यूजर का नामGingerAle
सवाल 4
जवाब 660


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! OH NO! Not Again!

If my train was load with toxic chemicals, I could wipe out a city; or 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 fluked me for me for my train driving skills.] I’d flunk me too!  



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.




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3 hours ago

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.




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



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10 hours ago

Good explanation.  I actually prefer "literal" but for you I'll cop to pedantic, negative connotation notwithstanding.


You don’t need to cop to pedantic; I do not think you are pedantic. You’re not pedantic in your solution presentations. (If you were, I’d be trolling the hell out of you.) LOL Although apparently, (sometimes) you tend toward the literal for certain colloquialisms (and Latin word-order in certain phrases), you do not seem to be a diehard literalist. I think this quite funny, and I’m sure Gracie Allen would find it funny and endearing. 


If someone asked you to bring them a cup of coffee, what would you bring to them? 

A cup of coffee beans, a cup of ground coffee beans, a cup of dehydrated coffee extract, or a cup of brewed or instant coffee?


When I was eleven, I once brought a literal cup of coffee (the mug was two-thirds full of instant coffee crystals) to my mum, after she asked me to bring her a cup of coffee. My mum was greatly amused. The idea came to me after my Great Uncle Cosmo explained certain food chemistries to me. One of the examples was that a typical cup of coffee was 1.13% to 1.88% soluble vegetable matter (extracted from approximately 12 grams of ground coffee beans) and the rest was water.  A cup of coffee is 98.75% water, but it’s called a cup of coffee. I never thought of it any other way until after my chemistry lesson from Cosmo.



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17 Jan 2022

Here, I answer a very similar question. If I choose four cards from a standard 52-card deck, with replacement, what is the probability that I will end up with all four Aces?


This solution post is based on the original:


The phrase “with replacement” is an explicit standard variation in the sampling method. In this case, it means the card (no matter what it is) is replaced after it is drawn.  It’s also important to note that the deck or selection process remains randomized after the card is replaced. And, for the ultra pedantic minded (Ron), the value of the drawn card is recorded before its replacement into the deck. The records are then analyzed statistically for the probability of drawing four aces in sequence.


This question is poorly written. The primary defect is the phrase “end up with four Aces,” which is a colloquialism as used here. The capital “A” in “Aces” is nonstandard, and gives ambiguous emphases to the word aces. The word “all” is not used in this question, so the interpretation is biased toward any combination of Aces where one or more may be repeated.


Rephrased as primer statistics question:

If I choose four cards with replacement from a standard 52-card deck, what is the probability that I will select four aces (of any suit)?

A success in this experiment is four aces without regard to the suit.


\(\large \left(\dfrac{4}{52}\right)^4 = \dfrac{1}{28561}\)



Compare to the probability of selecting four aces in four different suits, where any other combination is a failure.


\(\large \dfrac{4}{52}*\dfrac{3}{52}*\dfrac{2}{52}*\dfrac{1}{52} =\dfrac{3}{913952}\)




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16 Jan 2022