borinka99 0 Posted April 3, 2009 Share Posted April 3, 2009 In a drawer, there are 5 classical CD's and 4 Rock-n-Roll CD's. Armon randomly picks up 3 CD's. What is the probability that he selected 2 classical and 1 Rock-n-Roll CD?The answer is supposedly: 10/21 I'm trying to figure this out, can anyone explain? I tried doing the 5/9*4/8*4/7, but that's not right because it doesn't have to go a specific order, it just has to be 2 classical and 1 RnR. Link to post Share on other sites
grocery_mony 8 Posted April 3, 2009 Share Posted April 3, 2009 math is idiotic Link to post Share on other sites
Yoda 1 Posted April 3, 2009 Share Posted April 3, 2009 on the first draw u have 100% chance of getting a classical or RnR, i'm not solving out the rest, but figured i'd steer you in the right direction Link to post Share on other sites
Quacktastic 106 Posted April 3, 2009 Share Posted April 3, 2009 There are 84 different combinations of 3 CD's possible (9!/(9-3)!*3!)There are 40 different combinations of 2 classical and 1 rnrClassical (5!/(5-2)!*2!) = 10Rnr (4!/(4-1)!*1!) = 4You have 10 "sets" of 2 classical CD's and 4 "sets" of one rock n roll CD. Multiply the 2 together to get 40 different "sets" of the 3 CD's40/84 = 10/21Whoa, I just blacked out.... what happened? Link to post Share on other sites
Tactical Bear 3 Posted April 3, 2009 Share Posted April 3, 2009 In a drawer, there are 5 classical CD's and 4 Rock-n-Roll CD's. Armon randomly picks up 3 CD's. What is the probability that he selected 2 classical and 1 Rock-n-Roll CD?The answer is supposedly: 10/21 I'm trying to figure this out, can anyone explain? I tried doing the 5/9*4/8*4/7, but that's not right because it doesn't have to go a specific order, it just has to be 2 classical and 1 RnR. The possible outcomes for a "two classical, one rock" distribution are: CCR, CRC, RCCThe odds of each of these is below.CCR = (5/9 x 4/8 x 4/7) = 80/504 (Five out of nine times you'll select classic first. When you do, 4/8 times you'll select classic again, and 4/7 times you'll select rock on the final pick.)CRC = (5/9 x 4/8 x 4/7) = 80/504RCC = (4/9 x 5/8 x 4/7) = 80/504240/504 = 10/21There are easier ways to do it with, you know, math, but this is the easiest, most intuitive way to understand the problem. Lemme know if you need anymore help/explaining. Link to post Share on other sites
chrozzo 19 Posted April 3, 2009 Share Posted April 3, 2009 10/21yw! Link to post Share on other sites
borinka99 0 Posted April 3, 2009 Author Share Posted April 3, 2009 There are 84 different combinations of 3 CD's possible (9!/(9-3)!*3!)There are 40 different combinations of 2 classical and 1 rnrClassical (5!/(5-2)!*2!) = 10Rnr (4!/(4-1)!*1!) = 4You have 10 "sets" of 2 classical CD's and 4 "sets" of one rock n roll CD. Multiply the 2 together to get 40 different "sets" of the 3 CD's40/84 = 10/21Whoa, I just blacked out.... what happened?Nice, thanks!The possible outcomes for a "two classical, one rock" distribution are: CCR, CRC, RCCThe odds of each of these is below.CCR = (5/9 x 4/8 x 4/7) = 80/504 (Five out of nine times you'll select classic first. When you do, 4/8 times you'll select classic again, and 4/7 times you'll select rock on the final pick.)CRC = (5/9 x 4/8 x 4/7) = 80/504RCC = (4/9 x 5/8 x 4/7) = 80/504240/504 = 10/21There are easier ways to do it with, you know, math, but this is the easiest, most intuitive way to understand the problem. Lemme know if you need anymore help/explaining.Thanks!!! Link to post Share on other sites
borinka99 0 Posted April 3, 2009 Author Share Posted April 3, 2009 There are 84 different combinations of 3 CD's possible (9!/(9-3)!*3!)There are 40 different combinations of 2 classical and 1 rnrClassical (5!/(5-2)!*2!) = 10Rnr (4!/(4-1)!*1!) = 4You have 10 "sets" of 2 classical CD's and 4 "sets" of one rock n roll CD. Multiply the 2 together to get 40 different "sets" of the 3 CD's40/84 = 10/21Whoa, I just blacked out.... what happened?Hi, so just to clarify, when you write (9!/(9-3)!*3!), you actually mean 9!/((9-3)!*3!) right? So 9!/(6!3!)And, 9/(9-3)! is because you are taking away 3 cds 1 at a time just like 5!/(5-2)! is because you are taking away 2 classical? I'm just trying to figure out a "formula" so we can do this for different problems. Thanks for your help though! Link to post Share on other sites
gilbertology 0 Posted April 3, 2009 Share Posted April 3, 2009 The possible outcomes for a "two classical, one rock" distribution are: CCR, CRC, RCCThe odds of each of these is below.CCR = (5/9 x 4/8 x 4/7) = 80/504 (Five out of nine times you'll select classic first. When you do, 4/8 times you'll select classic again, and 4/7 times you'll select rock on the final pick.)When learning statistics and probability you shouldn't just memorize equations and plug in numbers. Don't worry about the ! equation stuff. All you have to do in most scenarios is think practically about what the question is asking and then figure out how to get the answer. The above is a great explanation for how to get the right answer to this particular question, and also explains why it is that it works.Stuff like this makes me want to get back into teaching math because solving questions like these are so easy to explain to people, but often teachers will confuse you and/or just give you the relevant equation and expect you to simply plug in the numbers. None of that stuff matters unless you understand why you arrived at the answer you did. Link to post Share on other sites
Quacktastic 106 Posted April 3, 2009 Share Posted April 3, 2009 Hi, so just to clarify, when you write (9!/(9-3)!*3!), you actually mean 9!/((9-3)!*3!) right? So 9!/(6!3!)And, 9/(9-3)! is because you are taking away 3 cds 1 at a time just like 5!/(5-2)! is because you are taking away 2 classical? I'm just trying to figure out a "formula" so we can do this for different problems. Thanks for your help though!That is correct.When learning statistics and probability you shouldn't just memorize equations and plug in numbers. Don't worry about the ! equation stuff. All you have to do in most scenarios is think practically about what the question is asking and then figure out how to get the answer. The above is a great explanation for how to get the right answer to this particular question, and also explains why it is that it works.Stuff like this makes me want to get back into teaching math because solving questions like these are so easy to explain to people, but often teachers will confuse you and/or just give you the relevant equation and expect you to simply plug in the numbers. None of that stuff matters unless you understand why you arrived at the answer you did.Show me how to think practically and not use equations when there are 45 classical CD's, 32 rock CD's, 24 rap CD's, 18 country CD's, 31 comedy CD's, and 20 blank CD's and I want to know the odds of drawing exactly 7 classical, 3 rock, 6 rap, 4 country, 6 comedy, and 2 blank CD's. Reading the explaination to understand the equation is great, but if you don't know the math behind it you will be stuck when bigger #'s come into play. Link to post Share on other sites
Sal Paradise 57 Posted April 3, 2009 Share Posted April 3, 2009 that rhinestonecowboy loves him some arguments! Link to post Share on other sites
RhinestoneCowboy 2 Posted April 3, 2009 Share Posted April 3, 2009 that rhinestonecowboy loves him some arguments!The what with the who now? Link to post Share on other sites
Sal Paradise 57 Posted April 3, 2009 Share Posted April 3, 2009 The what with the who now?WRONG. Link to post Share on other sites
leftygolfer 7 Posted April 3, 2009 Share Posted April 3, 2009 that rhinestonecowboy... LFT Link to post Share on other sites
RhinestoneCowboy 2 Posted April 3, 2009 Share Posted April 3, 2009 I cannot help that I am PASSIONATE ABOUT NUMBERS!!! Link to post Share on other sites
Sal Paradise 57 Posted April 3, 2009 Share Posted April 3, 2009 I cannot help that I am PASSIONATE ABOUT NUMBERS!!!me too! especially ones below 18 POW BANG BOOM! Link to post Share on other sites
RhinestoneCowboy 2 Posted April 3, 2009 Share Posted April 3, 2009 me too! especially ones below 18 POW BANG BOOM!SCHWING!!!! Link to post Share on other sites
LongLiveYorke 38 Posted April 3, 2009 Share Posted April 3, 2009 Just write a program and monte carlo the answer.That's basically what I do for everything nowadays. Link to post Share on other sites
runthemover 39 Posted April 3, 2009 Share Posted April 3, 2009 Just write a program and monte carlo the answer.That's basically what I do for everything nowadays.How do you solve a problem like Maria? Link to post Share on other sites
LongLiveYorke 38 Posted April 3, 2009 Share Posted April 3, 2009 How do you solve a problem like Maria?I think murder/suicide was the eventual answer to that question. Link to post Share on other sites
Tactical Bear 3 Posted April 3, 2009 Share Posted April 3, 2009 That is correct.Show me how to think practically and not use equations when there are 45 classical CD's, 32 rock CD's, 24 rap CD's, 18 country CD's, 31 comedy CD's, and 20 blank CD's and I want to know the odds of drawing exactly 7 classical, 3 rock, 6 rap, 4 country, 6 comedy, and 2 blank CD's. Reading the explaination to understand the equation is great, but if you don't know the math behind it you will be stuck when bigger #'s come into play.I posted the "non-mathy" answer, and I know how to figure this stuff out with a calculator. Your answer did very little to further explain how/why the answer is what it is, but my answer did very little to help him solve even slightly complex problems. That being said, the fact that the problem was as simple as it is (9 records; picking 3), isn't it reasonable to assume that the person learning the math isn't going to have to start doing binomial distributions and running regressions next week?I always err on the side of explaining the process, because for most people, math is as much about logic as it is about numbers. Understanding the how and why gives you the skills to apply your reasoning to slightly different problems, and after enough practice, you'll be able to intuit the next step without being instructed, with full confidence in your logic. Everybody can look the formula up in a prob/stats book (or online) -- and everybody absolutely should have these formulas in his bag somewhere, with the full understanding of how and when to use them -- but understanding the basics, for most people, will be the means to a much more gratifying end.Also: seriously, though, aren't Monte Carlo simulations sweet? I'm simulating a gambling bankroll -- with various long-term positions, hedge options, and the Kelly Criterion -- and trying to figure out optimal strategy for medium-term profit maximization. It's fun Link to post Share on other sites
CindyLou 11 Posted April 3, 2009 Share Posted April 3, 2009 How do you solve a problem like Maria?Tippy-toe! Tippy-toe! Link to post Share on other sites
Sal Paradise 57 Posted April 3, 2009 Share Posted April 3, 2009 Tippy-toe! Tippy-toe!I love jokes like that! Link to post Share on other sites
loogie 115 Posted April 3, 2009 Share Posted April 3, 2009 Armon had a drawer labeled "Classic Rock". In it was a math problem and a terrible joke. Link to post Share on other sites
borinka99 0 Posted April 3, 2009 Author Share Posted April 3, 2009 Show me how to think practically and not use equations when there are 45 classical CD's, 32 rock CD's, 24 rap CD's, 18 country CD's, 31 comedy CD's, and 20 blank CD's and I want to know the odds of drawing exactly 7 classical, 3 rock, 6 rap, 4 country, 6 comedy, and 2 blank CD's. Reading the explaination to understand the equation is great, but if you don't know the math behind it you will be stuck when bigger #'s come into play.Yea, I think it helps to first know the understanding behind the equation, and then the equation would help for future (and more complicated problems)I posted the "non-mathy" answer, and I know how to figure this stuff out with a calculator. Your answer did very little to further explain how/why the answer is what it is, but my answer did very little to help him solve even slightly complex problems. That being said, the fact that the problem was as simple as it is (9 records; picking 3), isn't it reasonable to assume that the person learning the math isn't going to have to start doing binomial distributions and running regressions next week?I always err on the side of explaining the process, because for most people, math is as much about logic as it is about numbers. Understanding the how and why gives you the skills to apply your reasoning to slightly different problems, and after enough practice, you'll be able to intuit the next step without being instructed, with full confidence in your logic. Everybody can look the formula up in a prob/stats book (or online) -- and everybody absolutely should have these formulas in his bag somewhere, with the full understanding of how and when to use them -- but understanding the basics, for most people, will be the means to a much more gratifying end.Also: seriously, though, aren't Monte Carlo simulations sweet? I'm simulating a gambling bankroll -- with various long-term positions, hedge options, and the Kelly Criterion -- and trying to figure out optimal strategy for medium-term profit maximization. It's funPutting both of the replies together (the logic one and the "equation") was perfect. Glad both of you guys answered haha. I'm usually pretty good with math but I couldn't figure this one out. Just didn't think enough. I started out with the 5/9 * 4/8 * 4/7 (which I knew was wrong because that meant that it would have to be picked in that exact order) but didn't go any further lol. Link to post Share on other sites
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