when a single card is drawn from a shuffled deck, find the odds in favor of getting a queen

- anonymous

when a single card is drawn from a shuffled deck, find the odds in favor of getting a queen

- katieb

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- dan815

there are 52 cards and 4 of them are queens so there is a 4/52 chance of getting a queen that is 1/13

- anonymous

I'm confused, that seems too easy

- dan815

because it is :P

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## More answers

- anonymous

thats a probability of positive occurrence question right?

- UsukiDoll

There are only 4 queens in the entire deck of cards
Queen of Diamonds
Queen of Hearts
Queen of Spades
QUeen of Clubs

- UsukiDoll

there are only 2 colors: red (diamonds and hearts) and black (clubs and spades)

- dan815

positive occurrence ?

- UsukiDoll

and each suit has 2 3 4 5 6 7 8 9 10 J Q K A

- UsukiDoll

Diamonds 2 3 4 5 6 7 8 9 10 J Q K A
Hearts 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

- anonymous

usuki, I appreciate your help but I know the suits and cards in a deck of cards, thank you!

- dan815

dont over think it, there are 52 cards, and you can pick any one of them on a draw, and 4 out of the 52 are queens so 4/52 chance

- dan815

which is 1/13

- anonymous

it just feels like what you typed is missing something. you're pulling a single card, which means you get 1 shot out of 52 to pull 1 of 4 queens

- UsukiDoll

for one suit
each of them does have 13 cards. one of them is the queen , so 1 in 13 chance.

- dan815

look at a dice

- anonymous

ok....

- dan815

there are 6 options on a dice

- dan815

whats the chance of getting a 1?

- dan815

its not a tricky question

- anonymous

it just occurred to me, based on the mult. choice answers I have, 1/13 can't be an answer.
A)1:12
B)51;1
C)1:51
D:12:1

- dan815

write 1/13 in a ratio

- dan815

1:12 = 1/13

- anonymous

OMG I'M AN IDIOT, SORRY YOU'RE RIGHT
DUH

- UsukiDoll

1 as in that queen
12 rest of the cards. ?!

- dan815

yes, there are 4 queens to 48 non queens s
4:48 = 1:12

- UsukiDoll

that last step in probability... tsk tsk I always forget that x.x

- anonymous

OK, I'LL CALL THAT ONE, HAVE A COUPLE MORE

- dan815

ok :)

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