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Math Help - Basic Probability Questions, Part 1

  1. #1
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    Basic Probability Questions, Part 1

    Note to this first part: I don't need help on these, I figured them out. Thanks, though

    ************

    I was hoping someone could help me with these, as I do not understand them at all. I work graveyard and so I end up missing class. My professor doesn't have much patience to work with us individually, and we have an exam on Wednesday.

    Thank you for any help you can provide.


    1.) A bag has 12 marbles of equal size - 5 red, 4 green, 3 blue. Two marbles are removed at random. Compute the following:

    a. the event that the first marble is blue and the second is red.
    b. the event that the 2 marbles are NOT both blue.
    c. if three marbles are selected in order what is the probability that all three marbles selected are blue?

    2.) Use combinations in answering each of the following questions and change your answers to the form 1/x where x is rounded to an integer: A seven card hand is dealt from an ordinary deck. What is the probability that...

    a. the hand contains two clubs and three diamonds?
    b. the hand contains exactly three aces?
    Last edited by amysteriousman; May 11th 2010 at 11:13 AM. Reason: Don't need help
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  2. #2
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    Hello, amysteriousman!

    2) Use combinations in answering each of the following questions
    and change your answers to the form \tfrac{1}{x} where x is rounded to an integer.

    A seven-card hand is dealt from an ordinary deck.
    There are: . {52\choose7} \;=\;133,\!784,\!560 possible hands.


    What is the probability that:

    (a) the hand contains exactly two Clubs and three Diamonds?
    There are: .13 Clubs, 13 Diamonds, and 26 Others.

    We want: . .2 Clubs, 3 Diamonds, and 2 Others.

    There are: . {13\choose2}{13\choose 3}{26\choose2} \:=\:7,\!250,\!100 ways.

    The probability is: . \frac{7,\!250,\!100}{133,\!784,\!560} \;=\;\frac{27,\!885}{515,\!556}  \;=\;\frac{1}{18.45...} \;\approx\;\frac{1}{18}




    b. the hand contains exactly three Aces?
    There are: .4 Aces, 48 Others.
    We want: . 3 Aces, 4 Others.

    There are: . {4\choose3}{48\choose4} \:=\:778,\!320 ways.

    The probability is: . \frac{778,\!320}{113,\!784,\!560} \;=\;\frac{9}{1547} \;=\;\frac{1}{171.88...} \;\approx\;\frac{1}{172}

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