A standard deck of playing cards contains 52 cards, equally divided among four suits (hearts, diamonds, clubs, and spades). Each suit has the cards 2 through 10, as well as a jack, a queen, a king, and an ace. If the 3 of spades is drawn from a standard deck and is not replaced, what is the probability that the next card drawn is a spade OR a king?
A. 1/17
B. 16/51
C. 4/17
D. 5/17

Question

anonymous · Answer

Let's see...

There would be 51 cards left, including 12 spades and 4 kings. One of the kings is a spade. 12 + 4 = 16, but since one of the kings is a spade, 16 - 1 = 15.

So the probability of drawing a spade OR king next is 15/51

accessdenied · Answer

A standard deck has 13 of each suit.

Hearts:      2 3 4 5 6 7 8 9 10 J Q K A
Diamonds: 2 3 4 5 6 7 8 9 10 J Q K A
Clubs:       2 3 4 5 6 7 8 9 10 J Q K A
Spades:    2 3 4 5 6 7 8 9 10 J Q K A

We removed the 3 of spades.  Our deck of cards now has 51 cards, including only 12 spades.  Here I mark the outcomes we want with brackets:

Hearts:      2 3 4 5 6 7 8 9 10 J Q [K] A
Diamonds: 2 3 4 5 6 7 8 9 10 J Q [K] A
Clubs:       2 3 4 5 6 7 8 9 10 J Q [K] A
Spades:    [2  4 5 6 7 8 9 10 J Q] K [A]  "spades or king" => cannot be king and spades

We are looking at two events that are not mutually exclusive (we can draw a card that matches both conditions, which would not fit the "or")
There is a 12/51 chance of drawing a spades and a 4/51 chance of drawing a king.  We cannot include the fourth king of spades as an option (1/51).
  4/51 + 12/51 - 1/51 = 15/51 
You'd just have to simplify 15/51