Probablity !!!! Confusing

daivinhtran · June 11th, 2011, 07:20 AM

10 black, 20 white, 40 red

The question is what is probability when we reach 5 of them, and we have 2 white, 2 red, 1 black.......
And some guy solved this question on youtube, and his answer was 12.2%
He solve like : (20C2 x 40C2 x 10C1)/ (70C5)
YouTube - ‪Probability - Hard (Part 1)‬&rlm;

I know he was right. BUt I don't understand it

I solve by another way like :
Doing it your way, you would calculate the probability for each of the 30 possible ways to draw 2W, 2R and one B. So, to draw WWRRB in that order, you get
(20/70)(19/69)(40/6

(39/67)(10/66)

Each iteration of 2W, 2R, and 1B has an identical probability. So, multiply by 30, and you get 12.2%.

Can somebody explain me about first way?
I really can't make sense!!

i KNOW WHAT 70c5 MEAN..... my ways, and his way 's so different..... It's not related with his way.... BUt it comes out same answer

soroban · June 11th, 2011, 10:19 AM

Hello, daivinhtran!

Quote:

There are: 20 white, 40 red, and 10 black marbles.
Five are drawn at random.
What is probability that we draw 2 white, 2 red, and 1 black?

And some guy solved this question on youtube, and his answer was 12.2%

He solved like this:[color=beige] .[/color] $\frac{_{20}C_2\,\times\,_{40}C_2\,\times\,_{10}C_1 }{_{70}C_5}$

I know he was right, but I don't understand it.

He reasoned it out like this . . .

$\text{W\!e draw 5 marbles from the available 70 marbles.}$
[color=beige]. . [/color] $\text{There are: }\:_{70}C_5\text{ possible outcomes.}$

We want 2 White from the available 20 Whites.
[color=beige]. . [/color] $\text{There are: }\:_{20}C_2\text{ ways.}$
We want 2 Red from the available 40 Reds.
[color=beige]. . [/color] $\text{There are: }\:_{40}C_2\text{ ways.}$
We want 1 Black from the available 10 Blacks.
[color=beige]. . [/color] $\text{There are: }\:_{10}C_2\text{ ways.}$

$\text{Hence, there are: }\:_{20}C_2\,\times\,_{40}C_2\,\times\,_{10}C_2\,\ text{ ways to get 2 White, 2 Red, and 1 Black.}$

$\text{Therefore: }\:P(\text{2 White, 2 Red, 1 Black}) \;=\;\frac{_{20}C_2\,\times\,_{40}C_2\,\times\,_{1 0}C_2}{_{70}C_5}$

daivinhtran · June 11th, 2011, 05:58 PM

Quote:

Originally Posted by soroban

Hello, daivinhtran!

Quote:

There are: 20 white, 40 red, and 10 black marbles.
Five are drawn at random.
What is probability that we draw 2 white, 2 red, and 1 black?

And some guy solved this question on youtube, and his answer was 12.2%

He solved like this:[color=beige] .[/color] $\frac{_{20}C_2\,\times\,_{40}C_2\,\times\,_{10}C_1 }{_{70}C_5}$

I know he was right, but I don't understand it.

He reasoned it out like this . . .

$\text{W\!e draw 5 marbles from the available 70 marbles.}$
[color=beige]. . [/color] $\text{There are: }\:_{70}C_5\text{ possible outcomes.}$

We want 2 White from the available 20 Whites.
[color=beige]. . [/color] $\text{There are: }\:_{20}C_2\text{ ways.}$
We want 2 Red from the available 40 Reds.
[color=beige]. . [/color] $\text{There are: }\:_{40}C_2\text{ ways.}$
We want 1 Black from the available 10 Blacks.
[color=beige]. . [/color] $\text{There are: }\:_{10}C_2\text{ ways.}$

$\text{Hence, there are: }\:_{20}C_2\,\times\,_{40}C_2\,\times\,_{10}C_2\,\ text{ ways to get 2 White, 2 Red, and 1 Black.}$

$\text{Therefore: }\:P(\text{2 White, 2 Red, 1 Black}) \;=\;\frac{_{20}C_2\,\times\,_{40}C_2\,\times\,_{1 0}C_2}{_{70}C_5}$

Can you give me more explaination to help me understand the relation between my and his way......

I can't make sense how related them are......

Erimess · June 11th, 2011, 08:11 PM

WOW. This is the THIRD posting of this same problem. I've stopped attempting to read it cause I don't feel like jumping back and forth from thread to thread to get all the various pieces. Have you EVER thought to post again on your same thread so that everything is in one place, so as not to confuse everyone?

Sigh.

filipd · June 11th, 2011, 11:49 PM

Quote:

Originally Posted by Erimess

WOW. This is the THIRD posting of this same problem. I've stopped attempting to read it cause I don't feel like jumping back and forth from thread to thread to get all the various pieces. Have you EVER thought to post again on your same thread so that everything is in one place, so as not to confuse everyone?

Sigh.

Hahaha I was starting to think that I am crazy. Those topics (Probability (HARD, confusing...) are everywhere.

June 11th, 2011, 07:20 AM	#1
daivinhtran Senior Member Joined: Mar 2011 Posts: 105 Thanks: 0	Probablity !!!! Confusing 10 black, 20 white, 40 red The question is what is probability when we reach 5 of them, and we have 2 white, 2 red, 1 black....... And some guy solved this question on youtube, and his answer was 12.2% He solve like : (20C2 x 40C2 x 10C1)/ (70C5) YouTube - ‪Probability - Hard (Part 1)‬&rlm; I know he was right. BUt I don't understand it I solve by another way like : Doing it your way, you would calculate the probability for each of the 30 possible ways to draw 2W, 2R and one B. So, to draw WWRRB in that order, you get (20/70)(19/69)(40/6(39/67)(10/66) Each iteration of 2W, 2R, and 1B has an identical probability. So, multiply by 30, and you get 12.2%. Can somebody explain me about first way? I really can't make sense!! i KNOW WHAT 70c5 MEAN..... my ways, and his way 's so different..... It's not related with his way.... BUt it comes out same answer
$daivinhtran is offline$

June 11th, 2011, 08:11 PM	#4
Erimess Senior Member Joined: Apr 2011 From: USA Posts: 782 Thanks: 1	Re: Probablity !!!! Confusing WOW. This is the THIRD posting of this same problem. I've stopped attempting to read it cause I don't feel like jumping back and forth from thread to thread to get all the various pieces. Have you EVER thought to post again on your same thread so that everything is in one place, so as not to confuse everyone? Sigh.
$Erimess is offline$

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