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Math Help - Probability Problem

  1. #1
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    Probability Problem

    Hello All, Please Help.

    1.) There are 3 white marbles, 4 blue marbles and 5 red marbles in a bag. If they are picked at random with replacement, find the probability of:

    a.) The first being a blue and the second a red. I got: 5/36



    b.) The probability they are both whites. I got 1/16

    and this one:

    If Pr(A) = 0.5
    Pr(B) = 0.4
    and Pr(A and B) = 0.1

    Find Pr(A' and B') i got: 0.2
    Find Pr(A' and B) i got: 0.3
    Find Pr(A or B) i got: 0.8

    THANKS!
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  2. #2
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    a)
    the probability that you get a blue then a red is:

     \frac{4}{12} \times \frac{5}{12} = \frac{15}{144}

    both whites are:

     \frac{3}{12} \times \frac{3}{12} = \frac{9}{144}

    and for b)

    I'm assuming A' means the complement of A? so we get

    P(A^c \cap B^c) = 1-P(A \cap B) = 1-0.1 = 0.9

    P(A^c \cap B) = 0.4-0.1 = 0.3

    P(A \cup B) = 0.5+0.4-0.1 = 0.8
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  3. #3
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    Quote Originally Posted by andrew2322 View Post
    Hello All, Please Help.

    1.) There are 3 white marbles, 4 blue marbles and 5 red marbles in a bag. If they are picked at random with replacement, find the probability of:

    a.) The first being a blue and the second a red. I got: 5/36



    b.) The probability they are both whites. I got 1/16

    and this one:

    If Pr(A) = 0.5
    Pr(B) = 0.4
    and Pr(A and B) = 0.1

    Find Pr(A' and B') i got: 0.2
    Find Pr(A' and B) i got: 0.3
    Find Pr(A or B) i got: 0.8

    THANKS!


    (Lucky you boldfaced the with replacement )
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  4. #4
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    Quote Originally Posted by lllll View Post
    a)
    the probability that you get a blue then a red is:

     \frac{4}{12} \times \frac{5}{12} = \frac{15}{144} Mr F says: *Ahem* That's 20/144 (= 5/36).

    both whites are:

     \frac{3}{12} \times \frac{3}{12} = \frac{9}{144} Mr F says: = 1/16.

    and for b)

    I'm assuming A' means the complement of A? so we get

    P(A^c \cap B^c) = 1-P(A \cup B) = 1-(0.5+0.4-0.1) = 0.2

    P(A^c \cap B) = 0.4-0.1 = 0.3

    P(A \cup B) = 0.5+0.4-0.1 = 0.8
    By the way, for b) a Karnaugh table is a fast and efficient way of doing this problem. You'll find a few examples in this forum. For example: http://www.mathhelpforum.com/math-he...3-prob-qn.html
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  5. #5
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    hey thanks guys

    Quote Originally Posted by lllll View Post
    a)
    the probability that you get a blue then a red is:

     \frac{4}{12} \times \frac{5}{12} = \frac{15}{144}

    both whites are:

     \frac{3}{12} \times \frac{3}{12} = \frac{9}{144}

    and for b)

    I'm assuming A' means the complement of A? so we get

    P(A^c \cap B^c) = 1-P(A \cap B) = 1-0.1 = 0.9

    P(A^c \cap B) = 0.4-0.1 = 0.3

    P(A \cup B) = 0.5+0.4-0.1 = 0.8
    so which one is right? Mr.F or IIII lol im so confused :P
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  6. #6
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    Quote Originally Posted by andrew2322 View Post
    so which one is right? Mr.F or IIII lol im so confused :P
    (4)(5) is 20, not 15. So clearly lllll made a small arithmetic slip up in part a.

    All your answers are correct. No need for confusion
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