FibonacciChick666
  • FibonacciChick666
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?
Mathematics
katieb
  • katieb
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Nnesha
  • Nnesha
there are 13 face cards right :P o,O hmm
FibonacciChick666
  • FibonacciChick666
Looking for others' interpretations here. I thought it should be \(\frac{13}{52}+\frac{9}{52}=24/52=12/26=6/13\)
FibonacciChick666
  • FibonacciChick666
I'm reading the problem wrong somehow(or book has a type-o)

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FibonacciChick666
  • FibonacciChick666
Yea, 13 face cards nosh
Nnesha
  • Nnesha
and 13 diamonds lol ??? no idea abt cards :P
amistre64
  • amistre64
some rules define aces as faces
FibonacciChick666
  • FibonacciChick666
They say the answer is 13/52
FibonacciChick666
  • FibonacciChick666
Made that mistake ( I defined an ace as a face first time around)
FibonacciChick666
  • FibonacciChick666
but this is my second go
FibonacciChick666
  • FibonacciChick666
Shouldn't the probability be greater than just diamonds if we have an or and not an XOR?
amistre64
  • amistre64
if we remove all the face cards, and the diamonds, that is our ... outcome space?
amistre64
  • amistre64
10+12 = 22 cards that are favored, out of 52 total
amistre64
  • amistre64
1thru9 jkq 9+12 = 18 ... out of 52 :)
FibonacciChick666
  • FibonacciChick666
wait, only KQJ are face cards.... I think
amistre64
  • amistre64
XOR doesnt seem to translate well into english prose
amistre64
  • amistre64
its 11pm .. and i forgot how to add ... im going to bed :)
FibonacciChick666
  • FibonacciChick666
lol, I am a bit confused by your argument. Let me attempt to understand
FibonacciChick666
  • FibonacciChick666
we should have 30 cards not an option
ganeshie8
  • ganeshie8
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FibonacciChick666
  • FibonacciChick666
3 suits, A-10 each
FibonacciChick666
  • FibonacciChick666
I added wrong above My conclusion should be 11/26
Nnesha
  • Nnesha
told you then i deleted the comment i thought this is impossible how fib can got it wrong lol
FibonacciChick666
  • FibonacciChick666
lol I can't add today, clearly
FibonacciChick666
  • FibonacciChick666
but still they are claiming 13/52 is the answer
FibonacciChick666
  • FibonacciChick666
@ganeshie8 Would you say 13/52 is incorrect?
FibonacciChick666
  • FibonacciChick666
I mean I think you proved it, but just want confirmation
ganeshie8
  • ganeshie8
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
FibonacciChick666
  • FibonacciChick666
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
FibonacciChick666
  • FibonacciChick666
Ok, so book is def. wrong and I can't add... Great. Thanks guys!

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