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
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answer #1
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answered by mjatthebeeb 3
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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
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answer #2
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answered by koolkat 3
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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
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answer #3
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answered by 13angus13 3
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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
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52
or 19/52 or 0.365
2007-01-31 12:36:41
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answer #4
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answered by leo 6
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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
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answer #5
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answered by Northstar 7
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