Pleaze help

Nice solution, however I also tried doing this problem and got a different answer. 

I viewed all the cards as different, so the number of ways to distribute the cards was 9!/3!/3!/3! = 1680. 

I think 9C3 is the number of ways to give out 3 cards to one person, not all 3.

SImilar to what you did, I then assumed everyone recieved a red card, 6 ways of doing this. 

The number of ways to distribute the remaining 6 cards was 6!/2!/2!/2! = 90.

90*6/1680 = 32.14%

=^._.^=

My solution above is WRONG.

I’ve made train wrecks of these types of problems before.

I didn’t realize I’d wrecked the train this time until I read Catmg’s comment:

I think 9C3 is the number of ways to give out 3 cards to one person, not all 3.

That is true, and that means this question requires Hypergeometric distribution counts and NOT just Binomial distribution counts to correctly solve it.

The main difference between Hypergeometric distribution and Binomial distribution is that in Binomial distributions the samples sets are replaced before the next sample is drawn; in Hypergeometric distributions the samples are not replaced before the next sample is drawn.  

This stands to reason: while any of the Binomial sets nCr(9,3) can exist, once a set is given to a person, the number and quality of sets remaining is greatly limited. For example: if person one receives three blue cards then person two cannot receive two yellows and a blue because there are no blue cards remaining to give.

Here is a solution using Hypergeometric distribution:

The number of possible dealt sets is  \(N=\dbinom{9}{3} * \dbinom{6}{3} * \dbinom{3}{3} = 1680 \\\)

For person one, there are  \(\dbinom{3}{1}\) ways to select one (1) of the three (3) red cards and then there are \(\dbinom{6}{2}\)  ways to choose two more (non red) cards.

For person two, there are  \(\dbinom{2}{1}\) ways to select one (1) of the two (2) remaining red cards and then there are \(\dbinom{4}{2}\)  ways to choose two more (non red) cards.

For person three, select the remaining red card and the two remaining (non red) cards. There is one (1) way to do this.

Then

\(n = 3  \dbinom{6}{2} \cdot 2\dbinom{4}{2}  \hspace {1em} \small | \text{ Where n =  the number of hands with a red card.}\\ \Large \rho_{\small \text{(3 persons with one red card)}} \normalsize = \dfrac {n} {N} = 3! \cdot \dfrac{\dbinom{6}{2}\dbinom{4}{2}}{\dbinom{9}{3}\dbinom{6}{3}} =  \dfrac{9}{56} \approx 16.07\%\\\)

GA

--. .-

Everything about your solution seems right to me. 

However, at the very end I got the answer 9/28 which is about 32.14%

6C2 = 15

4C2 = 6

9C3 = 84

6C3 = 20

3! = 6

(6*15*6)/(84*20) = (3*3*3)/(84) = 9/28

=^._.^=

Here is a much easier way to solve this. The method does not use Hypergeometric selection

There are \((3^3) = 27\) arrangements of success. Divide the number of successes by the total number of sets giving \((3^3) / (nCr(9,3) =  \dfrac {9}{28} \approx 32.14\%\)

This matches your calculation, Catmg. 

I’m glad you are paying attention to my presentations and train wrecks! 

If my train was loaded with toxic chemicals, I could wipe out a city. Or, in this case, at least send a student down the wrong track.   

The calculation is exactly twice the value I presented for the Hypergeometric solution.

The formula is correct, but I must have made a mistake in the calculator input. I always check complex equations three times, but still, it escapes me...

 I saved the input used for the calculation: (3!*nCr(6,2)*nCr(3,2))/((nCr(9,3)*nCr(6,3))

...I used a three (3) instead of a four (4) in the second binomial.

[My great uncle Cosmo was a locomotive engineer (and an electrical engineer), he would have flunked me for me for my train driving skills.]   

-----------

Here’s a graphic, demonstrating the arrangements for the above solution.

\(\hspace {.3em}\left[ {\begin{array}{ccc} \scriptsize \hspace {.3em} P_1 & \scriptsize P_2 & \scriptsize P_3 \hspace {.2em} \\ \end{array} } \right] \small \hspace {.2em} \text {Persons Horizontal, sets Vertical. }\small \text{ Though identified by number, persons are indistinguishable. }\\ \left[ {\begin{array}{ccc} R & R & R \\ X & X & X \\ X & X & X \\ \end{array} } \right] \text {First arrangement of “R} \scriptsize{s} \normalsize \text {" where“X” is any other card}\\ \left[ {\begin{array}{ccc} R & R & X \\ X & X & R \\ X & X & X \\ \end{array} } \right] \text {Second arrangement} \\ \left[ {\begin{array}{ccc} R & R & X \\ X & X & X \\ X & X & R \\ \end{array} } \right] \text {Third arrangement} \\ \left[ {\begin{array}{ccc} R & X & R \\ X & R & X \\ X & X & X \\ \end{array} } \right] \text {Fourth arrangement} \\ \, \\ \textbf {. . .} \hspace {1em} \textbf {. . .} \hspace {1em} \textbf {. . .} \\ \, \\ \left[ {\begin{array}{ccc} X & X & X \\ X & X & X \\ R & R & R \\ \end{array} } \right] \text {27^{th} arrangement} \\ \)

From this graphic, it’s easy to see the (3^3) = 27 arrangements of success.

GA

--. .-

That is a very cool way to visualize the problem. 

One thing I love about math is how many different ways there are to solve the question. :))

=^._.^=

WTFF!! Your solution may be free, but it’s also absurd and moronic. (Just because It’s Free doesn’t mean you should be smoking it!)  

The only white balls I see at the moment are in a specimen jar of formalin solution on a shelf in a cabinet. They’ve been in the specimen jar for over 21 years, and they’ve not changed much over that time.  The balls are kitty balls and were originally attached to my cat, known on the forum by his stage name, D.C. Thomas Copper.  I sometimes called him the balless wonder.  My cat was not amused by this, but my dog, Mr. Peabody, thought it quite funny. They were always busting each other’s balls. Mr. Peabody had the advantage though, because he knew where D.C. Thomas’ balls are.   

D.C. Copper is now 22-years-old, still alive and somewhat active; although he mostly cat-naps. Occasionally, after his dinner, he reviews the forum and offers suggestions for potential ball-busting troll posts.   ...Like this one 

GA

--. .-

GA,

I am only 13 and don't smoke.

Interesting story though.  :)

I am refering to Web2.0calc being free to use!

<(it'sfree)>

So are we all, well, me, Catmg and Ginger all in agreement that the prob is 9/28? 

I've just run a Monte-Carlo simulation with 20000 trials and get a probability of 0.32125 (this will vary slightly each time it's calculated because of the use of random numbers of course).  This is close to 9/28 (≈0.32143).

Thanks very much Alan, It is nice to get confirmation that our answers are most likely correct. :))