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Mathematics 15 Online
OpenStudy (uri):

Stats : From a deck of 52 cards,a card is drawn at random.Find the probability that it is: A) a red card B) a face card C)a diamond card D) either a ten,or a queen E) a face card,or a 5,or a 4 of spade.

OpenStudy (uri):

What? How?

OpenStudy (anonymous):

no of card by total cards i hope....

hartnn (hartnn):

yes, first find how many red cards are there in total then divide it by total number of cards P =#red cards / total # of cards

OpenStudy (anonymous):

imean .. as first one ... no of red cards / total no of cards ...

OpenStudy (anonymous):

no of face card by total no of cards ;....

OpenStudy (uri):

How to find how many red cards are there?

hartnn (hartnn):

if you know what a deck is, you will know how many red and black cards are there and how many face cards also

OpenStudy (anonymous):

got to know the chart ..... if u play card ....

OpenStudy (uri):

-____________- I don't know? playing cards is HARAM.

OpenStudy (anonymous):

hmm ... neither mee but for doing the sum gota know the chart ...

hartnn (hartnn):

red cards = diamond cards + heart cards

OpenStudy (uri):

@hartnn So there are always 13 red cards,hartnn? so 13/52?

hartnn (hartnn):

there are 13 heart cards and 13 diamond cards both are considered as red so, total # of red cards = 26 P= 26/52=....?

OpenStudy (uri):

Wait 13 diamond,13 hearts so 26?

OpenStudy (anonymous):

hope 26/52

OpenStudy (uri):

Lol yes got it. @hartnn

OpenStudy (uri):

so 26/52 =0.5

hartnn (hartnn):

yes, 0.5 is correct :) Face cards - Jacks, Queens, and Kings are called "face cards" because the cards have pictures of their names. so, in 52, how many face cards will be there ?

hartnn (hartnn):

note: there are total 4jacks, 4 queens and 4 kings, one of each suit.

OpenStudy (uri):

12/52

OpenStudy (uri):

@hartnn

hartnn (hartnn):

12/52 is correct. how are face cards 36 in number ?

OpenStudy (uri):

yay so 0.230769

OpenStudy (anonymous):

kk calculated ace too...

hartnn (hartnn):

C. is simple total 13 diamonds P=... ?

OpenStudy (uri):

13/52

hartnn (hartnn):

yes, correct d) there are 4 tens one of each suit same way, there are 4 queens , one of each suit P=....?

hartnn (hartnn):

'or' means addition

OpenStudy (uri):

0.25 for diamond cards,and what does *one on each suit* means?

hartnn (hartnn):

there are 4 suits in the deck hearts spades diamonds and clubs

hartnn (hartnn):

look at the wiki link for details

hartnn (hartnn):

so, 4 tens are 1 ten of heart 1 ten of spade 1 ten of diamond and 1 ten of club same for every other card!

hartnn (hartnn):

can you get d) now ?

OpenStudy (uri):

Okay so i have a choice i can either go for *a ten* or a *queen* I go for queen so,4/52 = 0.0769

hartnn (hartnn):

'or' in probability means we add the individual probabilities 4/52 for queen + 4/52 for a ten (4+4)/52 will be final answer for d)...just simplify

OpenStudy (uri):

Oh then that concept is confusing but okay *8/52?*

OpenStudy (uri):

0.15384

hartnn (hartnn):

yes, 8/52 ...simplify that. same for last part just add total number of face cards + total # of 5's and total # of 4 of spade

OpenStudy (uri):

Face cards are 12,total 5's are 4 and spade's are 4? So,12+4+4 =20/52 =0.3846153.

hartnn (hartnn):

there are total 4 4's one 4 of heart, one 4 of diamonds one 4 of spades and one 4 of clubs

hartnn (hartnn):

so how many # of 4 of spades ?

OpenStudy (uri):

Huh? o_O 4?

hartnn (hartnn):

Face cards are 12,total 5's are 4 <---correct there is only \(\huge 1\) 4 of spade

OpenStudy (uri):

*A 4 of spades?* so only 1 spade? :p

hartnn (hartnn):

there are total 4 4's one 4 of heart, one 4 of diamonds one 4 of spades <------here , only 1 and one 4 of clubs in the question "4 of spade" is asked, which is unique in all there are 13 spades

OpenStudy (uri):

Don't understand :p

hartnn (hartnn):

which part ? only '4 of spade' ? because # of faces=12 .....correct. # of 5's =4 ....correct (one of each suit) # of 4 of spade =.... ?

OpenStudy (uri):

I don't understand # of 4 of spade

hartnn (hartnn):

ok, there are 13 cards in each suit A,2,3,4......9,10, J, Q, K A,2,3,4......9,10, J, Q, K of hearts A,2,3,4......9,10, J, Q, K of diamonds A,2,3,4......9,10, J, Q, K of spades A,2,3,4......9,10, J, Q, K of clubs total =52 out of this, how many '4' of spades are there ?

OpenStudy (jack1):

there's 1x ace of spades in a deck 1x 2 of spades 1x 3 of spades 1x 4 of spades ######## <---- 1x 5 of spades... etc

OpenStudy (jack1):

how many of this card are there in a deck...?

OpenStudy (jack1):

EXACTLY that card...

hartnn (hartnn):

^ only \(\huge \color {red}1\) 4 of spade, unique!

OpenStudy (uri):

Okay so there are always 13 spades...? :3

hartnn (hartnn):

yes, total 13 spades but that info you don't need here

hartnn (hartnn):

E) a face card,or a 5,or a 4 of spade. a face ---->12 a 5----->4 a 4 of spade --->1 P=(12+4+1)/52

OpenStudy (uri):

Okay...:) So i have a Question when there was *A face card* why we didn't use 1?

hartnn (hartnn):

because there are in all 12 face cards one Jack of heart one Jack of diamond one Jack of club one Jack of spade total 4 jacks similarly, 4 queens and 4 kings total face cards = 12

OpenStudy (uri):

Oh...:D so aren't spades 13 in total? :P

hartnn (hartnn):

yes, there are 13 spades in all A,2,3,4......9,10, J, Q, K of spades <-----total =13 what made you ask this ?

OpenStudy (uri):

WHY ARE WE USING 1 THEN!! THAT MADE ME ASK.. :3

OpenStudy (whpalmer4):

1 = A

hartnn (hartnn):

because the question doesn't ask of only spades only spades = 13 its ask for a "4 of spade" which is unique

OpenStudy (uri):

Okay :P so we use *1?* so face cards is 12,5's are 4 so 12+4+1/52 =17/52 =0.32692

hartnn (hartnn):

if thats blurred http://www.geeksforgeeks.org/wp-content/uploads/cards.png

OpenStudy (uri):

Aren't you smart? :3

hartnn (hartnn):

O.o is that a part of the question ? (answer : P=1 :P)

OpenStudy (uri):

P =1? answer of what? :3

hartnn (hartnn):

of 'Aren't you smart?' nvm if you didn't get it.......

OpenStudy (jack1):

lol

OpenStudy (uri):

So you mean you are *A 4 of spade smart*? :P

OpenStudy (jack1):

probability = 100%

terenzreignz (terenzreignz):

Drawing a single card? Boring... :P

OpenStudy (uri):

Ha! Thanks @hartnn Bhaiya :D

hartnn (hartnn):

always welcome ^_^

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