Tough probability

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punitkaur
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Topic: Tough probability
PostThu Nov 05, 2009 4:14 pm
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There are three secretaries who work for four departments. If each of the four departments has one report to be typed out, and the reports are randomly assigned to a secretary, what is the probability that all three secretaries are assigned at least one report?

Ans is 8/9

I have a basic doubt here. When calculating total outcomes or ways in which 4 reports can be assigned to 3 secretaries... how to decide whether it should be 3^4 or
4 x 3 x 2

for latter method - 1st secretary can be assigned a report in 4 ways, second can be assigned 3 ways and 3rd in 2 ways.

I know I am going wrong somewhere and need someone to clarify this doubt of mine.
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grockit_jake
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PostFri Nov 06, 2009 1:00 pm
Since each report is assigned randomly, you can look at each report as its own probability:

Since there are 3 options of where each report can do and 4 total reports, your total number of options is 3*3*3*3 = 3^4.

I would look at the term "at least 1" as "NOT Zero" or "1 - ZERO"

The probability of secretary A not receiving an individual report is 2/3. So the chances of NOT receiving any of the 4 reports (since they are assigned independantly) is (2/3)^4.
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PostFri Nov 06, 2009 11:37 pm
grockit_jake wrote:
Since each report is assigned randomly, you can look at each report as its own probability:

Since there are 3 options of where each report can do and 4 total reports, your total number of options is 3*3*3*3 = 3^4.

I would look at the term "at least 1" as "NOT Zero" or "1 - ZERO"

The probability of secretary A not receiving an individual report is 2/3. So the chances of NOT receiving any of the 4 reports (since they are assigned independantly) is (2/3)^4.
I am not clear on how this would lead to 8/9 as the ans... can you please explain?

also am not clear on where am going wrong with the approach of using 1- probability of any one of the secretaries getting no reports:

p(no reports)= (2/3)^4
P(Sec 1 gets no reports) or P(Sec 2 gets no reports)or P(Sec 1 gets no reports) = 3* ((2/3)^4)= 16/27

p(at least one report) = 1-16/27 = 11/27.
Where am I going wrong?

really appreciate your help.
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palvarez
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PostSat Nov 07, 2009 12:48 am
Let X, Y, Z be 3 secretaries.

X + Y + Z = 4

How many integer solutions does this equation have?

6c2 = 15

The question is asking to find the number of solutions such that X, Y, Z >= 1.

In this case, find all int solutions for X + Y + Z = 1 (after taking 1 out of each secy).

solutions = 3c2 = 3.

Answer is: 3/15 = 1/5
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PostSat Nov 07, 2009 1:01 am
I think 3^4 doesn't cut it.

3^4 does work when repetition and order are important like in picking up a code on a combination lock or picking a number in n-ary number system.

In this scenario, there is no repetition (once ya assign a letter to a particular secretary, you aint gonna assign the same to another secretary).

Order matters, because assigning dept1's letter to secy A is not same as dept1's letter to secy B.
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PostMon Nov 09, 2009 6:20 am
Hi palvarez,

Thanks for the explaination. I am unable to understand it.

How can we derive the answer with this formular:

P(reports) = 1 - P (no reports)..

Your help is much appreciated...

Regards,
Viju
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PostMon Nov 09, 2009 12:51 pm
viju9162 wrote:
Hi palvarez,

How can we derive the answer with this formular:

P(reports) = 1 - P (no reports)..

Your help is much appreciated...

Regards,
Viju
If we are looking for a minimum of 1 report going to each secretary than the total of what you want to happen + what you dont want to happen = 1.

With this equation finding the probability of no reports is critical because then when you subtract it from 1 you are left with AT LEAST 1 report or more being assigned which is what it is looking for.

There are always more than one way to answer these questions but this is going to be your quickest way.
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PostMon Nov 09, 2009 6:38 pm
would be helpful to have an instructor (or anyone else) give a detailed explanation with the final answer - still seems to be a lot of confusion over this question.

Thank you!
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achar_varun
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PostThu Dec 03, 2009 9:21 am
@punitkaur:

what you're getting wrong here is that you are assuming that if the 1st secretary gets a report he/she will not get another report... the question does not say that... it says that the reports are distributed among the 3 secretaries. It could be that 1st one got all 4 or she got none.

So,

Number of ways 1st report can be distributed among 3 secs =3
------------2nd report ditto--------------------------------------------=3
similarly for 3rd and 4th report.

So, total ways = 3*3*3*3.

Hope this clears the confusion.
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PostFri Dec 04, 2009 2:12 am
I have been staring at this problem for sometime and the answer I get is 4/9..

This is not an attempt to solve it using the simplest way..Now I am curious and just want to get the right answer!!!
Lets say secretaries are called a,b,c. no. of ways of distributing the letters are:

a b c
------
4 0 0
0 4 4 set 1
4 0 4
-------
1 3 0
0 1 3 set 2
0 3 1
1 0 3
3 1 0
3 0 1
-------
1 2 1
1 1 2 set 3
2 1 1
------
2 2 0
2 0 2 set 4
0 2 2

no. of ways for set 1=3
no. of ways for set 2= 6* 4C3=24
no. of ways for set 3 = 3*4C2*2=36( because there are two ways of distributing the single 2 letter(s) after 2 are picked)
no. of ways for set 4=3*4C2=18

Probability = set3/(set1+set2..+set4)=36/81=4/9.
What am I missing? Help please!
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PostFri Dec 11, 2009 12:39 am
Got it!

ok you guys are have right check this out

total possible ways to distribute the papers are 3*3*3*3 or 81

so (81-x)/81 when x is equal to the total ways that at least one secretary will not get a paper

so lets list them for you. but remember order doesn't matter

4 0 0
3 1 0
3 0 1
2 2 0
2 0 2
1 3 0
1 0 3
1 2 1
0 3 1

total = 9 possibilities that at least one doesn't get a paper

81-9 = 72

72/81 = 8/9
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PostSat Dec 19, 2009 9:58 pm
Where did you get this problem from ?
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PostThu Dec 24, 2009 7:27 pm
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
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PostSun Jan 03, 2010 1:04 pm
Gazpt got it right...another way to look at it is:

1 - (probability not all 3 secretaries are assigned one report in 4 departments)

- So, we are looking for either 1 of them being assigned reports for all 4 departments (That will be 3 choices)
- or 1 of them assigned a report for 2 departments and another assigned reports for the other 2 departments (3 choices)
- or 1 gets reports assigned for 3 departments and another assigned reports for 1 department (3 choices again)

Order does not matter. You can draw the sample sample and use S1, S2, S3 and try it out.

S1 S1 S1 S1
S2 S2 S2 S2
S3 S3 S3 S3
S1 S1 S1 S2
S2 S2 S2 S3
S3 S3 S3 S1
S1 S2 S1 S2
S2 S3 S2 S3
S3 S1 S3 S1

Our total number of outcomes are 3C1 * 3C1 * 3C1 * 3C1 which is 3* 3 * 3 * 3 = 81 (Here we are just bothered that 1 out of the 3 gets a report in each department)

So that will give you 1 - (9/81) = 8/9

Hope this helps !!
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PostTue Jan 05, 2010 12:52 pm
Hi,

I am quite sure that 4/9 is the right result.

@sameer568: To my mind you are wrong. Indeed there is only one way of getting to the result S1 S1 S1 S1 (i.e. the first secretary is assigned to all four reports), however there are actually 4 ways of getting to the result S1 S1 S1 S2. Even though order doesn't matter for the result, it does matter for calculating the probabilities. Intuitively that comes to light when you consider that the second secretary could be assigned to the 1st, 2nd, 3rd or 4th report.

--> When you flip a coin 2 times and you are only interested in the overall result (H = Head, T = Tail), i.e. either TT or HH or HT, you won't assign a probability of 1/3rd for each of the events, but rather 1/4 respectively for each event, namely TT, HH, HT and TH, however, ultimately HT equals TH, so that this event has the probability of 1/2.

In the secretary case, in my mind you should approach the problem using the multinomial coefficient.
(E.g. English Wikipedia: http://en.wikipedia.org/wiki/Multinomial_theorem
or German: http://de.wikipedia.org/wiki/Multinomialkoeffizient).

So what you do is: Total number of assignment orders are: 3^4 = 81, because every time a report is assigned (4 times), there are 3 secretaries which can be chosen.

Now we are searching for the possibility that one of the secretaries gets assigned to 2 reports, the other two to 1 report respectively. I.e. we want to know how many alternatives there are to split a set of the cardinality 4 (number of reports) into three sets of the cardinality 2, 1, and 1 (numbers of assigned reports per secretary) respectively. Using the concept of the multinomial coefficient (actually an expansion of the binomialcoefficient concept) you get the following result.

First, there are 3 alternatives of splitting the set of cardinality 3 into 2 subsets of the cardinalities 2 and 1, namely one secretary with 2 reports (the set with cardinality 1 because it contains 1 secretary) - and two secretaries with 1 report each, comprised by the set with cardinality 2 (because it contains 2 secretaries). The three alternatives are: (2,1,1) , (1,2,1) and (1,1,2), which can also be calculated using the multinomial coefficient 3C(2, 1) = 3!/(2!*1!) = 3

Now we calculate the number of alternatives for one of these events as a result, e.g. for (2, 1, 1):

--> There are 12 alternatives of splitting a set of cardinality 4 into 3 subsets of the cardinalities 2, 1 and 1 respectively, so that the first secretary gets 2 reports, the other secretaries 1 each (i.e. event (2,1,1)):

--> 4C(2, 1, 1) = 4!/(2!*1!*1!) = 12

(Intuition: By dividing by 2!*1!*1! you correct for the number of permutations within the subsets, as these are irrelevant.)

There are also 12 alternatives respectively for the events (1,2,1) and (1,1,2). So there are 36 alternatives of 81 that fulfill the condition that each secretary is assigned at least 1 report.

The possibility for the event is thus: 36/81 = 4/9


It's hard to explain that concept this way, but I think it is right. Important to understand is that even though order doesn't matter for the result it matters for calculating the probabilities!
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