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About probability without replacement.

  1. Apr 15, 2012 #1
    1. The problem statement, all variables and given/known data

    2 cards are drawn from a deck of 52 playing cards at random without replacement.
    (a) Find the probability that both cards drawn are of the same suit.
    (b) Find the probability that both cards drawn are face cards given that they are of the same suit.

    2. Relevant equations



    3. The attempt at a solution

    I have no problem with the part (a) but part (b) makes me confused.
    Do I have to use conditional probability to work out the solution of part (b)?
    If that's the case, how to determine the event involved? In other words, how to do?

    Thank you.
     
    Last edited: Apr 15, 2012
  2. jcsd
  3. Apr 15, 2012 #2

    Ray Vickson

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    Prove that you have no problem with (a) by showing your solution. That will also be the basis for solving (b). As you surmise, (b) does involve conditional probability, and you solve it by using the definitions and formulas for conditional probability that are in your textbook or course notes. Of course, before you can calculate anything you need to understand what is being asked, so you need to translate the words into a mental picture. Just imaging drawing out two cards over and over again, many times, and taking note of the results. Now try to put question (b) into the context of your experiment.

    Note added in editing: I assume by the word "unit" you mean the same face-value, so two Queens would be OK, but not a King and a Queen. I have never before heard of the word "unit" in connection with cards.

    RGV
     
    Last edited: Apr 15, 2012
  4. Apr 15, 2012 #3

    HallsofIvy

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    In order that the two cards be face cards, the first card must be one of Jack, Queen, King. What is the probability of that? How many face cards are there in the 52 card deck. Now the second card must be the same as the first. What is the probability of that? How many of that same rank are left in the remaining 51 cards?
     
    Last edited: Apr 15, 2012
  5. Apr 15, 2012 #4
    sorry , wrongly typed, not same "unit", but same "suit"
     
  6. Apr 15, 2012 #5

    HallsofIvy

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    For the first problem, then, the first card drawn can be anything. There are then 51 cards left in the pack, 12 of them of the same suit as the first.

    For the second problem, the probability that the first card is a face card is 12/52= 3/13. There are then 3 face cards left of the 11 cards left in that same suit.
     
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