Question

A standard deck of cards contains 52 cards. One card is selected from the deck.
​(a)
Compute the probability of randomly selecting a diamond or spade.
​(b)
Compute the probability of randomly selecting a diamond or spade or club.
​(c)
Compute the probability of randomly selecting a queen or club.

Answer 1

Would it be (13/52)(13/52)?? Or (1/52)(1/52)

Answer 2

^that is part a

Answer 3

Wait, how did you get the 48 if I'm only selecting a diamond or a spade though?

Answer 4

there are 4 diamonds of spaded in a deck of 52 cards

Answer 5

P(diamond or spade)=P(diamond)+P(spade)

Answer 6

Ahh ok.

Answer 7

Wait, I'm confused. Like there's 10 spades and diamonds without counting the jacks and queens.

Answer 8

there are 13 spades and 13 diamonds

Answer 9

Yes, but I wasn't counting all of the spades and diamonds that are included with the Jacks and Queens

Answer 10

why?

Answer 11

Well aren't I supposed to be counting ONLY spades and diamonds?

Answer 12

queen of spades is a spade

Answer 13

http://math.ucr.edu/home/baez/mathematical/cards.png

Answer 14

Well okay. I've had that pulled up already but thanks.

Answer 15

there are 13 hearts, 13 clubs, 13 diamonds and 13 spades

Answer 16

yes, I do know that. And randomly selecting a diamond or spade would be 4?

Answer 17

4?

Answer 18

*2

Answer 19

I still don't know what you are doing

Answer 20

half the deck is a spade or diamond

Answer 21

that is 26/52=1/2

Answer 22

Ok, that makes total sense now. Better off than telling me random numbers and getting me all confused. So for part A it'd be:
1/2 right? since spades and diamonds is half the deck right?

Answer 23

yes

Answer 24

Ok. hang on. I'm gonna try to figure out B on my own.

Answer 25

Before I go on it'd be 39/52 right? Since there's 13 of each.. If not, I'd like an explanation on what I'm supposed to do.

Answer 26

that is correct

39/52=3/4

Answer 27

So, .75. And for C selecting a queen or club would be:
(4/52)(12/52)
Correct?

Answer 28

Wait, I have a question.

Answer 29

P(A or B)=P(A)+P(B)-P(A and B)

Answer 30

Would I still be counting the spade that is within the queen?

Answer 31

Cuz there's 4 queens and 13 spades...

Answer 32

there are...and one of those is the queen of spades

Answer 33

So would be counting it then right?

Answer 34

we would ...just don't count it twice

Answer 35

Ahh, ok. Gotcha. So setting it up.. I'd have:
(4/52)(12/52)
right?

Answer 36

no...you do not multiply

Answer 37

Subtract?

Answer 38

Or add, I don't know.

Answer 39

well it's one of those

Answer 40

Well, you have me this formula sooo.... I'm guessing subtract then..
P(A or B)=P(A)+P(B)-P(A and B)

Answer 41

P(queen or club)=P(queen)+P(club)-P(queen and club)

Answer 42

OMG, this is so confusing. I don't know.

Answer 43

P(queen)=?

Answer 44

Queen= 4/52
Club = 12/52

Answer 45

P(club)=13/52

Answer 46

P(queen and club)=?

Answer 47

I thought we weren't counting the spades twice.

Answer 48

we count each unrestricted and we subtract any we double counted. That is what the formula does

Answer 49

OHh ok. that makes sense so 17/52?

Answer 50

almost

P(Q or C)=P(Q)+P(C)-P(Q and C)

\[=\frac{4}{52}+\frac{13}{52}-\frac{1}{52}=\frac{16}{52}\]

Answer 51

Oh, could't I have subtracted that in the beginning and still come out to the same answer?

Answer 52

yes...but you can't write P(club)=12/52 since the real answer is P(club)=13/52

Answer 53

Oh, ok. I see where that's coming from. Anyways thank you soo much. I doo appreciate it.