Question

One card is selected from a standard deck of 52 cards. What is the probability that the card is either a diamond or a face card?

Answer 1

there are 13 face cards right :P o,O hmm

Answer 2

Looking for others' interpretations here. I thought it should be \(\frac{13}{52}+\frac{9}{52}=24/52=12/26=6/13\)

Answer 3

I'm reading the problem wrong somehow(or book has a type-o)

Answer 4

Yea, 13 face cards nosh

Answer 5

and 13 diamonds lol ??? no idea abt cards :P

Answer 6

some rules define aces as faces

Answer 7

They say the answer is 13/52

Answer 8

Made that mistake ( I defined an ace as a face first time around)

Answer 9

but this is my second go

Answer 10

Shouldn't the probability be greater than just diamonds if we have an or and not an XOR?

Answer 11

if we remove all the face cards, and the diamonds, that is our ... outcome space?

Answer 12

10+12 = 22 cards that are favored, out of 52 total

Answer 13

1thru9 jkq

9+12 = 18 ... out of 52 :)

Answer 14

wait, only KQJ are face cards.... I think

Answer 15

XOR doesnt seem to translate well into english prose

Answer 16

its 11pm .. and i forgot how to add ... im going to bed :)

Answer 17

lol, I am a bit confused by your argument. Let me attempt to understand

Answer 18

we should have 30 cards not an option

Answer 19

3 suits, A-10 each

Answer 20

I added wrong above My conclusion should be 11/26

Answer 21

told you then i deleted the comment
i thought this is impossible how fib can got it wrong lol

Answer 22

lol I can't add today, clearly

Answer 23

but still they are claiming 13/52 is the answer

Answer 24

@ganeshie8 Would you say 13/52 is incorrect?

Answer 25

I mean I think you proved it, but just want confirmation

Answer 26

A : diamonds
B : face cards

n(A) = 13
n(B) = 12
n(A \(\cap\) B) = 3

outcomes in favor = n(A \(\cup \) B)
= n(A) + n(B) - n(A \(\cap\) B)
= 13 + 12 - 3
= 22

total outcomes = 52

so p(A \(\cup\) B) = 22/52

Answer 27

here is their argument:
"The prob. that a card is either a diamond or a face card is determined as follows:
Let D be the event the card is a diamond. Let F be the event the card is a face card. Then, Pr[D}=13/52 and Pr[F]=12/52. We want to compute the probability of the event DUF
Pr[DUF]=Pr[D]+Pr[F]-Pr[D(intersect)F]=13/52+12/52-3/13=13/52

Answer 28

Ok, so book is def. wrong and I can't add... Great.

Thanks guys!