English Deutsch Français Italiano Español Português 繁體中文 Bahasa Indonesia Tiếng Việt ภาษาไทย
All categories

5 answers

If this is one question: P(spade OR Face Card) in one draw:

There are 13 unique spades in a deck of 52 cards
Each suit has 3 Face cards, therefore there are 12 Face cards
However, 3 of them are already spades, so you don't want to double count those.

Therefore, there are 13 spades and 9 non-spade Face Cards.

Your probability is then: (13+9)/52 = 22/52 or 11/26 = 42.3% chance

2007-01-31 12:37:24 · answer #1 · answered by mjatthebeeb 3 · 1 0

How many cards out of 52 are either spades or picture cards? The probability is that number over 52.

2007-01-31 12:40:08 · answer #2 · answered by koolkat 3 · 0 0

use P(a) + P(b) - P(a intersection b) ----i used the word intersection instead of the symbol, inverted U)

P(a) = 13 cards(spades)
P(b) = 12 picture cards(K,Q and J; 4 each)
P(a intersection b) = 3( K of spades, Q of spades and J of spades)

that is 13/52 + 12/52 - 3/52 = 22/52 = 11/26

2007-01-31 12:48:14 · answer #3 · answered by 13angus13 3 · 0 0

spade is 1/4
picture card is 9/52


we have 13 spades, and 9 picture cards. But 3 of the picture cards are spades, so we have
13+9-3
---------
52

or 19/52 or 0.365

2007-01-31 12:36:41 · answer #4 · answered by leo 6 · 0 0

s = spade
c = picture card

p(s or c) = p(s) + p(c) - p(s and c)
= 13/52 + (4*3)/52 - 3/52 = (13 + 12 - 3)/52 = 22/52

2007-01-31 12:42:27 · answer #5 · answered by Northstar 7 · 0 0

fedest.com, questions and answers