probability

mom · 2013-06-10 02:44:25

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.

SteveB · 2013-06-10 03:12:52

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)

mom · 2013-06-10 03:35:10

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.

bobbym · 2013-06-10 03:56:57

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?

mom · 2013-06-10 04:27:26

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

bobbym · 2013-06-10 04:31:46

3 / 13 is correct. Very good!

mom · 2013-06-10 04:35:05

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.

bobbym · 2013-06-10 04:50:36

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!

Bob · 2013-06-10 05:47:58

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. ]

Bob

anonimnystefy · 2013-06-10 06:29:35

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

mom · 2013-06-10 12:44:33

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.

bobbym · 2013-06-10 13:48:06

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?

mom · 2013-06-10 14:53:49

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!!!

bobbym · 2013-06-10 15:02:30

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.

mom · 2013-06-10 15:10:15

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

thanks again

OzMark · 2013-06-11 15:19:22

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)

Bob · 2013-06-11 19:04:31

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

bobbym · 2013-06-11 21:08:24

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

OzMark · 2013-06-11 21:20:12

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)

Bob · 2013-06-12 00:07:13

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

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

Math Is Fun Forum

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

probability

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

Re: probability

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

Re: probability

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

Re: probability

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

Re: probability

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

Re: probability

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

Re: probability

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

Re: probability

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

Re: probability

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

Re: probability

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

Re: probability

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

Re: probability

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

Re: probability

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

Re: probability

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

Re: probability

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

Re: probability

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

Re: probability

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

Re: probability

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

Re: probability

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

Re: probability

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