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#1 2013-06-10 02:44:25

mom
Member
Registered: 2012-04-25
Posts: 94

probability

I am completely lost and I need to know how to work out these problems. I have the following:

Suppose that 9 female and 5 male applicants have been successfully screened for 5 positions. If the 5 positions are filled at random from the 14 finalists, what is the probability of selecting at least 4 females?

The experiment consists of selecting 5 finalists from 14. Since the order in which these 5 finalists are selected is immaterial, the samples are combinations of 14 finalists taken 5 at a time. Find the total number, N, of samples in the sample space.

N=C(14,5)
  =2002

I know that is the answer for this step of the problem but I don't know how 2002 was reached. Please help.

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#2 2013-06-10 03:12:52

SteveB
Member
Registered: 2013-03-07
Posts: 557

Re: probability

mom wrote:

I am completely lost and I need to know how to work out these problems. I have the following:

N=C(14,5)
  =2002

I know that is the answer for this step of the problem but I don't know how 2002 was reached. Please help.

14! / (9! x 5!) = 2002

14! = 14x13x12x11x10x9x8x7x6x5x4x3x2x1
9! = 9x8x7x6x5x4x3x2x1
5! = 5x4x3x2x1

14! / (9! x 5!) = (14x13x12x11x10) / (5x4x3x2x1) = 2002

So we have 14 people and we are choosing 5 of them in any order in terms of the 5 and the remaining 9.
Imagine that we have 14 people and we wish to choose one. There are 14 ways of doing this.
Then choose one of the remaining 13, then one of the remaining 12, and so on down to 10.
There are five reorderings of the five chosen people that we have counted too many times.
The division by 5! is to allow for this. The number obtained is 2002.
Hence there are 2002 ways of choosing 5 people from 14.
(Assuming that the order of the 5 people is not important eg. {1,2,3,4,5} = {5,3,4,1,2} etc.)

Last edited by SteveB (2013-06-10 08:27:34)

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#3 2013-06-10 03:35:10

mom
Member
Registered: 2012-04-25
Posts: 94

Re: probability

thank you so much!! I was doing 5!/14!(5-9)! and it was not coming out. This is difficult for me and I am sure that I will have more questions. Thanks again. At least now, I can figure the sample size.

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#4 2013-06-10 03:56:57

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 85,315

Re: probability

Hi mom;

Suppose that 9 female and 5 male applicants have been successfully screened for 5 positions. If the 5 positions are filled at random from the 14 finalists, what is the probability of selecting at least 4 females?


In mathematics, you don't understand things. You just get used to them.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

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#5 2013-06-10 04:27:26

mom
Member
Registered: 2012-04-25
Posts: 94

Re: probability

Yes!!! Thanks so much for your help. I was able to calculate the probability of 9f, 7m for 5 positions with 16 finalists - the probability of at least 4f. I got it right!!! 3/13
smile

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#6 2013-06-10 04:31:46

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 85,315

Re: probability

3 / 13 is correct. Very good!


In mathematics, you don't understand things. You just get used to them.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

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#7 2013-06-10 04:35:05

mom
Member
Registered: 2012-04-25
Posts: 94

Re: probability

bobby,

thank you. I am quite proud of myself and could not have accomplished it without the help of this site. I love being able to understand what I am doing. smile

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#8 2013-06-10 04:50:36

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 85,315

Re: probability

Hi mom;

I love being able to understand what I am doing.

Understanding is one of those things that is highly overrated. It is much better to be a good guesser!

For instance, behold the great Carnac! When the big brains got stuck they would put the question in an envelope and the great Carnac would attune himself to the answer.

As Yoda would say the force is strong in him!

View Image: carnac-the-magnificent-king-of-late-night-demotivational-poster-1262745528.jpg

In mathematics, you don't understand things. You just get used to them.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

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#9 2013-06-10 05:47:58

bob bundy
Moderator
Registered: 2010-06-20
Posts: 6,336

Re: probability

Why has the Great Carnac got an old fashioned telephone dial for a necklace?  Is it so he can ring for help when the vibes are not good?

[historical note for the young-uns:  A telephone dial was a circular object used to send pulses to the telephone system to establish a connection with another phone.  I have included a picture below.  The difficulty with those old models was getting them in your pocket;  oh yes, and the long trailing wires as you walked down the street.  roflol ]

Bob

View Image: telephone.jpg

You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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#10 2013-06-10 06:29:35

anonimnystefy
Real Member
From: The Foundation
Registered: 2011-05-23
Posts: 15,507

Re: probability

Ah, I remember those. I liked playing with the dial.


“Here lies the reader who will never open this book. He is forever dead.

“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

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#11 2013-06-10 12:44:33

mom
Member
Registered: 2012-04-25
Posts: 94

Re: probability

I don't have to understand everything but when I am receiving a grade on the material, it is nice to know what I am doing and how it works. I honestly could care less about the probability of 3 girls being chosen for a position out of 8 with 7 males and 5 positions with 15 interviews.......

I appreciate the help. smile

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#12 2013-06-10 13:48:06

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 85,315

Re: probability

Hi mom;

I was just joking a little bit. Glad you understand the material.

So no one knows why the great Carnac would be using an old phone?


In mathematics, you don't understand things. You just get used to them.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

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#13 2013-06-10 14:53:49

mom
Member
Registered: 2012-04-25
Posts: 94

Re: probability

I do on that one and I did appreciate the joke haha just stressed out over this math. I posted another question as I am stuck yet again. UGH!!!

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#14 2013-06-10 15:02:30

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 85,315

Re: probability

I know the feeling. Just relax, it will all work out or the planet will explode. Either way there is no real sense in worrying.


In mathematics, you don't understand things. You just get used to them.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

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#15 2013-06-10 15:10:15

mom
Member
Registered: 2012-04-25
Posts: 94

Re: probability

thanks. No worries here. I am moving on till I can get help with that problem.

thanks again

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#16 2013-06-11 15:19:22

OzMark
Member
Registered: 2013-06-10
Posts: 9

Re: probability

Hi

I get 45/143             (bobbym you get 54/143, maybe just a transcription transpose error)

Which equals 0.3147 (bobbym=0.3776)

Ive calculated the entire set of outcomes and summed them to 1.0000, so im pretty sure I'm right.
Sorry I dont use the nPr or nCr functions, I think its too easy to make mistakes with them.
I do probability from first principals, so my terminology is not conventional

I  hope my spreadsheet makes sense, here it is with all the outcomes

imgur.com/DKhMcUu

Oh and here is how I obtained the any order multipliers (dont know what they are really called)

imgur.com/6CtcOsU

Last edited by OzMark (2013-06-11 15:31:16)

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#17 2013-06-11 19:04:31

bob bundy
Moderator
Registered: 2010-06-20
Posts: 6,336

Re: probability

hi OzMark

Mom wrote:

what is the probability of selecting at least 4 females?

I agree with your spreadsheet.  The question asks for 'at least' 4 females so you need to do 45/143 + 9/143 = 54/143.

Bob


You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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#18 2013-06-11 21:08:24

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 85,315

Re: probability

Hi OzMark;

I get 45/143             (bobbym you get 54/143, maybe just a transcription transpose error)

I used the hypergeometric distribution which is for sampling without replacement. The answer is 54 / 143


In mathematics, you don't understand things. You just get used to them.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

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#19 2013-06-11 21:20:12

OzMark
Member
Registered: 2013-06-10
Posts: 9

Re: probability

Hi Bob Bundy,

Thanks, your right, I only had the case : 4 females 1 male,
I should have had the cases: 4 females 1 male and 5 females 0 males
which is 45/143 + 9/143 = 54/143
This is great, Im learning here

So what is the etiquitte now
Should I redo the spreadsheet, or make a second one, or
leave it (which might be the best reflection of this dialogue and give other readers a bit to think about)
?

Again Thanks,
Mark

Last edited by OzMark (2013-06-11 21:25:47)

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#20 2013-06-12 00:07:13

bob bundy
Moderator
Registered: 2010-06-20
Posts: 6,336

Re: probability

hi OzMark

Your spreadsheet numbers are correct so I don't see why it shouldn't stand.  Personally, I try to steer clear of  're-writing history' so, if I've made a mistake I just tell people what I should have said and leave it like that.  I'm afraid there's a few of those dotted around my posts shame 


It's clear to anyone else reading the thread and your way of tackling the problem was interestingly different from what others had done.

When you've been a member for a short while, you'll be able to add images directly into a post.

Bob


You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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