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Math Help - simple probability question

  1. #1
    Junior Member
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    simple probability question

    Could someone help me with this question ; it would be much appreciated

    Ivan and Jo are chess players. In any game the probabilities of Ivan beating Jo, Jo beating Ivan or the game resulting in a draw are 0.6. 0.1 or 0.3 respectively. They play a series of three games. Calculate the probability that:
    a) The three games are drawn

    b) They win alternate games with Ivan winning the first one
    c) Exactly 2 of the 3 games are drawn
    d) Joe does not win a game

    I'm not sure if this is right but for
    a) i just did 0.3 ^ 3
    and for
    b) 0.6 * 0.1 * 0.6

    So yeah, if anyone could assist me on these, it would be great
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  2. #2
    MHF Contributor Unknown008's Avatar
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    a) P(\text{Three games are draw}) = P(\text{first game draw}) \times P(\text{second game draw}) \times P(\text{third game draw})

    So, yes, it's right!

    b) P(\text{alternate wins, Ivan wins first}) = P(\text{Ivan wins}) \times P(\text{Jo wins}) \times P(\text{Ivan wins})

    Yes, right.

    c) P(\text{2 games draw}) = P(DD\bar{D}) + P(D\bar{D}D) + P(\bar{D}DD)

    where D represents the even that the game is a draw.

    d) P(\text{Jo does not win}) = P(\text{Jo loses all game})   + P(\text{Jo loses 1 game and draws 2})+ P(\text{Jo loses 2 games and draw 1}) + P(\text{Jo draws all game})
    Last edited by Unknown008; November 2nd 2010 at 01:24 AM.
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