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Math Help - Probability

  1. #1
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    Smile Probability

    There are two players playing named John and Peter. The one who wins three games first is the winner of the match. So the max no. of games needed is 5. And John has probability of winning is 0.6 and assume that all games are independent.

    (a)Given that John has lost the first two games, Find the probability so that he can still be the winner of the match?
    (b) Given that John lost the first game, Find the probability that John will still be the winner?
    (c) Find the probability that the number of games needed is exactly 5?

    Sorry for so many questions but I have tried a lot and am unable to answer!
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  2. #2
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    Quote Originally Posted by santatrue View Post
    There are two players playing named John and Peter. The one who wins three games first is the winner of the match. So the max no. of games needed is 5. And John has probability of winning is 0.6 and assume that all games are independent.
    (a)Given that John has lost the first two games, Find the probability so that he can still be the winner of the match?
    (b) Given that John lost the first game, Find the probability that John will still be the winner?
    (c) Find the probability that the number of games needed is exactly 5?
    For part (a), what is the probability of this string, PPJJJ?

    For part (b), what is the probability of each of these strings, PJJJ, PPJJJ, PJPJJ, PJJPJ? [Add these four.]

    For part (c), what is the probability of each of these strings, PPJJJ, PJPJJ, PJJPJ, JPPJJ, JPJPJ, JJPPJ, JJPPP, JPJPP, JPPJP, PJJPP, PPJJP, PJPJP? [Add these 12.]
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  3. #3
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    Quote Originally Posted by Plato View Post
    For part (a), what is the probability of this string, PPJJJ?

    For part (b), what is the probability of each of these strings, PJJJ, PPJJJ, PJPJJ, PJJPJ? [Add these four.]

    For part (c), what is the probability of each of these strings, PPJJJ, PJPJJ, PJJPJ, JPPJJ, JPJPJ, JJPPJ, JJPPP, JPJPP, JPPJP, PJJPP, PPJJP, PJPJP? [Add these 12.]
    isnt is suppose to be Conditional Probability or you can also do this way
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  4. #4
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    Quote Originally Posted by santatrue View Post
    isnt is suppose to be Conditional Probability
    I used Conditional Probability!
    Given John looses the first two means that Peter wins the first two.
    Look at it this way, Conditional Probability just restricts our attention to a subset of the outcome space.
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  5. #5
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    Quote Originally Posted by santatrue View Post
    There are two players playing named John and Peter. The one who wins three games first is the winner of the match. So the max no. of games needed is 5. And John has probability of winning is 0.6 and assume that all games are independent.

    (a)Given that John has lost the first two games, Find the probability so that he can still be the winner of the match?
    (b) Given that John lost the first game, Find the probability that John will still be the winner?
    (c) Find the probability that the number of games needed is exactly 5?

    Sorry for so many questions but I have tried a lot and am unable to answer!
    (a) The Propability is \frac{PPJJJ}{PP} = JJJ

    (b) JJJPx4 =  4(0.6)^3(0.4)

    (c)  6(0.6)^3(0.4)^2+6(0.6)^2(0.4)^3
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  6. #6
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    Quote Originally Posted by Plato View Post
    I used Conditional Probability!
    Given John looses the first two means that Peter wins the first two.
    Look at it this way, Conditional Probability just restricts our attention to a subset of the outcome space.
    So shouldnt it be for the first question to find for JJJ only , why PP as he has already won
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  7. #7
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    Quote Originally Posted by santatrue View Post
    So shouldnt it be for the first question to find for JJJ only , why PP as he has already won
    He was Wrong
    If
    A = The Probability that Paul wins the first 2 games.
    B = The probaility that John wins overall.

    Then the probability of B given A is:
    P(B|A) = \frac{P(A and B)}{P(A)}= \frac{PPJJJ}{PP} = \frac{(0.4)^2(0.6)^3}{(0.4)^2} =  0.6^3
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  8. #8
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    Quote Originally Posted by santatrue View Post
    So shouldnt it be for the first question to find for JJJ only , why PP as he has already won
    Sorry that I did not make it clear that you do indeed divide by the given probability. I see how I misslead you.
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