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More probability, help with cards.

  1. May 12, 2012 #1
    1. The problem statement, all variables and given/known data

    What is the probability that a hand of 8 cards dealt from a shuffled pack contains:

    b)exactly three cards in at least one of the suits.




    2. Relevant equations



    3. The attempt at a solution

    So

    1)choose a suit (4 ways)
    2)choose 3 cards from the 13 cards in the suit (C(13,3))
    3) choose remaining 5 cards from the remaining 3 suits. (C(39,5)).

    so we have 4C(13,3)C(39,5)

    Hence the probability is (4C(13,3)C(39,5))/C(52,8)

    where C(52,8) is the range of the possible set.

    But the top line is wrong, why?
     
  2. jcsd
  3. May 12, 2012 #2

    tiny-tim

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    hi charmedbeauty! :smile:
    you're double-counting …

    if in 1)and 2) you choose 3 hearts, in 3) you could choose 3 diamonds (+ 1 club + 1 spade)

    if in 1)and 2) you choose 3 diamonds, in 3) you could choose 3 hearts (+ 1 club + 1 spade) …

    same thing! :wink:
     
  4. May 12, 2012 #3
    Thanks Tiny Tim for the help...
    I realised that and figured out the sltn using inclusion/exclusion with 4 subsets...
    but is there any easier way to do this??
    I figured it out but I knew the answer, but it seemed a very difficult(for me) sltn... is there any easier way?

    also for part c) exactly three cards in exactly one of the suits.

    Should I just use inclusion /exclusion again??
     
  5. May 12, 2012 #4

    tiny-tim

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    (4? 8/3 < 3, so don't you only need to do it once?)

    sorry, but there's usually no quick way to do these questions :redface:
    nooo, there is a quicker way :smile:

    (what do the other cards have to be?)
     
  6. May 12, 2012 #5
    not of the same suit

    so initially choose a suit 4C(13,3) with 3 cards from selected suit

    then fill rest of hand

    C(39,5) from the remaining suits


    but now I have overcounted again


    so -3C(13,5)

    take away the rest of the possibilities of choosing 5 cards from 4 suits, ie we could only have one suit so take away the other three.

    better?
     
  7. May 12, 2012 #6

    tiny-tim

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    if only one suit has 3 cards, the distribution must be 3,2,2,1 :wink:
     
  8. May 12, 2012 #7
    I don't follow
    the distribution of what? suits?

    I only one suit has three cards

    there are 5 places remaining to fill with 3 suits.

    so it could be 5 from one suit

    4 from one suit and 1 from another

    2 from one suit, 2 from the other, and one from the last suit

    and they are the only combinations possible since we cant introduce another three cards from the same suit.

    I dont see why I should only choose the intersection of the remaining suits.?

    why cant I have my 8 card hand consisting of 3 from one suit, and 5 from another?
     
  9. May 12, 2012 #8

    tiny-tim

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    oh i misunderstood

    i thought 3 was the maximum
     
  10. May 12, 2012 #9
    ok so is this sltn correct...

    4C(13,3)C(39,5)-3C(13,3)

    ??
     
  11. May 12, 2012 #10

    tiny-tim

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    no, you need to subtract the number of ways of getting exactly 3 in 2 suits :wink:
     
  12. May 12, 2012 #11

    Ray Vickson

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    You could count like this: you need either (1) three cards from just one of the suits; and (2) three cards from each of two suits.

    In (1): if, for example, your three cards are hearts, you need the other five cards from spades, clubs and diamonds, choosing no more than two from each of these. That means that you do, in fact, choose two from each of two suits and one from the other suit. How many ways are there to do that? There are C(13,3) ways of choosing three hearts. There are C(3,2) = 3 ways of choosing the two suits that have doubles, and for each such way there are C(13,2)^2 ways of choosing the doubles; then there are 13 ways of choosing the remaining card. Now, of course, the three cards are not hearts, you need to do the same calculations for each of the other choices. Altogether, the number of ways in case (1) is 4*3*13*C(13,3)*C(13,2)^2.

    You can do a similar calculation in case (2). Adding them up gives you the numerator.

    RGV
     
    Last edited: May 12, 2012
  13. May 13, 2012 #12
    ok so i need to subtract -3C(13,3)C(26,3)?
     
  14. May 13, 2012 #13
    ok so i need to subtract -3C(13,3)C(26,3)?
     
  15. May 13, 2012 #14

    tiny-tim

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    (just got up :zzz:)
    (did you mean 3C(13,3)C(26,2)? anyway …)

    that only gives you 6 (or 5) cards :wink:
     
  16. May 13, 2012 #15
    Im confused about C(26,2)

    I thought it should be C(26,3) Since you said "no, you need to subtract the number of ways of getting exactly 3 in 2 suits".

    isn't that C(26,3)??
     
  17. May 14, 2012 #16

    tiny-tim

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    oh i see!

    no i meant eg exactly three hearts and exactly three diamonds :smile:
     
  18. May 14, 2012 #17
    oh ok, so its C(25,2).

    thanks for the help tinytim
     
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