SOLUTION: A standard deck of cards contains 4 cards of each number type: 2 through 10, jack, queen, king, and ace. Each of the 4 cards is of a different suit: spades, hearts, diamonds, or cl
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Question 1008079: A standard deck of cards contains 4 cards of each number type: 2 through 10, jack, queen, king, and ace. Each of the 4 cards is of a different suit: spades, hearts, diamonds, or clubs. Spades and clubs are black; hearts and diamonds are red. Jacks, queens, and kings are considered face cards.
a) In a standard deck of cards, how many cards are black or show an even number?
b) If you pick a card from a standard deck, what is the probability that the card is black or shows an even number?
c) If I pick a card from a standard deck, what is the probability that the card is NOT black NOR shows an even number?
I appreciate any help.
Found 2 solutions by Boreal, stanbon:
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
Number black are 26
number even are 20
that's 46.
But I double counted black cards that are even, and there are 10 of them.
the answer is 36/52=9/13
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That is also the answer to b.
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Neither black nor even is the complement or 1- the probabilty.
1-(9/13)=(4/13)
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
A standard deck of cards contains 4 cards of each number type: 2 through 10, jack, queen, king, and ace. Each of the 4 cards is of a different suit: spades, hearts, diamonds, or clubs. Spades and clubs are black; hearts and diamonds are red. Jacks, queens, and kings are considered face cards.
a) In a standard deck of cards, how many cards are black or show an even number?
black = 26
even number:: 4*5 = 20
black and even:: 2*5 = 10
Ans: black or even = 26 + 20 - 10 = 36
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b) If you pick a card from a standard deck, what is the probability that the card is black or shows an even number?
Ans: 36/52
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c) If I pick a card from a standard deck, what is the probability that the card is NOT black NOR shows an even number?
Ans: 1 - (36/52)
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Cheers,
Stan H.
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