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Probability function of a discrete random variable

  1. Mar 18, 2010 #1
    1. The problem statement, all variables and given/known data Ten cards are face down in a row on a table. Exactly one of them is an ace. You turn the cards over oen at a time, moving from left to right. Let X be the random variable for the number of cards turned before the ace is turned over. What is the probability function for X?



    2. Relevant equations
    P(a|b)=P(a&b) / P(B)


    3. The attempt at a solution P(X=0) = P(1st card is the ace)=1/10
    P(X=1)=P(2nd card is the ace|1st card is not ace) * P(1st card is not ace)=9/10 * 1/9
    P(X=2)=9/10 * 8/9 * 1/8 = 1/10
    so p(x)=1/10 for x=0,1,...,9
    I am having trouble understanding how the book arrived at the solution. For P(X=1), it appears to me that they manipulated the equation P(a|b)=P(a&b) / P(B) to be P(a|b) * P(b) = P(a&b). So they are solving for P(a&b). But isn't the key to solve for P(a|b)?
     
  2. jcsd
  3. Mar 18, 2010 #2

    jav

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    In this case, you are trying to solve for the probability that card X is an ace (without knowing any of the other cards). If you solve for the P(a|b) you are solving for the probability of event a occurring given that you know b has occurred.

    ie you want to solve for P(X is an ace and 1,2,...,X-1 are not an aces). That is what they mean when they say solve for P(a n b).
     
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