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Quite hard discrete prob. distribution q.

  1. Jul 14, 2004 #1
    i dont even have any idea how to start. plz gives some hint.

    each Supabrek packet contains a card from a set of n picture cards. A packet selected at random has the same prob. of containing any one of the n cards.

    The Smith family already possess k ( < n ) different picture cards. They purchase a few more packets, selected one by one randomly until they find a card that is different from the k cards they already have. If this happens when they bought R packets, show that

    P ( R = r ) = { (n-k)/(n) } { (k/n)^(r-1) }


    thanks.
     
  2. jcsd
  3. Jul 14, 2004 #2

    AKG

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    The probability of such an event is equal to:

    (Probability that the first card is one of the k) x (Probaility that card 2 is one of the k) x ... x (Probability that card R-1 is one of the k) x (Probability that card R not one of the k).

    Since each card is independent of the previous:

    (Prob that card 1 is one of the k) = (Prob that 2 is one of the k) = ... = (Prob R-1 is one of k).

    The probability that any one of these cards is one of the k is obviously (k/n) because k out of every n cards is one of the k cards they already have. Note that this probability is multiplied together R-1 times in calculating the final result. now, the probability that the Rth card (or any card for that matter) is not in the k cards is (n-k)/n, obviously. So, the final answer is:

    [tex]\left ( \frac{n - k}{n} \right ) \left ( \frac{k}{n} \right )^{R - 1}[/tex]
     
  4. Jul 14, 2004 #3
    Denian, what level are you at? Do you learn any distributions (e.g. geometric distribution) at school?
     
  5. Jul 14, 2004 #4
    I was confused as to exactly what P(R) meant when I first read this. It seems that they are asking for: the probability that after purchasing R packets, one packet has a card different from the k cards in possession.

    Here is my thought process. You want two things to happen (events): (1) one of those R packets contains a different card AND (2) the rest R - 1 packets contain cards already in possession. What is the probability of (1)? k cards are in possession, leaving n - k cards left. The sample space is n so the probability of (1) is (n - k)/n. What is the probability of (2)? The probability that one of the R - 1 packets has a card already in possession is k/n ('opposite' of what (1) is). All R - 1 packets have the same probability of having one of the k cards and since you want all of them to have one of the k cards, the probability of this is (k/n)R - 1.

    Thus, the probability of (1) and (2) is (n - k)/n (k/n)R - 1.
     
  6. Jul 14, 2004 #5
    thank you AKG and Hoon!!!!


    no.
     
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