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Probability, balls in an urn

  1. Mar 27, 2013 #1
    Hi, if in an urn there are 3 red balls and 2 white balls and we draw 2 balls from the urn without replacement.
    If we assume that at each ball in the urn is equally likely to be chosen, what is the probability that both balls are red?

    I know the solution is [tex] \frac{\binom{3}{2}}{\binom{5}{2}}[/tex], but i want to show the elements of the sample space, for example are they the elements: [tex]r_1, r_2, r_3, w_1, w_2[/tex]?
    If i split the events in two disjoint events as R_1={the first ball is red} and R_{the second ball is red} what are the elements of these sets?
     
    Last edited: Mar 27, 2013
  2. jcsd
  3. Mar 28, 2013 #2
    R_1 and R_2 are not disjoint.
     
  4. Mar 28, 2013 #3
    The sample space in this question contains all distinct ways of choosing two of the five balls, with 5C2 elements. Your event (in which both balls are red) has 3C2 elements, the number of distinct ways of choosing two of the three red balls.
     
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