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Math Help - Help with probability problems!...

  1. #1
    Senior Member
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    Exclamation Help with probability problems!...

    1. The problem statement, all variables and given/known data
    1. Suppose we roll 10 fair 6-sided dice. What is the probability that there are exactly two 2's showing?
    2. Suppose we are dealt five cards from a standard 52-card deck. What is the probability that
    a) we get all 4 aces and the king of spades
    b) all 5 are spades
    c) we get no pairs (all are different values)
    d) a full house (3 of a kind and 2 of a kind)


    2. Relevant equations

    This one is correct:
    there are 2 pots
    in pot1 there is 5 red balls and 7 blue balls
    in pot2 there is 6 red balls and 12 blue balls
    3 balls are chosen randomely from each pot
    chances of all 6 balls to be same color = P(A)
    chances of all 6 balls to be red = P(B)
    chances of all 6 balls to be blue = P(C)

    P(A) = P(B or C)
    =P(B) + P(C) -0
    P(B)=|B|/|S| = |B|/(12choose3)(18choose3) =
    (5choose3)(7choose0)(6choose3)(12choose0)/(12choose3)(18choose3) = 5/4488
    P(C)=|C|/|S| = |C|/(12choose3)(18choose3) =
    (5choose0)(7choose3)(6choose0)(12choose3)/(12choose3)(18choose3) = 35/816
    P(A) = 5/4488 + 35/816 - 0 = 395/8976




    3. The attempt at a solution

    1.
    number of outcomes = 6^10 = 60466176
    10!/2!8! = 45
    so i get
    45/60466176

    2.
    number of outcomes = 52x51x50x49x48 / 5x4x3x2x1 = 2598960
    a) Here i have two different methods, i don't know if both are wrong or one is right...
    method 1
    (4choose1 * 4choose4 ) / 2598960 = 1/649740
    method 2
    ( (13choose1) * (13choose1) * (13choose1) * (13choose1) ) / 2598960 = ~0.066
    b)
    (13choose5)/2598960 = ~4.95x10^(-4)
    c)
    ( (4choose1)*(4choose1)*(4choose1)*(4choose1)*(4choo se1) ) /2598960 = ~3.95x10^(-4)
    d)
    (4choose3)*(4choose2) /2598960 = 1/108290
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  2. #2
    Super Member

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    Hello, Sneaky!

    2. Suppose we are dealt five cards from a standard 52-card deck.

    There are: . \displaystyle _{52}C_5 \;=\;{52\choose5} \;=\;\frac{52!}{5!\,47!} \;=\;2,598,960\text{ possible Poker hands.}



    What is the probability that

    a) we get all 4 Aces and the King of Spades

    There is one way to get the four Aces \{A\heartsuit,\:A\spadesuit,\:A\diamondsuit,\:A\cl  ubsuit\}
    . . and one way to get the K\spadesuit.

    \text{Therefore: }\;P(\text{4 Aces} \wedge K\spadesuit) \;=\;\dfrac{1}{2,\!598,\!960}




    b) all 5 are Spades

    There are: . \displaystyle {13\choose5} \:=\:1287\text{ ways to get 5 Spades.}

    \text{Therefore: }\:P(\text{5 Spades}) \;=\;\dfrac{1287}{2,\!598,\!960} \;=\;\dfrac{33}{66,\!640}




    c) We get no pairs (all are different values)

    The 1st card can be any card: . 52\text{ choices.}

    The 2nd card must not match the 1st card: . 48\text{ choices.}

    The 3rd card must not match the 1st or 2nd card: . 44\text{ choices.}

    The 4th card must not match the first 3 cards: . 40\text{ choices.}

    The 5th card must not match the first 4 cards: . 36\text{ choices.}


    But we have imparted an order to the five cards.

    Since their order does not matter, there are:

    . . \dfrac{52\cdot48\cdot44\cdot40\cdot36}{5!} \;=\; 1,\!317,\!888 \text{ hands with no pairs.}


    \text{Therefore: }\;P(\text{no pairs}) \;=\;\dfrac{1,\!317,\!888}{2,\!598,\!960} \;=\;\dfrac{2112}{4165}




    d) a Full House (3 of a kind and 2 of a kind)

    There are 13 choices for the value of the Triple.
    . . There are: . {4\choose3} = 4\text{ ways to get the Triple.}

    There are 12 choices for the value of the Pair.
    . . There are: . {4\choose2} = 6\text{ ways to get the Pair.}

    Hence, there are: . 13\cdot4\cdot12\cdot6 \:=\:3744\text{ Full Houses.}


    \text{Therefore: }\;P(\text{Full House}) \;=\;\dfrac{3744}{2,\!598,\!960} \;=\;\dfrac{6}{4165}
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  3. #3
    Senior Member
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    How can i complete the first question?
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