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Thread: help a workin girl out~please? =( math

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    Default help a workin girl out~please? =( math

    i dont know if this is appropriate or not lol..but its worth a shot right? =/

    well i have a math test on monday and i am like CONFUSED and starting to panic. im already not doing well in this class and desperately need a good grade...i know i should have gone to see a school tutor earlier but was so busy with work and all...so i come to you guys =)

    the class im taking is finite math and right now we're on probability..
    can someone please, if they could, help explain how to work through these problems.

    1. Suppose 55% of the students at College own an iPod, 69% own a cell phone, and 35% own both. What is the probability a randomly selected student owns neither item?


    2. There are 6 freshmen, 6 sophomores, 8 juniors, and 7 seniors from which to choose a committee of 4 members. What is the probability the committee contains at most 3 freshmen?



    if anyone... please.. =( please..thank you..

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    Default Re: help a workin girl out~please? =( math

    I can't help, but if you post the questions on yahoo answers, you'll probably get more and faster replies.

    Feature costumes for sale!

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    Default Re: help a workin girl out~please? =( math

    Quote Originally Posted by xfatrabbitx View Post
    1. Suppose 55% of the students at College own an iPod, 69% own a cell phone, and 35% own both. What is the probability a randomly selected student owns neither item?
    It often helps to think of an example with 100 students.
    Then 55 students own an iPod,
    69 students own a cell phone,
    35 students own both.

    So, of the 55 students that own an iPod, 35 own also a cell phone. Thus 20 ( =55-35 ) of the 55 iPod-students own only an iPod and no cell phone.

    Similarly, 34 ( = 69-35 ) of the 69 cell-phone-students own only a cell phone and no iPod.

    Thus of the 100 students,
    35 own an iPod and a cell phone,
    20 own only an iPod (and no cell phone) and
    34 own only a cell phone (and no iPod).

    Thus 11 ( =100-35-20-34 ) students own neither item.
    That's 11%.

    If you like a more mathematical approach, start with:
    Let x be the number of all students.
    Then 0,55 x students own an iPod,
    and so on. (with the same result, of course!)

    Let me remark, that to answer that type of questions, Venn-diagrams are sometimes used (). But I don't like them.

    I'll answer the second question in a minute.
    Last edited by ::maxwell::; 02-24-2008 at 08:16 AM. Reason: mistake by copy and paste

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    Default Re: help a workin girl out~please? =( math

    (Wrong answer)
    Last edited by ::maxwell::; 02-24-2008 at 01:32 PM. Reason: wrong answer

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    Default Re: help a workin girl out~please? =( math

    okay-- I was able to do the first one on my own, but that second one's a doozy.

    I personally still didn't understand how you did it-- wah! maybe I'll try again later.

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    Default Re: help a workin girl out~please? =( math

    Ugh! I hate probability. That's what screwed me up on my last stats test. Sucks sucks sucks.
    " Remember during each test there is some girl in Australia jealous of you who wants to do what you're doing."- Lilithmorrigan

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    Shopping and Fucking? Life is short, but youth is shorter. Ride the wheels off, I say." - FeministStripper

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    Default Re: help a workin girl out~please? =( math

    Holy shit. This reminds me again why I went to beauty school....

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    Default Re: help a workin girl out~please? =( math

    Quote Originally Posted by Pan Dah View Post
    I would interpret "the committee contains at most 3 freshmen" as equivalent to "The number of freshmen on the committee is <= 3" which would then mean "the number of freshmen on the committee != 4".
    You're right.
    Also, I think ~99.91% is the correct answer.

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    Default Re: help a workin girl out~please? =( math

    wow... =X thank you tho...better read over your guys posts like a bajillion times...

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    Default Re: help a workin girl out~please? =( math

    rofl I've taken a few math classes in my time and I can't solve any of these.

    School is so useless

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    Default Re: help a workin girl out~please? =( math

    Ok I've very bored so I gave the 2nd problem a try. I'll probly be wrong but I've never taken a course in probability and suck at maths so give me a break.

    Anyways since you're going to pick 3 times, and it's a total of 27 students with 3 freshmen. you've got something that looks like this:

    6/27=22%
    5/26=19%
    4/24=16%

    the average of those #'s is 19%.

    Now since you're going to pick 3 times, divide that by 3 and you get the final answer: 6%

    So the chance that a group will be composed of 3 freshmen is 6%. a 1% chance of a group containing 3 freshmen seems awful low.

    OP you should get your hands on a solution manual.

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    Default Re: help a workin girl out~please? =( math

    Pan Dah: OK, how about a problem along the lines of a stripclub scenario. Lets say that 3 different customers walk into club at the same time. The customers historical tendency to buy a lapdance within 10 minutes of entering a club are as follows: Cust A: 1 out of 20 times Cust B: 2 out of 20 times Cust C: 4 out of 20 times.

    Question #1- What are the chances that dancer walking up (within 1st 10 minutes )to 1st random of 3 customers will get a "Yes" answer to buying a lapdance?

    #2- What would likelihood of at least 1"Yes" answer be if 2 different dancers were to walk up to only 2 of 3 random customers.

    #3- What would be likelihood of 3 different dancers walking up to "their" open man, and at least 1 customer says "yes"? How about percentages of "all 3 " custies saying "Yes" on initial approach? My guess for last part is 0.1%(1 in 1000).

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    Default Re: help a workin girl out~please? =( math

    ok, let's give it a shot. i be edumacated!!! sort of...

    the first problem.
    35% of students have both items. 55% have an ipod. 69% have a cell.
    let's keep it simple and have the number of students be 100.
    so out of the 55 and 69 of who owns each item, 35 students own both.
    so we take 55-35= 20 students have only an ipod
    69-35= 34 students have only a cell
    34+20= 54 students who have either a cell or ipod
    54+35 (# of students who have both)= 89 students
    100-89= 11 students who have neither item

    ok, for the second problem. we have a total of 27 students and 4 positions. how many TOTAL combinations are there? i know that to find the total number of combinations, you would take 27 to the 4th power. the answer for that is 531441 total combinations.
    27^4= 531441

    let's break this problem down here. it says 'AT MOST 3 freshman'. so i'm going to include the probability of having 1, 2 or 3 freshman.

    let's see how many combinations we have with 1 freshman.
    i took 21 (minus the 6 freshman) and cubed it (the 3rd power), giving me 9261 combinations.
    now, we have 6 freshman for that one position, giving me 6 different combinations of this (6 to the 1st power). so i multiplied 6 by 9261 and got 55566 combinations.
    21^3=9261
    9261 x 6= 55566

    next step, let's see the number of combinations with 2 freshman.
    i took 21 and squared it (2nd power), giving me 441 possible combinations.
    now, we have 6 freshman left over for those 2 spots, so i squared 6 which is 36 for the number of combinations for those two spots.
    i multiplied 441 by 36 and got 15876 for the total amount of combinations with 2 freshman.
    21^2= 441
    6^2= 36
    441 x 36= 15876

    let's see the number of combinations with 3 freshman.
    well, there's only one spot left for those 21 other students, so the answer is 21. we have 3 spots for the 6 freshman. so i took 6 and cubed it, giving me 216 combos for those 3 freshman. i multiplied 216 and 21 and got 4536 combinations with 3 freshman.
    6^3= 216
    216 x 21= 4536

    so back to the first part, we have a total of 531441 combinations. when we add up all the different combinations of having 1, 2 and 3 freshman in that position, we get 75978 combinations. to find the percentage of 75978 out of 531441, we divide.
    55566 + 15876 + 4536= 75978
    75978/531441= .14297

    this means there is a 14.297% chance of the commitee having AT THE MOST 3 freshman.

    i have a kinda crappy memory, but i think i got it right?

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    Default Re: help a workin girl out~please? =( math

    Quote Originally Posted by scarymary77 View Post
    ok, let's give it a shot. i be edumacated!!! sort of...

    the first problem.
    35% of students have both items. 55% have an ipod. 69% have a cell.
    let's keep it simple and have the number of students be 100.
    so out of the 55 and 69 of who owns each item, 35 students own both.
    so we take 55-35= 20 students have only an ipod
    69-35= 34 students have only a cell
    34+20= 54 students who have either a cell or ipod
    54+35 (# of students who have both)= 89 students
    100-89= 11 students who have neither item

    ok, for the second problem. we have a total of 27 students and 4 positions. how many TOTAL combinations are there? i know that to find the total number of combinations, you would take 27 to the 4th power. the answer for that is 531441 total combinations.
    27^4= 531441

    let's break this problem down here. it says 'AT MOST 3 freshman'. so i'm going to include the probability of having 1, 2 or 3 freshman.

    let's see how many combinations we have with 1 freshman.
    i took 21 (minus the 6 freshman) and cubed it (the 3rd power), giving me 9261 combinations.
    now, we have 6 freshman for that one position, giving me 6 different combinations of this (6 to the 1st power). so i multiplied 6 by 9261 and got 55566 combinations.
    21^3=9261
    9261 x 6= 55566

    next step, let's see the number of combinations with 2 freshman.
    i took 21 and squared it (2nd power), giving me 441 possible combinations.
    now, we have 6 freshman left over for those 2 spots, so i squared 6 which is 36 for the number of combinations for those two spots.
    i multiplied 441 by 36 and got 15876 for the total amount of combinations with 2 freshman.
    21^2= 441
    6^2= 36
    441 x 36= 15876

    let's see the number of combinations with 3 freshman.
    well, there's only one spot left for those 21 other students, so the answer is 21. we have 3 spots for the 6 freshman. so i took 6 and cubed it, giving me 216 combos for those 3 freshman. i multiplied 216 and 21 and got 4536 combinations with 3 freshman.
    6^3= 216
    216 x 21= 4536

    so back to the first part, we have a total of 531441 combinations. when we add up all the different combinations of having 1, 2 and 3 freshman in that position, we get 75978 combinations. to find the percentage of 75978 out of 531441, we divide.
    55566 + 15876 + 4536= 75978
    75978/531441= .14297

    this means there is a 14.297% chance of the commitee having AT THE MOST 3 freshman.

    i have a kinda crappy memory, but i think i got it right?

    *blink* *blink*

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    Default Re: help a workin girl out~please? =( math

    Financial mathematician here

    These answers sound good to me. Of course, the real point is being able to think through all of this, but having the correct answer is a good start. Being able to see why the incorrect ones are wrong would be good too.

    Separating out people with only a phone, only an ipod, both or neither:
    http://en.wikipedia.org/wiki/Venn_diagram

    Choosing 4 people who are all freshmen (choose them one at a time):
    http://www.ma.utexas.edu/users/parker/sampling/repl.htm

    Hmm. Possibly a bunch of books / PDFs with examples around the internet if you google.



    Quote Originally Posted by Pan Dah View Post
    So far I'm with you...

    This is where we diverge.

    I would interpret "the committee contains at most 3 freshmen" as equivalent to "The number of freshmen on the committee is <= 3" which would then mean "the number of freshmen on the committee != 4". Then looking at the complementary event, the probability of 4 freshmen is:
    6/27 x 5/26 x 4/25 x 3/24 = .0008547

    so the probability of 3 or less freshmen = 1 - .0008547 = .9991453 = ~99.91%

    Maybe.
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    Default Re: help a workin girl out~please? =( math

    Just a little insight; no solutions here. The first shows probability with non-independen variables. The second shows independence, since one cannot be a member of more than one class. These both show clearly selection without replacement, that is once a person is selected, s/he is out of the running for being selected again. For the second problem that means you have to run the problem again for everytime a selection takes place. The first one is just a little thinking along the Venn diagram idea.

    By the way, a good book seems to be all I need to solve problems like these. It's been awhile since I took my stat/prob courses, which can get very confusing if not taught well. So don't feel too bad if it doesn't stick with you, just find a good book and work it through. What I learned mostly is about lottery players' incentives.
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    Default Re: help a workin girl out~please? =( math

    Quote Originally Posted by Pan Dah View Post
    I'm sorry maxwell deleted his solution. Even though we now agree he misread the question slightly, he included some good explanations of his steps, and I used those in my revision so my post alone isn't all that helpful.
    Well, I thought I would be too confusing to have such a long, but wrong answer as a first solution to the second problem.

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    Default Re: help a workin girl out~please? =( math

    OKAY! pandah you're correct..on the right path..thank you mary though, for attempting and typing up that detailed explanation... you guys have solved the first one...and i do understand it..so kudos to all of us,....

    the second one is solved like this..

    since we're looking for the probability of AT MOST 3 freshman, we do have to account for 0, 1, 2,3 freshman...

    so without the lengthy explanation...

    (5nCr0 x 21nCr8 + 5nCr1 x 21nCr7 + 5nCr2 x 21nCr6)/ 26 nCr8

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    Default Re: help a workin girl out~please? =( math

    Quote Originally Posted by xfatrabbitx View Post
    so without the lengthy explanation...

    (5nCr0 x 21nCr8 + 5nCr1 x 21nCr7 + 5nCr2 x 21nCr6)/ 26 nCr8
    I don't want to confuse you, but I don't think that's right.
    Perhaps it was only a typo, but if you want to use binomial coeeficients, you're solution should be

    (6nCr0 x 21nCr4 + 6nCr1 x 21nCr3 + 6nCr2 x 21nCr2 + 6nCr3 x 21nCr1)/ 27nCr4=
    .9991453 (in accordance with Pan Dan's result.)

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    Default Re: help a workin girl out~please? =( math

    Quote Originally Posted by ::maxwell:: View Post
    I don't want to confuse you, but I don't think that's right.
    Perhaps it was only a typo, but if you want to use binomial coeeficients, you're solution should be

    (6nCr0 x 21nCr4 + 6nCr1 x 21nCr3 + 6nCr2 x 21nCr2 + 6nCr3 x 21nCr1)/ 27nCr4=
    .9991453 (in accordance with Pan Dan's result.)
    oops you're right.. i didint realize the original problem and just took the answer off my test question..which was similar but not the same numbers.. okay..

    THATS ^ the correct answer =)

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