Probability Question

[borinka99 0]

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.

[grocery_mony 8]

math is idiotic

[Yoda 1]

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

[Quacktastic 106]

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?

[Tactical Bear 3]

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.

[chrozzo 19]

10/21yw!

[borinka99 0]

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!!!

[borinka99 0]

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!

[gilbertology 0]

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.

[Quacktastic 106]

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.

[Sal Paradise 57]

that rhinestonecowboy loves him some arguments!

[RhinestoneCowboy 2]

that rhinestonecowboy loves him some arguments!

The what with the who now?

[Sal Paradise 57]

The what with the who now?

WRONG.

[leftygolfer 7]

that rhinestonecowboy... LFT

[RhinestoneCowboy 2]

I cannot help that I am PASSIONATE ABOUT NUMBERS!!!

[Sal Paradise 57]

I cannot help that I am PASSIONATE ABOUT NUMBERS!!!

me too! especially ones below 18 POW BANG BOOM!

[RhinestoneCowboy 2]

me too! especially ones below 18 POW BANG BOOM!

SCHWING!!!!

[LongLiveYorke 38]

Just write a program and monte carlo the answer.That's basically what I do for everything nowadays.

[runthemover 39]

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?

[LongLiveYorke 38]

How do you solve a problem like Maria?

I think murder/suicide was the eventual answer to that question.

[Tactical Bear 3]

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

[CindyLou 11]

How do you solve a problem like Maria?

Tippy-toe! Tippy-toe!

[Sal Paradise 57]

Tippy-toe! Tippy-toe!

I love jokes like that!

[loogie 115]

Armon had a drawer labeled "Classic Rock". In it was a math problem and a terrible joke.

[borinka99 0]

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 fun

Putting 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.