Calculating Probabilities for Drawing Specific Cards

[Physics_wiz]

You have a 52 cards deck and you draw 3 random cards from it. What are the probabilities of the following (order doesn't matter in all of them):

1. Drawing two specific cards (suit doesn't matter) with the third one being any other card. For example 6 9 x or 9 6 x or x 6 9, etc..

2. Drawing two specific cards (ie. 4&5, 8&3, etc..) with the third one being any other card and having all three cards be of different suits.

2. Drawing two specific cards of the same suit (ie. 6 of spades and 9 of spades, 4 of hearts and 7 of hearts, etc..) with the third one being any other card with a different suit.

I'm more interested in knowing how to calculate those probabilities than in just knowing the probabilities, so if you can show me how to calculate all 3 of them that would be great!

Edit: for the first condition, the third card (x) can be one of the two cards selected (ie. 6 9 9 is valid).

 

[neurocomp2003]

find the probablity of drawing a card or # then do the ...bah can't remember the rule name...i'm thinking bayes but that's prolly wrong

 

[O Great One]

In the following, I'm assuming that the two specific cards are of different ranks.

1. Drawing two specific cards (suit doesn't matter) with the third one being any other card. For example 6 9 x or 9 6 x or x 6 9, etc..
I'm assuming that the third card can be the same rank as one of the first two cards so: (4*4*50)/52C3 = 0.036199.

2. Drawing two specific cards (ie. 4&5, 8&3, etc..) with the third one being any other card and having all three cards be of different suits. I'm assuming that the third card must be of different rank than the first two cards so: (4*3*22)/52C3 = 0.011945701.

3. Drawing two specific cards of the same suit (ie. 6 of spades and 9 of spades, 4 of hearts and 7 of hearts, etc..) with the third one being any other card with a different suit. I'm assuming that the third card must be of a different rank than the first two cards so: 33/52C3 = 0.001493212.

I hope this helps. :>