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possibility problem

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by vikram4689 » Thu Jun 23, 2011 7:25 pm
Options for companies to get writers task = 3*3*2*1

Total options for selecting writers = 3*3*3*3

Prob = 3*3*2*1*/3*3*3*3 = 2/9.

What the OA ?
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by holyxie » Thu Jun 23, 2011 7:47 pm
the OA is 4*3*2*3/3^4
so is 8/9
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by vikram4689 » Thu Jun 23, 2011 8:43 pm
Well how come OA is thinking 3*3 in 4*3*2*3.

For the 1st task a company has 3 options.
For the 2st task a company has 2 options.
For the 2st task a company has 1 option.
For the 2st task a company has 3 options. (because out condition is fulfilled and any one get this task).

Any explanation given by OA source
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by holyxie » Thu Jun 23, 2011 10:20 pm
it only says to separate the tasks to three piles,which is 2, 1, 1
then the formula looks like this C(4,2)*C(2,1)*C(3,1)/3^4.
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by sansky » Thu Jun 23, 2011 11:14 pm
Consider the problem this way....As every writer has to get atleast one job, then you will have one writer with 2 jobs..So u have 2,1,1 job situation.

The no. of ways the first writer can get the job is C(4,2)
The no. of ways the second writer can get one of the remaining two jobs is C(2,1)
There is only one way the third writer can get the remaining job..

So your numerator is C(4,2)* c(2,1)*1

But you have have three different ways in which the scenario of 2,1,1 can occur...So the numerator multiplied by 3 will give the total number of favorable outcomes...and 3^4 is the total no of outcomes...

Hence the OA solution of 2*3*2*3/3^4
Last edited by sansky on Sun Jul 03, 2011 9:15 pm, edited 1 time in total.
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by Frankenstein » Thu Jun 23, 2011 11:26 pm
Hi,
What is C(4,2) according to you guys?
Is it 4!/2!.2! or not?
If it is so, how is C(4,2)*C(2,1)*C(3,1) = 4*3*2*3 ?
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by newgmattest » Fri Jun 24, 2011 12:07 am
As every writer has to get at least one job, then you will have one writer with 2 jobs..So u have 2,1,1 job situation.

The no. of ways the first writer can get the job is C(4,2).
The no. of ways the second writer can get one of the remaining two jobs is C(2,1).
There is only one way the third writer can get the remaining job.

So your numerator is C(4,2)* C(2,1)*1

But you have have three different ways in which the scenario of 2,1,1 can occur.So the numerator multiplied by 3. also note for that 2 job, you can interchange in 2 ways, so multiply by 2 as well. Total number of favorable outcomes = 2*3*C(4, 2)*C(2, 1)*C(1, 1).
Total no of outcomes = 3 ^ 4

P = 2*3*C(4, 2)*C(2, 1)*C(1, 1)/(3^4) = 8/9
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by Jim@Knewton » Fri Jun 24, 2011 1:59 am
Very interesting discussion and you guys have provoked me to add my two cents here:

Q: 3 writers work for 4 different companies, every company has one task. What is the possibility for every write to get one task?

Let the 4 companies (tasks) be a,b,c & d

Let the 3 writers be w1, w2 and w3

Total: Then total possible ways = (a can be given to w1 / w2 / w3 = 3 ways)*(b = 3 ways)*(c= 3 ways)*(d=3 ways)

=> Total number of possibilities = 3*3*3*3 = 3^4 = 81

Ways to distribute number of tasks amongst w1, w2, w3:
1. 4,0,0 (4 tasks to w1/ w2 / w3 and 0 to the other two...)
2. 3,1,0
3. 2,2,0
4. 2,1,1

We are interested in situation 4

For a = 3 possibilities, b = 2 and c= 1 and since the d can go to any of the writers, hence task d = 3 possibilities

=> Number of possible ways for situation 4. counted so far = 3*2*1*3 = 18

BUT there could be 4 different tasks for the 4th task (instead of 4th: d= 3 possibilities, it could also be 4th = a or b or c as the 4th task)

=> Advanced Count = (3*2*1*3)*4 = 72

BUT this 72 contains a series of double counts ( ad = da and so on, in the "2" of the 2,1,1 sequence)

=> Final Count = 72/ 2 = 36

[ If you understand this logic, you can directly use the shorter combination formulae: Count = 4c2*2c1*3c1 = 36 and many other ways (as is the case with a lot of probability questions), depending on how you want to count it!]

Hence Probability = 36/81 = 4/9


Similarly probability for all distributions:
1. 4,0,0 = 3/81 = 1/27 (Note: This means 4 tasks to anyone of the w's (not just w1) and 0 to any of the other 2)
2. 3,1,0 = 24/81 = 8/27
3. 2,2,0 = 18/81 = 2/9
4. 2,1,1 = 36/81 = 4/9

It's been 2 days since I slept (one of the life rewards for getting an MBA!), so pardon the typos and any other deviations/errors you may find here :-)

I hope that we are all enjoying the odds in the possibilities of probability!!!
Best, Jim
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by Frankenstein » Fri Jun 24, 2011 3:42 am
Hi,
Good work, Jim. Most of the previous posters were getting the logic right and writing C(4,2)*C(2,1)*C(3,1). But because of temptation of the answer everyone was making it 4*3*2*3.
That is the reason for my earlier post of how C(4,2)*C(2,1)*C(3,1) = 4*3*2*3 ?

Can the OP post the source of this question?
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by holyxie » Fri Jun 24, 2011 7:01 am
Yeah but the source of the question is from a mathematics book i got from China. I dont think it is gonna be helpful for you guys cuz the whole book is written in Chinese. But i can share some good questions with you guys in the future.
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by vikram4689 » Fri Jun 24, 2011 6:02 pm
Cant get this part.
BUT there could be 4 different tasks for the 4th task (instead of 4th: d= 3 possibilities, it could also be 4th = a or b or c as the 4th task)

=> Advanced Count = (3*2*1*3)*4 = 72

BUT this 72 contains a series of double counts ( ad = da and so on, in the "2" of the 2,1,1 sequence)

=> Final Count = 72/ 2 = 36
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by Jim@Knewton » Fri Jun 24, 2011 6:19 pm
Sure Vikram,
I'll try to elaborate:
In addition to what is mentioned in my post above:

if the 2,1,1 arrangement has "d" as the fourth task, then one of the assignments could look like this:
w1...w2...w3
a.....b....c
d
and when "a" is the fourth task (implicit because, in the present method, we multiply by 4), then another assignment could look like this:
w1...w2...w3
d.....b....c
a

In both case "d,a" are assigned to w1 but counted twice in the total of 72. This is the case with the other combination pairs as well - would be too laborious to list them out but once you understand the mechanism for one pair we should be able to extend it to others conceptually. Since order does not matter (d,a = a,d), we divide by 2 to find the unique number of combinations.

And, like I said before - many ways to solve probability questions, so you probably have your own mechanisms ....but the final answers are objective :-) and should therefore match!
Hope this helps...
Best, Jim
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by GMATGuruNY » Sat Jun 25, 2011 2:37 pm
holyxie wrote:3 writers work for 4 different companies, every company has one task. what is the possibility for every writer to get one task?
I received a PM asking me to comment.
Let's call the 4 tasks A, B, C and D.

First we need to count how many ways the tasks can be assigned.
Since each task can be assigned to any one of the 3 writers:
Number of choices for A = 3.
Number of choices for B = 3.
Number of choices for C = 3.
Number of choices for D = 3.
Multiplying, we get:
3*3*3*3 = 81 possible assignments.

Now we need to determine how many ways the tasks can be assigned so that every writer is assigned at least 1 task.
For every writer to be assigned at least 1 task, one writer must be assigned a pair of tasks, while the other 2 writers are each assigned 1 task.

Case 1:
One writer is assigned AB, one writer is assigned C, and one writer is assigned D.
Number of ways to arrange the 3 elements AB, C and D = 3! = 6.

Remaining cases:
In the arrangements above, AB can be replaced with any pair of tasks.
Thus, the result above (6) needs to be multiplied by the number of pairs that can replace AB.
Number of pairs that can be made from the 4 tasks = 4C2 = 6.

Multiplying the two results, we get:
Number of ways to assign the tasks so that every writer is assigned at least 1 task = 6*6 = 36.

P(every writer gets at least one task) = 36/81 = 4/9.
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by vikram4689 » Sat Jun 25, 2011 7:16 pm
Hi Mitch,

newgmattest's approach seems to be similar to your approach. But the answers are different. can you please explain this
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