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Math Help - Probability

  1. #1
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    Probability

    Two students, Karl and Hanna, play a game in which they take it in turns to select a card, with replacement from a well-shuffled pack of 52 playing cards. The first person to select an ace wins the game. Karl has the first turn.

    ai.) Find the probability that Karl wins on his third turn.
    aii.) Show that the probability that Karl wins prior to his (n+1)th turn is \frac{13}{25}(1-\frac{12}{13}^{2n})
    aiii.) Hence find the probability that Karl wins the game.
    b.) If Karl and Hanna play this game seven times, find the probability that Karl will win more games than Hanna.


    Help I really do not know how to do this.
    I know that the probability of drawing an ace is 1/13.
    I thought this is also a binomial distribution as results are independent( with replacement ) and also, there are only two outcomes (draw an ace or not)

    So to attempt at ai.), I used the calculator, binompdf(3,1/13,3) but it is not the answer and I do not know what to insert for the x-value really, I thought it must be 3 since they ask for the probability Karl wins for his third turn. But then perhalps, they wanted to include winning the second and first turns too so I thought it will be P(x more than and equal to 3) so I should type in the calculator binomcdf(3,1/13,3) where I'm calculating P(x less than equal to 3) but the calculator only accepts more than equal to inequality signs and not less than equal to...

    I really am stuck for this question, can someone please help me clear this? Thank you so much and really appreciate your kindness.
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  2. #2
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    Re: Probability

    Yes, the probababilty of drawing an ace on any one draw is 1/13. The proability of drawing any card except an ace is 12/23. In order that "Karl wins on his third turn", karl must NOT draw an ace on his first two draws but draws an ace on the third: the probability of that is (12/23)(12/23)(1/13). Of course, Hanna must not draw an ace on the first or second draws. The probability of that is (12/23)(12/23). Therefore, the probability of Karl drawing an ace on the third draw of those is the product: [(12/23)(12/23)(1/13)][(12/23)(12/23)]= (12/23)^5(1/13)
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  3. #3
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    Re: Probability

    Thank you! (: Is this then a binomial distribution?
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