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Math Help - *Probability Problem - please help!*

  1. #1
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    *Probability Problem - please help!*

    I need the answer/brief explanation for this question by tonight! Any replies would be greatly appreciated!

    In a particular game, a fair die is tossed. If the number rolled is either a four or five, you win $1. If the number rolled is a six, you win $4. If the number rolled is a one, two, or three, you win nothing. You are going to play the game twice.

    The probability that you win at least $1 both times is
    (a) 1/2
    (b) 4/36
    (c) 1/36
    (d) 1/4
    (e) 3/4
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  2. #2
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    Re: *Probability Problem - please help!*

    Write down all of the numbers that you can roll that result in you winning at least $1 for a single roll of the die. Then, figure out the probability of getting a successful roll for the first game. Then, multiply it by the probability of getting a successful roll for the second game.

    So, if you were asked the probability that you would win exactly $1 both times, you have two numbers you can roll out of six each time you roll the die. So, the probability is \dfrac{2}{6}\cdot \dfrac{2}{6} = \dfrac{1}{9}.

    Now, alter the problem when you are looking for the probability that you must win at least $1 both times. How many numbers can you roll out of six?
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  3. #3
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    Re: *Probability Problem - please help!*

    @SlipEternal, Okay so is this correct?:

    Numbers that you can roll that result in winning at least $1 for a single roll: four ($1), five ($1), and six ($4)
    The probability of that is therefore 3/6. So then you do 3/6 x 3/6 = 9/36 = 1/4. The answer is D.

    Can you tell me why my original answer was incorrect too?
    I first found the probability of winning nothing and then subtracted 1 from it. So, numbers that you can roll that result in winning nothing for a single roll: one, two, or three
    The probability of that is therefore 3/6. So then I did 1 - (3/6 x 3/6) = 3/4 or .75
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  4. #4
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    Re: *Probability Problem - please help!*

    Quote Originally Posted by newyorkcitylivin View Post
    @SlipEternal, Okay so is this correct?:

    Numbers that you can roll that result in winning at least $1 for a single roll: four ($1), five ($1), and six ($4)
    The probability of that is therefore 3/6. So then you do 3/6 x 3/6 = 9/36 = 1/4. The answer is D.

    Can you tell me why my original answer was incorrect too?
    I first found the probability of winning nothing and then subtracted 1 from it. So, numbers that you can roll that result in winning nothing for a single roll: one, two, or three
    The probability of that is therefore 3/6. So then I did 1 - (3/6 x 3/6) = 3/4 or .75
    The two games are independent. You can win at least a dollar on the first game or not and you can win at least a dollar on the second game or not. So, there are four outcomes relating to winning at least a dollar each game. They are:

    1. You do not win anything either game.
    2. You win at least a dollar the first game, but not the second.
    3. You do not win anything the first game, but you win at least a dollar the second.
    4. You win at least a dollar both games.

    1 minus the probability that you do not win anything either game gives the probability that you are in one of the outcomes 2, 3, or 4. You only wanted the probability of outcome 4.
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  5. #5
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    Re: *Probability Problem - please help!*

    @SlipEternal, OH wow that makes so much sense now. I see why the answer is 1/4 and not 3/4. Ahhh and to think this was the only question I missed on my AP Stats probability test. Thank you so much!
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