Tough probability

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punitkaur
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Topic: Tough probability
PostThu Nov 05, 2009 3:14 pm Reply with quote
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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PostFri Nov 06, 2009 12:00 pm Reply with quote
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 10:37 pm Reply with quote
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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PostFri Nov 06, 2009 11:48 pm Reply with quote
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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palvarez
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PostSat Nov 07, 2009 12:01 am Reply with quote
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 5:20 am Reply with quote
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 11:51 am Reply with quote
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 5:38 pm Reply with quote
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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