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Basic Probability Theory (Equaly Likely Principle)

  1. Jun 3, 2014 #1
    1. The problem statement, all variables and given/known data
    Five cards numbered 1 to 5 are shuffled and placed face down on a table. Two of the cards are picked at random. [Hint: find all of the possible outcomes of this experiment which form the sample space S and use the Equally Likely Principle.]

    Find the probability of the following events:

    A, the number 1 is selected;

    B, the smaller of the two numbers selected is 2;

    C, the difference of the numbers is even


    2. Relevant equations
    The probability of an event A is defined as the proportion of outcomes in the sample space S that correspond to A:
    P(A)=n(A)/n(S)



    3. The attempt at a solution
    I am very new to this, I wasn't sure of something and needed it clearing up before I can give a definite answer for each part. Does anyone know if, in this problem, out of the two cards that are picked is picking a 1 and 2, the same as picking a 2 and 1 (if you know what I mean).

    For part A then: On the basis of the equally likely principle, if they are considered the same then the total possible outcomes are 10 and then the answer would be 2/10=1/5. If they are not considered the same then the total possible outcomes becomes 20 and the answer would become 2/20=1/10.

    For Part B I am not sure it matter if they are the same or not but from what I can gather, the total possible outcomes are 3 (2 and 3, 2 and 4, 2 and 5). so the answer would be 2/3.

    For Part C if they are the same then the total possible outcomes is 4 so the answer would be 2/4=1/2. If they are not the same then the total possible outcomes would be 8 and therefore the answer would be 2/8=1/4.

    The only downside to my "go" at this problem is that to find the possible number of outcomes for things I had to pen them all down and cross them off. If there was a better way to do that I would appreciate some advice on that :) .

    Thanks :)
     
  2. jcsd
  3. Jun 3, 2014 #2

    PeroK

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    Why not get yourself a pack of cards and experiment and see what happens?
     
  4. Jun 3, 2014 #3
    Well, you could pick the 2 up first and then the 1 up second and you could also pick the 1 up first and then 2 second. But once you have them in your hands, they are the same two cards regardless of order. That is what is confusing me.
     
  5. Jun 3, 2014 #4

    PeroK

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    You can solve the problem either way, as long as you are consistent:

    You can consider 20 (5x4) possibilities for 2 cards; of which 4 have 1 as the first card and 4 have 1 as the second card.

    Or, you consider only 10 combinations; of which only 4 have a 1.

    It's up to you; whatever makes the calculations simpler and more logical.
     
  6. Jun 3, 2014 #5
    Ah right ok, thanks. So I guess my answers are incorrect seeing as I got different answers depending on which way I looked at it.
     
  7. Jun 3, 2014 #6

    PeroK

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    Doing a problem two different ways is a good way to confirm you've got the correct answer. But, of course, if you get different answers, then at least one of them must be wrong.

    It's still a good idea to try it out with real cards and see what happens.
     
  8. Jun 3, 2014 #7

    Ray Vickson

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    Doing the problem two different ways is good, provided that you do both ways correctly. Physically, you really do pick one card first, then pick a second card afterwards. Therefore, in (A) the two outcomes (first,second) = (1, x) or (x,1) (where x = 2, 3, 4 or 5) describers exactly what constitutes the event "pick a 1". In (B) the event "smaller is 2" = (2,x) or (x,2), where x = 3, 4 or 5. In (C) the event "difference is even" = {(a,b): |b-a|=0 or |b-a| = 2 or |b-a| = 4}.
     
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