Solving coefficients in regards to the probability of drawing cards

Valenzuelaca

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Joined
Mar 30, 2017
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1
I was wondering is anyone could help me check if my math is correct here. I'm trying to calculate the probability of my drawing a certain combination of cards from a deck. Question 1: I have a deck of 60 cards, within that deck are 4 copies of card "x". I'd like to figure out the probability of drawing 2 of card "x", being able to change the total number of cards drawn until I find a where the probability becomes greater than 50%. I think my equation is suppose to be (4C2)(56C"y")/(60C"z") where y represent the total number of cards drawn minus the 2 I want, and z represents the total number of cards drawn. Is this correct? Every time I calculate the answer is muh lower than I expect so I'm not sure if I'm fusing something up.
Question 2. I'd like to be able to do the same thing as above but change the need to draw multiple instances of multiple cards. Is there a way to do this as well?
Thanks
 

Ishuda

Elite Member
Joined
Jul 30, 2014
Messages
3,345
I was wondering is anyone could help me check if my math is correct here. I'm trying to calculate the probability of my drawing a certain combination of cards from a deck. Question 1: I have a deck of 60 cards, within that deck are 4 copies of card "x". I'd like to figure out the probability of drawing 2 of card "x", being able to change the total number of cards drawn until I find a where the probability becomes greater than 50%. I think my equation is suppose to be (4C2)(56C"y")/(60C"z") where y represent the total number of cards drawn minus the 2 I want, and z represents the total number of cards drawn. Is this correct? Every time I calculate the answer is muh lower than I expect so I'm not sure if I'm fusing something up.
Question 2. I'd like to be able to do the same thing as above but change the need to draw multiple instances of multiple cards. Is there a way to do this as well?
Thanks
At least two, no replacement:
1st draw = 4 chances to draw card x
2nd draw = 3 chances to draw card x
3rd draw = 58 cards
4th draw = 57 draw
So there are 4 * 3 * 58 * 57 = 4! * (60-2)C2 ways to draw 2 cards of a copy of 4 available out of 60 cards. Now generalize that and realize that is just one arrangement of drawing the two cards. How many arrangements can you have.
 
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