Question

A standard deck of cards has four suits (hearts, diamonds, spades, and clubs) each with 13 cards in it (1-10, jack, queen, and king). Face cards are all the jacks, queens, and kings. The red cards are any cards in the hearts or diamonds suits, while black cards are any cards in the spades or clubs suits.

When drawing a card randomly from a deck, what is the probability that it will be either a black card or a face card?

Type your answer as a fraction reduced to the simplest form, like this: 7/11

Answer 1

So there are 4 suits...each has 13 cards in it....13 * 4 = 52

We want...either a black card....or a face card.....we know the face cards are jacks...queens...kings

so there are 4 suits...and 3 face cards in each suit so 4 * 3 = 12 face cards...

we also want a black card if not a face card....there are *not counting face cards now*

52 - 12 = 40 cards that are not face cards....there are only 2 suits that are red and 2 that are black...so we have

40 / 2 = 20 possible black cards...

so altogether

we have

20 black cards
and 12 face cards

20 + 12 = 32

this is out of ALL cards so

\[\frac{ 32 }{ 52 }\]

simplify this down to get...???

Answer 2

8/13?

Answer 3

Let A be the event card drawn is black
B the event card drawn is face card
\[P(A cup B)=P \left( A \right)+P \left( B \right)-P \left( A and B \right)\]
\[P \left( A \cup B \right)=\frac{ 26 }{ 52 }+\frac{ 12 }{ 52 }-\frac{ 6 }{52 }=\frac{ 32 }{52 }=\frac{ 8 }{13 }\]