How many ways are there to choose 3 cards from a standard deck of 52 cards, if all three cards must be of different suits? (Assume that the order of the cards does not matter.)
WARNING: I am not toooooo good at this probability stuff....but here is my answer
you draw one card....it is a ' given ' it WILL be ONE of the suits
Now you have 51 cards left, 39 of which are a 'different' suit. SO you have a 39/51 chance of getting a different suit.
1 x 39/51
Now you have 50 cards left.... only 26 are a different suit than the first two drawn, so you have a 26/50 chance of a different suit
1 x 39/ 51 x 26/50 = .3976 or 39.76% chance of getting THREE different suits (using a standard poker deck of 52)
Anyone else get the same answer???
For the first card, we can pick any card = 52 choices
For the second card, we have 39 ways to pick a card of a different suite from the first one.
And for the last card, we have 26 ways to pick a card that is different from the first two suites.
So....the total number of ways = 52 * 39 * 26 = 52,728 ways
There are 4C3 ways to choose which suit goes first and stuff so thats 4.
There are 13^3 = 2197 ways to choose the cards.
2197*4 = 8788