Consider a standard deck of playing cards (52 cards, no jokers). If you draw two cards, what is the probability that the cards are:
a. two spades?
13 spades, choose 2 13C2 78 1
------------------- = ------ = ---- = ----- = .0588235294
52 cards, choose 2 52C2 1326 17
b. two kings?
4 kings, CHOOSE 2 4C2 6 1
------------------- = ------ = ---- = ----- = .0045248869
52 cards, CHOOSE 2 52C2 1326 221
c. two mismatched cards (do not share suit or face value)?
(4 suits CHOOSE 2) and (13 face values POSITION 2)
For instance if you chose the 2 suits as clubs and hearts, and then picked
7 and queen for the face values, then you have to POSITION those on the
clubs or hearts, For you could either have the 7 of clubs and the queen of
hearts, where you have POSITIONed them one way, or the 7 of hearts and the
queen of clubs where you POSITION them another way.
"And" means "multiply". So, we have
(4C2)(13P2) (6)(156) 12
----------- = --------- = ---- = .7058823529
C(52,2) 1326 17
d. two queens?
That's just like b, 2 kings, above:
4 queens, CHOOSE 2 4C2 6 1
------------------- = ------ = ---- = ----- = .0045248869
52 cards, CHOOSE 2 52C2 1326 221
e. any pair (same face value)?
(13 face values CHOOSE 1) AND THEN (4 suits CHOOSE 2)
(13C1)(4C2) (13)(6) 78 1
----------- = -------- = ---- = ----
52C2 1326 1326 17
f. two matching cards (have either same suit or same face value)?
We have to break that into two mutually exclusive cases
1. Cases where they have the same suit but two different face values.
(4 suits CHOOSE 1)x(13 face values CHOOSE 2)
(4C1)(13C2)
2. Cases where they have the same face value but two different suits.
(13 face values CHOOSE 1)x(4 suits CHOOSE 2)
(13C1)(4C2)
Since it's either case 1 OR case 2 we add. [Often with OR you add;
but with AND you multiply].
(4C1)(13C2)+(13C1)(4C2) (4)(78)+(13)(6) 390 5
----------------------- = ---------------- = ------ = ----
52C2 1326 1326 17
Edwin
