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Math Help - Difficult Probabilty Word Problems

  1. #1
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    Cool Difficult Probabilty Word Problems

    I got these obscure word problems today, I have a while to do them, but I wasn't given much direction and am not sure even where to start looking. They deal in probabilities and I believe it's going to involve some Calculus as my group was given these problems while taking calculus two.

    I'm not asking for a solution, I was just wondering if someone thinks they know how they could be done, if they could perhaps offer a little direction in an approach.

    Here they are:

    Jimmy and Dave have decided to create they're own duel. They decide to shoot one at a time firing one shot per turn at each other until one person is hit. Jimmy is the weaker shot with a 2/5 chance each shot of hittin Dave. Dave has a 2/3 chance of hitting Jimmy each shot. Jimmy, the weaker shot, will shoot first. What is the probability that Jimmy will win this game?
    (This one seems impossibly complicated.
    A board game consists of four positions labeled A, B, C, and D. Whenever you reach position A or B, you roll some dice whose results determine what position you will move to next. Ah, but these dice are loaded.

    From position A, there is a 1/4 chance of moving to position B, a 1/12 chance of moving to position C, and a 1/6 chance of moving to D.


    From position B, there is a 1/3 chance of staying at B, there is a 1/3 chance of moving to position A, a 1/6 chance of moving to position C, and a 1/6 chance of moving to D.

    Whenever you reach position C or D the game is over and you win some cake.

    Suppose you begin the game at position A, what is the probability you end the game at position C
    Please someone more schooled in probability please steer me in the right direction? I need to figure these out sometime in the future at least.
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  2. #2
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    The chance that Jimmy wins is simply a sum that forms a series.

    The probability that Jimmy hits in the first shoot is 2/5. The probability that he hits in the second shot is dependend on him missing in the first (1-2/5) and on Dave missing on the second (1-2/3) while Jimmy then must hit in the thirs. The prob for hitting on the third shot is dependend on Jimmy missing in the first and second shot and on Dave missing on the first and second while Jimmy must hit in the third round and so forth. This gives you a sum pf the probability of all events which you must set up for a yet unspecified n where n is an index variable to indicate all cases of hits in any periods before or in n but not later. Let this n then converge to infinity (then there is no later event) and see that the sum converges to a specific value. This is your total probability that Jimmy wins.

    For 2: Again, just add up all the events that yield the outcome you are looking for. Your probabilty of reaching C is the prob of moving from A to B to C plus the probability of moving from A to C. Formulate this in a sum wich considers this possible infinite loop and see to which value your series converges.

    Try to solve this and ask again if you have any problems.
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  3. #3
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    Hello, CoryG89!

    Jim and Dave have decided to create their own duel.
    They decide to shoot one shot per turn at each other until one person is hit.
    Jim is the weaker shot with a 2/5 chance of hitting Dave.
    Dave has a 2/3 chance of hitting Jim.

    Jim, the weaker shot, will shoot first.
    What is the probability that Jim will win this game?

    We have these probabilities:

    . . \begin{Bmatrix}<br />
P(\text{Jim hits}) &=& \frac{2}{5}  \\ \\[-3mm] P(\text{Jim miss}) &=& \frac{3}{5}\end{Bmatrix} \quad\begin{Bmatrix}P(\text{Dave hits}) &=& \frac{2}{3} \\ \\[-3mm] P(\text{Dave miss}) &=& \frac{1}{3} \end{Bmatrix}


    Jim could win on his first shot: . P(\text{Jim wins on 1st shot}) \:=\:\frac{2}{5}


    If Jim misses his first shot: . P(\text{Jim miss}) = \frac{3}{5}
    . . then Dave must miss his first shot: P(\text{Dave miss}) = \frac{1}{3}
    . . then Jim must hit on his second short: P(\text{Jim hits}) = \frac{2}{5}
    Hence: . P(\text{Jim wins on 2nd shot}) \:=\:\frac{3}{5}\cdot\frac{1}{3}\cdot\frac{2}{5} \;=\;\left(\frac{1}{5}\right)\frac{2}{5}


    For Jim to win on his third shot, the follwing must happen:
    . . Jim misses his 1st shot: . \frac{3}{5}
    . . Dave misses his 1st shot: . \frac{1}{3}
    . . Jim misses his 2nd short: . \frac{3}{5}
    . . Dave misses his 2nd shot: . \frac{1}{3}
    . . Jim hits on his 3rd shot: . \frac{2}{5}
    Hence: . P(\text{Jim wins on 3rd shot}) \:=\:\frac{3}{5}\cdot\frac{1}{3}\cdot\frac{3}{5}\c  dot\frac{1}{3}\cdot\frac{2}{5} \:=\:\left(\frac{1}{5}\right)^2\frac{2}{5}


    We find that: . P(\text{Jim wins on 4th shot}) \;=\;\left(\frac{1}{5}\right)^3\frac{2}{5} . . . and so on.


    \displaystyle \text{So: }\;P(\text{Jim wins}) \;=\;\frac{2}{5} + \left(\frac{1}{5}\right)\frac{2}{5} + \left(\frac{1}{5}\right)^2\frac{2}{5} + \left(\frac{1}{5}\right)^3\frac{2}{5} + \hdots

    . . . . . . . . . . . . . =\;\dfrac{2}{5}\,\underbrace{\bigg[1 + \frac{1}{5} + \left(\frac{1}{5}\right)^2 + \left(\frac{1}{5}\right)^3 + \hdots \bigg]}_{\text{geometric series}}


    The geometric series has first term a = 1,\,\text{ common ratio }r = \frac{1}{5}
    . . Its sum is: . S \:=\:\dfrac{1}{1-\frac{1}{5}} \:=\:\dfrac{1}{\frac{4}{5}} \:=\:\dfrac{5}{4}


    Therefore: . P(\text{Jim wins}) \;=\;\dfrac{2}{5}\cdot\dfrac{5}{4} \;=\;\dfrac{1}{2}

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