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Thread: Quick probability question- need assistance!

  1. #1
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    Default Quick probability question- need assistance!

    Here it is-

    Alice and Bob play a card game. Alice wins 70% of the time and there are no ties. If they play 5 games what is the probability that Alice wins at least three more games than Bob?


    My answer:

    5C4 * (1/7)

    +

    5C5 * (1/7)

    It doesnt seem 100% correct, but Im not too sure here...all help is appreciated!

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    If the probability is 70%, how come you are using 1/7?

    What's the answer? Maybe I can work backwards. This is a tough probability question.

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    Also you have to inlcude three in your calculations as she wins AT LEAST 3 MORE games than Bob.

    I used the binomcdf function in the catalog of your calculator and entered 1-binomcdf(5, 0.70, 2) which gave me 0.83692 for an answer.

    5 being the number of trials or games, 0.70 is the probability of success, and 2 is the desired number of successes(in this case using binomcdf we want to take everything away from 1 that we do not want, we do not want the probability of Alice winning 0, 1, or 2 games--by subtracting these from one we are left with the probability of Alice winning 3, 4, or 5 games). binomcdf adds up all the probabilities of Alice winning 0, 1, or 2 games and subtracts them from 1, which leaves you with the probability of Alice winning 3, 4, or 5 games more than Bob.

    Hope that makes some sort of sense. Has your teacher not showed you the binompdf and binomcdf functions yet?

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    Figured it out. What I meant was 7/10 not 1/7.

    It worked out to the following:

    5C5 * (7/10)^5 * (3/10)^0

    +

    5C4 * (7/10)^4 * (3/10)^1

    (Or something like that, cant remember).

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    Also bignerd, you misread the question as it states that Alice wins "THREE MORE games" than Bob, meaning that with 5 games total, there are only two possible choices- Alice wins 4 games (4-1=3) or Alice wins 5 games (5-0=5).

    But thanks for the attempt, turned out it was a simple formula handed out on a sheet that I didnt look at!


    (Edit: I had written in this post "Three OR More" which was wrong and not the way I answered the Q- Fact is, question states three more)
    Last edited by rinny; 07-31-2010 at 10:38 AM.

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    If Alice wins three or more games, that means she could win three games, four games or five games.

    Winning "three or more" games is not the same as "more than three" games which according to your math above, is what you understood the question as and if you answer it in that fashion you will be incorrect.

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    No, read it again. It is not at least THREE OR MORE games than Bob, but rather at least THREE MORE games than bob. Nowhere in the question does it say three OR more.

    There is no OR, it is THREE MORE meaning all options out of 5 that she wins three MORE than Bob.

    A B Is this an option?
    0 5 No (0-5=-5 and -5<=3)
    1 4 No (1-4=-3 and -3<=3
    2 3 No (2-3= -1 and -1<=3)
    3 2 No (3-2=1 and 1<=3)
    4 1 YES (4-1=3 and 3>= 3)
    5 0 YES (5-0=5 and 5>=3)

    The condition is: Is the difference between Alices and Bobs game greater than or equal to 3? The only possibilities are Alice wins 4 or 5.
    Last edited by rinny; 07-31-2010 at 10:34 AM.

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