretaries : 1,2,3
As per the ques , one of the secretaries will have to type for 2 depts and the other two for 1 dept each.
Which means 2 dept. will give their report to 1 secy and the other two depts will give their report to the other 2 secys (1 each).
No. of ways to club the depts in the form of 2,1,1 = 6. (AB, C, D) (AC, B,D) (AD, B, C) (BC, A, D) (BD, A, C) (CD, A,B) 
No. of ways of distributing the reports to 3 secys = 3!.
Therefore total no. of favorable cases = 6x3! = 36.
Total possible cases = 3^4 ( each dept can give its report to any of the secys).

Probability = 36/(3^4) = 4/9

Thanks
Rajat

suramya26 · Answer

why cant the total number of possiblities be
4^3?
I have used the following approach for it.
Kindly please tell me where i am wrong
assuming s1,s2,s3 be the secretaries and r1,r2,r3,r4 be the reports
if the reports are o be distributed among secretaries then
s1 can have 4 options of reports
similiarly s2,s3,s4 all will have 4 options
hence according to me 4x4x4=4^3=64 should be total ways
also please let me know the main concept of n^r takes place...

KarishmaB · Answer

This concept cannot be used here.

Say there are 4 report options for s1.
s1 gets r1.

Now what about s2? Are there still 4 report options for her?

Also, each report has to be allotted to a secretary. Each secretary needn't be allotted a report. It is possible that s1 gets no report or gets 2 or 3 or even all 4 (when considering total number of cases).

foryearss · Answer

Dear  VeritasKarishma  , I am really confused ,can you please let me know where's my mistake : 
total ways : 3^4
now , let's choose 3 reports in 4C3 and give them to the three secretaries in 3! ways : 
we are left with one report : it hase 3 choices to go (each for a secretary)
so the wanted ways : 4C3 * 3! * 3 = 4 *3*2* 3 
so I'm getting 8/9

in another approach : I'm getting the correct answer : 
let's choose 2 reports 4C2 and give them to one secretary *3 , then the 2 left reports will go to the 2 left secretaries in 2 ways 
so I get : 4C2 *3 *2 =2 *3 *3 *2
so p = 4/9

can you please let me know why my first is wrong
Thanks

KarishmaB · Answer

There is double counting in your method 1:
Say there are 4 reports R1, R2, R3, R4 and 3 secys S1, S2, S3

You select 3 of the 4 reports (R1, R2 and R3) and give them to S1, S2 and S3 respectively. Then you give R4 to S1. 
or
You select 3 of the 4 reports (R4, R2 and R3) and give them to S1, S2 and S3 respectively. Then you give R1 to S1.

In both cases, S1 has R1 and R4, S2 has R2 and S3 has R3. But you counted them as two different cases. Hence this approach is not correct.

rainingrelics · Answer

Hello Ma'am,

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).

Upon adding the possibilities the answer would be 60, which is wrong. I'd just like to understand why.

There are three secretaries who work for four departments.

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There are three secretaries who work for four departments.

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