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  1. #1
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    Probabilities Plz Help

    Hey
    If the odds in favor of Boris beating Elena in a chess game are 5 to 4, what is the probability that Elena will win an upset victory in a best-of-five chess tournament?

    Can someone give me an answer with explanation Plz.

    Thanx
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  2. #2
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    Hello, ballin_sensation!

    If the odds in favor of Boris beating Elena in a chess game are 5 to 4, what is the
    probability that Elena will win an upset victory in a best-of-five chess tournament?
    The probability of Elena winning a game is \frac{4}{9} . . . of losing, \frac{5}{9}


    There are three ways in which she can win three games.

    (1) She wins the first three games:
    . . . \left(\frac{4}{9}\right)^3 \:=\:\frac{64}{9^3}

    (2) She wins two of the first three games, then wins the fourth game.
    . . . {3\choose2}\left(\frac{4}{9}\right)^2\left(\frac{5  }{9}\right)\cdot\frac{4}{9} \;=\;\frac{960}{9^4}

    (3) She wins two of the first four games, then wins the fifth game.
    . . . {4\choose2}\left(\frac{4}{9}\right)^2\left(\frac{5  }{9}\right)^2\cdot\frac{4}{9} \;=\;\frac{9600}{9^5}


    Therefore: . P(\text{Elena wins}) \;=\;\frac{64}{9^3} + \frac{960}{9^4} + \frac{9600}{9^5} \:=\:\frac{23,424}{59,049} \;=\;0.396687497 \;\approx\;40\%

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  3. #3
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    Quote Originally Posted by ballin_sensation View Post
    Hey
    If the odds in favor of Boris beating Elena in a chess game are 5 to 4, what is the probability that Elena will win an upset victory in a best-of-five chess tournament?

    Can someone give me an answer with explanation Plz.

    Thanx
    This seems like a binomial probabilility so

    the prob of x successes in n trials is

    P(x)=\binom{n}{x}P^{x}(1-P)^{n-x}

    where P is the prob of success

    since we want an upset victory she needs to win 3,4, or 5 games

    so we can add P(3) +P(4)+P(5) or take 1 -P(0)-P(1)-P(2)

    Now lets set up the formula:

    The odds are 5 to 4 against her so P=\frac{4}{9}
    and they are playing the best of 5 so

    P(x)=\binom{5}{x}\left( \frac{4}{9}\right)^{x} \left(\frac{5}{9} \right)^{5-x}

    P(3)=\binom{5}{3}\left( \frac{4}{9}\right)^{3} \left(\frac{5}{9} \right)^{5-3}=\frac{5!}{3!2!} \cdot\left( \frac{4}{9}\right)^{3} \left(\frac{5}{9} \right)^{5-3}= 10 \frac{(4^3)(5^2)}{9^5} \approx .2710

    P(4)=\binom{5}{4}\left( \frac{4}{9}\right)^{4} \left(\frac{5}{9} \right)^{5-4}=\frac{5!}{4!1!} \cdot\left( \frac{4}{9}\right)^{4} \left(\frac{5}{9} \right)^{5-4}= 5 \frac{(4^4)(5^1)}{9^5} \approx .1083

    P(5)=\binom{5}{3}\left( \frac{4}{9}\right)^{5} \left(\frac{5}{9} \right)^{5-5}=\frac{5!}{5!0!} \cdot\left( \frac{4}{9}\right)^{5} \left(\frac{5}{9} \right)^{5-5}= 1 \frac{(4^5)(5^0)}{9^5} \approx .0173

    so P(3)+P(4)+P(5) \approx .3966
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  4. #4
    Behold, the power of SARDINES!
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    Dang!

    Too slow HAHA
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  5. #5
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    Thanx for the help guys, really appreciate it.
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