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The following cards are dealt to three people at random, so that everyone gets the same number of cards. What is the probability that everyone gets a red card?

[asy] unitsize(1 cm);real cardwidth = 0.8, cardheight = 1.2, spacebetweencards = 0.4;int i;for (i = 0; i <= 2; ++i) { filldraw(shift(i*(cardwidth,0) + i*(spacebetweencards,0))*((0,0)--(cardwidth,0)--(cardwidth,cardheight)--(0,cardheight)--cycle), red);} for (i = 3; i <= 5; ++i) { filldraw(shift(i*(cardwidth,0) + i*(spacebetweencards,0))*((0,0)--(cardwidth,0)--(cardwidth,cardheight)--(0,cardheight)--cycle), yellow);} for (i = 6; i <= 8; ++i) { filldraw(shift(i*(cardwidth,0) + i*(spacebetweencards,0))*((0,0)--(cardwidth,0)--(cardwidth,cardheight)--(0,cardheight)--cycle), blue);} label(

 

I have tried, but am stuck! 

 Jan 17, 2023
 #1
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With 3 red cards, 3 yellow cards, and 3 blue cards, there are a total of 9 cards. Since each person is dealt the same number of cards, and the number of red cards is less than the total number of cards, the probability that everyone gets a red card is 1/3.

 Jan 17, 2023
 #2
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Here is my attempt:
Each person gets: 9 / 3 =3 cards


 Person 1 draws their first card, then their second card and their 3rd card. Then, person 2 draws their first card, then their second card and their 3rd card. Finally, person 3 draws their first card, then their second card and their 3rd card. That gives us 9 separate draws, of which 3 are red. That means that there are  (9 C 3) ways for 3 red card to be drawn overall.


How many ways are there for every person to get a red card? The number of red cards is the same as the number of people, so each gets exactly 1 red card. For person 1, that could be their first card or their 2nd  or their 3rd . So there are 3 ways for person 1 to have 1 red card. That is true for person 2 and person 3 as well. 


Therefore, the number of ways for everyone to have a red card is: 3 * 3 * 3 =27


 Therefore, the probability is: 27 / (9 C 3)=27 / 84 =9 / 28

 Jan 17, 2023
 #3
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The answer is actually 5/27, but thanks for the help!

Guest Jan 18, 2023

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