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Probability 3

  1. Feb 18, 2006 #1
    Could anyone please help me with making a start.... and a finish to this little exercise. Many thanks Nat.

    Four people A, B, C and D are to play in a small table-tennis tournament played on a simple knock-out basis: their names are drawn at random to play in two pairs, then the two winners play in the final. The probability that a player beats a player with a later letter is 2/3. All matches are independent. Find the probabilities that:

    1) A wins the tournament

    2) C and D meet in the final

    3) B and C meet at some stage

    Must I use a tree diagram?
     
  2. jcsd
  3. Feb 18, 2006 #2

    arildno

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    1) A must win two fights, irrespective of the pairing arrangements. What is the probability of that?

    2) This does depend on the probability that C and D does NOT meet each other in the first fight, and that BOTH win their first fight (irrespective of who beats A or B)
     
  4. Feb 18, 2006 #3

    Is 1) p(A) = 2/3 * 2/3 = 4/9
     
  5. Feb 18, 2006 #4

    arildno

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    Correct! :smile:
     
  6. Feb 18, 2006 #5
    Is 2) p(C & D to meet in final) = 2/3 * 2/3 * 2/3 * 2/3 = 16/81
     
  7. Feb 18, 2006 #6

    arildno

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    No.
    1. There is 2/3 chance they'll not meet in the first round. 1/3 of those times, C will go to the finals, and 1/3 of those times again D will also go to the finals to meet C.
     
  8. Feb 18, 2006 #7
    I see so...

    P(C&D meet in final) = 2/3 * 1/3 * 1/3 = 2/27
     
  9. Feb 18, 2006 #8

    arildno

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    3*3*3 equals 27 last time I checked..
     
  10. Feb 18, 2006 #9
    lol, i did spot it straight away and corrected it :shy:

    Right then...

    for number 3 :rolleyes:
     
    Last edited: Feb 18, 2006
  11. Feb 18, 2006 #10

    arildno

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    Now, for 3) calculate separately the probabilities for the disjoint events a) B&C meets in the first round b) B&C meet in the finals.

    Either a) or b) may happen..
     
  12. Feb 18, 2006 #11
    a) p(B&C meets in the first round) = 1/3

    b) p(B&C meet in the finals) = 2/27

    So the p(B&C meet at some stage) = 1/3 * 2/27 = 2/81
     
  13. Feb 18, 2006 #12

    arildno

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    Incorrect!
    a) is correct, but why do you think b) is correct??

    Furthermore, you are to have probabilities of disjoint events EITHER of which occuring will mean that b&c meets at some stage. Should you multiply the probabilities together in that case?
     
    Last edited: Feb 18, 2006
  14. Feb 18, 2006 #13

    Ok so

    a) p(B&C meets in the first round) = 1/3

    b) p(B&C meet in the finals) = ?
    Well I thought this was correct as for question b) the p(C&D to meet in final is 2/27 so why would p(B&C meet in final) would be different?

    So the p(B&C meet at some stage) = 1/3 + ? = ?
     
  15. Feb 18, 2006 #14

    arildno

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    b) will be different in that in the first case, for example, C must always beat a BETTER player in order to go to the finals.
    There is 1/3 chance that the initial pairing is (A&B, C&D) What is the chance that this was the set up AND that both B and C wins (and thus proceed to the finals to meet each other)?

    Furthermore, it is a 1/3 chance that the initial pairing was (A&C,B&D) What is the chance that this was the set up AND that both B and C wins
    (and thus proceed to the finals to meet each other)?
     
  16. Feb 18, 2006 #15
    b) P = 2/3 * 1/3 * 1/3 * 1/3 = 2/81

    So p(B&C meet at some stage) = 1/3 + 2/81 = 29/81
     
  17. Feb 18, 2006 #16

    arildno

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    No, no, no!
    Think again please.
     
  18. Feb 18, 2006 #17
    Let me see...


    There is 1/3 chance that the initial pairing is (A&B, C&D) What is the chance that this was the set up AND that both B and C wins (and thus proceed to the finals to meet each other)? this gives 1/3 * 1/3 * 1/3 = 1/27

    Furthermore, it is a 1/3 chance that the initial pairing was (A&C,B&D) What is the chance that this was the set up AND that both B and C wins
    (and thus proceed to the finals to meet each other)? this gives 1/3 * 1/3 * 2/3 = 2/27
     
  19. Feb 18, 2006 #18

    arildno

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    Almost correct, but the probability for the first case is also 2/27, since C beats D 2/3 of the time.

    Thus, the TOTAL probability of B&C meeting in the finals is 2/27+2/27=4/27

    And the total probability of B&C meeting at some stage is therefore....?
     
  20. Feb 18, 2006 #19
    p(B&C meeting at some stage) = 1/3 + 4/27 = 13/27
     
  21. Feb 18, 2006 #20

    arildno

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    You're done! :smile:
     
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