Pls Help me solve this Math (Probability) problem

[vulcanman]

Of

employees,

are to be assigned an office and

are to be assigned a cubicle. If

of the employees are men and

are women, and if those assigned an office are to be chosen at random, what is the probability that the offices will be assigned to

of the men and

of the women?

Its been a while since I did Probability Theory ... trying to keep pace with my High School kid

So here is how I solved it .. (I know its wrong ... but I need someone to tell me why).

For starters ... it does not matter which particular man or woman is selected.

Given that ...
We fill the first office with a man ... probability? 3/5
Now we fill the second office with a man ... probability 2/4
Now we fill the third office with woman ... probability 2/3

Combined Probability: 3/5 * 2/4 * 2/3 = 0.2 (Edited Post for correction)

Anybody ?

 

[busydude]

Combined Probability: 3/5 + 2/4 + 2/3 = 1.766

How can probability be more than 1?

 

[vulcanman]

How can probability be more than 1?

Oops ... I meant to multiply.

 

[crashtestdummy]

Probability should be 2/3.

Guaranteed to have at least one man in an office. That leaves two spots, with two men and two women. Regardless of who fills the next spot, there will be two of the other gender and one of the same gender left.

 

[Jaepheth]

Outcome A = All offices are assigned to men
Outcome B = 2 men and 1 woman receive an office
Outcome C = 1 man and 2 women receive offices


What is the probability of outcome B
3/5*2/4*2/3+2/5*3/4*2/3+3/5*2/4*2/3
= 0.6*0.5*0.6666 + 0.4*0.75*0.6666 + 0.6*0.5*0.6666
=0.2+0.2+0.2

= 0.6

= 3/5


Crashtestdummy's answer differs from mine by 1/5 because his decision tree removes paths F→F→M (2/5*1/4 = 1/10) and F→M→F (2/5*3/4*1/3 = 1/10). That is to say; if the person handing out offices decides that since 1 man has to receive an office they might as well assign a man an office first; it removes two possible ways in which 2 offices can be assigned to the women and thus effects the end probability. In other words; that decision tree means one of the offices is effectively a male only office that neither woman has a shot at getting. Hope that makes sense.

 

[vulcanman]

Outcome A = All offices are assigned to men
Outcome B = 2 men and 1 woman receive an office
Outcome C = 1 man and 2 women receive offices


What is the probability of outcome B
3/5*2/4*2/3+2/5*3/4*2/3+3/5*2/4*2/3
= 0.6*0.5*0.6666 + 0.4*0.75*0.6666 + 0.6*0.5*0.6666
=0.2+0.2+0.2

= 0.6

= 3/5

FANTASTIC ! So the error in solution was that I only considered one scenario of selection ... you considered all three possibilities of filling those seats.

But here is my counter question - What would be the probability of one of the offices being occupied by a man?

 

[GotIssues]

FANTASTIC ! So the error in solution was that I only considered one scenario of selection ... you considered all three possibilities of filling those seats.

But here is my counter question - What would be the probability of one of the offices being occupied by a man?

Exactly 1? That would be 2/5 * 1/4 * 3/3 + 3/5 * 1/2 * 1/3 = 20%

 

[crashtestdummy]

Crashtestdummy's answer differs from mine by 1/5 because his decision tree removes paths F→F→M (2/5*1/4 = 1/10) and F→M→F (2/5*3/4*1/3 = 1/10). That is to say; if the person handing out offices decides that since 1 man has to receive an office they might as well assign a man an office first; it removes two possible ways in which 2 offices can be assigned to the women and thus effects the end probability. In other words; that decision tree means one of the offices is effectively a male only office that neither woman has a shot at getting. Hope that makes sense.

Yup, you're right. It was a lazy shortcut on my part, though it actually only differs by 1/15.

 

[Jaepheth]

But here is my counter question - What would be the probability of one of the offices being occupied by a man?

Your possible outcomes are:
MMM
MMF
MFM
MFF
FMM
FMF
FFM

Where a genders position indicates office I, II, and III respectively.

For office I:
MMM
MMF
MFM
MFF
Which can be simplified to "What is the probability the first office goes to a man?" (MXX)
3/5 = 0.6

Similarly, for office II:
MMM
MMF
FMF
FMM

3/5*2/4 + 2/5*3/4 = 0.6
(We only have to calculate to the second node because it doesn't matter who gets office III)

For office III:
MMM
MFM
FMM
FFM
3/5*2/4*1/3+3/5*2/4*2/3+2/5*3/4*2/3+2/5*1/4*1 = 0.6


So for all three offices the probability of being assigned a male occupant is 0.6

Yup, you're right. It was a lazy shortcut on my part, though it actually only differs by 1/15.

Well that's embarrassing

 

[Fayd]

Of

employees,

are to be assigned an office and

are to be assigned a cubicle. If

of the employees are men and

are women, and if those assigned an office are to be chosen at random, what is the probability that the offices will be assigned to

of the men and

of the women?

Its been a while since I did Probability Theory ... trying to keep pace with my High School kid

So here is how I solved it .. (I know its wrong ... but I need someone to tell me why).

For starters ... it does not matter which particular man or woman is selected.

Given that ...
We fill the first office with a man ... probability? 3/5
Now we fill the second office with a man ... probability 2/4
Now we fill the third office with woman ... probability 2/3

Combined Probability: 3/5 * 2/4 * 2/3 = 0.2 (Edited Post for correction)

Anybody ?

it's a hypergeometric probability problem.

(3 c 2)(2 c 1)/(5 c 3) = 6/10.

jaepheth, you put way too much effort into this.

 

[Jaepheth]

jaepheth, you put way too much effort into this.

Probably.

 

[blackdogdeek]

for a more brute force solution, these are the 10 possible outcomes:

M1M2M3
M1M2F1
M1M3F1
M2M3F1
M1M2F2
M1M3F2
M2M3F2
M1F1F2
M2F1F2
M3F1F2

Only 6 satisfy your criteria.

 

[MotionMan]

How can probability be more than 1?

I was helping my daughter (and wife) with a math problem over the phone. It had to do with fractions (5th grade math). It took me about 10 minutes just to convince them that the fraction of the pole that was painted green could not be 7/5.

LOL.

MotionMan