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Math Help - Probability drawn cards

  1. #1
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    Probability drawn cards

    A stack of cards contains four Aces, three Kings, two Queens and six other cards. Seven cards are randomly drawn from the stack, without replacement.
    (a) Find the probability that the drawn cards contain at least one Ace.

    no problem.. Pr = 1 - Pr(0 \ aces) = 1- \frac{ \left(\begin{array}{cc}4\\0\end{array}\right) \left(\begin{array}{cc}11\\7\end{array}\right)}{\l  eft(\begin{array}{cc}15\\7\end{array}\right)}=1-\frac{330}{6435}=0.9487

    (b) Find the probability that the drawn cards contain exactly two Aces if it is known that they contain no Kings or Queens.

    Now im not sure what it means by 'if it is known that they contain no Kings or queens. Should I just use
    \frac{ \left(\begin{array}{cc}4\\2\end{array}\right) \left(\begin{array}{cc}6\\5\end{array}\right)}{\le  ft(\begin{array}{cc}15\\7\end{array}\right)}
    I am thinking this doesn't really make sense, since it is known that it doens't contain any kings or queens...

    Thanks,
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  2. #2
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    Quote Originally Posted by Robb View Post
    A stack of cards contains four Aces, three Kings, two Queens and six other cards. Seven cards are randomly drawn from the stack, without replacement.
    (a) Find the probability that the drawn cards contain at least one Ace.

    no problem.. Pr = 1 - Pr(0 \ aces) = 1- \frac{ \left(\begin{array}{cc}4\\0\end{array}\right) \left(\begin{array}{cc}11\\7\end{array}\right)}{\l  eft(\begin{array}{cc}15\\7\end{array}\right)}=1-\frac{330}{6435}=0.9487

    (b) Find the probability that the drawn cards contain exactly two Aces if it is known that they contain no Kings or Queens.

    Now im not sure what it means by 'if it is known that they contain no Kings or queens. Should I just use
    \frac{ \left(\begin{array}{cc}4\\2\end{array}\right) \left(\begin{array}{cc}6\\5\end{array}\right)}{\le  ft(\begin{array}{cc}15\\7\end{array}\right)}
    I am thinking this doesn't really make sense, since it is known that it doens't contain any kings or queens...
    Therefore, how about \frac{ \left(\begin{array}{cc}4\\2\end{array}\right) \left(\begin{array}{cc}6\\5\end{array}\right)}{\le  ft(\begin{array}{cc}10\\7\end{array}\right)}?
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  3. #3
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    Hello, Robb!

    Plato is absolutely correct . . .


    A stack of cards contains four Aces, three Kings, two Queens and six other cards.
    Seven cards are randomly drawn from the stack, without replacement.

    (b) Find the probability that the seven cards contain exactly two Aces
    if it is known that they contain no Kings or Queens.

    The stack had: . \{\text{4 Aces, 3 Kings, 2 Queens, Others}\}

    Since the seven cards had no Kings or Queens, the cards were drawn from:
    . . \{\text{4 Aces},\;{\color{red}\rlap{///////}}\text{3 Kings},\;{\color{red}\rlap{////////}}\text{2 Queens},\;\text{6 Others}\} . = \;\{\text{4 Aces, 6 Others}\}

    There are: . {10\choose7} possible outcomes . . . and {4\choose2}{6\choose5} ways to get exactly two Aces.

    Got it?

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  4. #4
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    Cheers guys, Just wasn't exactly sure how to deal with the fact we knew it wasn't K or Q

    Thanks,
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