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FibonacciChick666

  • one year ago

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?

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  1. Nnesha
    • one year ago
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    there are 13 face cards right :P o,O hmm

  2. FibonacciChick666
    • one year ago
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    Looking for others' interpretations here. I thought it should be \(\frac{13}{52}+\frac{9}{52}=24/52=12/26=6/13\)

  3. FibonacciChick666
    • one year ago
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    I'm reading the problem wrong somehow(or book has a type-o)

  4. FibonacciChick666
    • one year ago
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    Yea, 13 face cards nosh

  5. Nnesha
    • one year ago
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    and 13 diamonds lol ??? no idea abt cards :P

  6. amistre64
    • one year ago
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    some rules define aces as faces

  7. FibonacciChick666
    • one year ago
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    They say the answer is 13/52

  8. FibonacciChick666
    • one year ago
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    Made that mistake ( I defined an ace as a face first time around)

  9. FibonacciChick666
    • one year ago
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    but this is my second go

  10. FibonacciChick666
    • one year ago
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    Shouldn't the probability be greater than just diamonds if we have an or and not an XOR?

  11. amistre64
    • one year ago
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    if we remove all the face cards, and the diamonds, that is our ... outcome space?

  12. amistre64
    • one year ago
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    10+12 = 22 cards that are favored, out of 52 total

  13. amistre64
    • one year ago
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    1thru9 jkq 9+12 = 18 ... out of 52 :)

  14. FibonacciChick666
    • one year ago
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    wait, only KQJ are face cards.... I think

  15. amistre64
    • one year ago
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    XOR doesnt seem to translate well into english prose

  16. amistre64
    • one year ago
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    its 11pm .. and i forgot how to add ... im going to bed :)

  17. FibonacciChick666
    • one year ago
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    lol, I am a bit confused by your argument. Let me attempt to understand

  18. FibonacciChick666
    • one year ago
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    we should have 30 cards not an option

  19. ganeshie8
    • one year ago
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    |dw:1444012968296:dw|

  20. FibonacciChick666
    • one year ago
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    3 suits, A-10 each

  21. FibonacciChick666
    • one year ago
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    I added wrong above My conclusion should be 11/26

  22. Nnesha
    • one year ago
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    told you then i deleted the comment i thought this is impossible how fib can got it wrong lol

  23. FibonacciChick666
    • one year ago
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    lol I can't add today, clearly

  24. FibonacciChick666
    • one year ago
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    but still they are claiming 13/52 is the answer

  25. FibonacciChick666
    • one year ago
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    @ganeshie8 Would you say 13/52 is incorrect?

  26. FibonacciChick666
    • one year ago
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    I mean I think you proved it, but just want confirmation

  27. ganeshie8
    • one year ago
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    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

  28. FibonacciChick666
    • one year ago
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    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

  29. FibonacciChick666
    • one year ago
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    Ok, so book is def. wrong and I can't add... Great. Thanks guys!

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