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Math Help - Playing Card Probability

  1. #1
    Junior Member
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    Playing Card Probability

    How do I go about solving this?

    4 cards are drawn from a pack of 52 playing cards without replacment. What is the probability that two are black and two are red?
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  2. #2
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    The number of ways of getting 2 red cards from the 26 is \left(\begin{array}{cc}26\\2\end{array}\right) = 325
    And similiarly for black cards.

    The total number of different ways of taking 4 cards from 52 is \left(\begin{array}{cc}52\\4\end{array}\right) = 270725

    So the probability of getting 2 red cards, and 2 black cards, is;

    <br />
\frac{ \left(\begin{array}{cc}26\\2\end{array}\right) \left(\begin{array}{cc}26\\2\end{array}\right)}{\l  eft(\begin{array}{cc}52\\2\end{array}\right)}=\fra  c{325*325}{270725}=0.3901
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  3. #3
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    Thank you very muck for your help, however I do not fully understand the method you used to solve it... as in I have not come across the bracket things before
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  4. #4
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    Hello, Paulo1913

    Does this version look familiar to you?



    The number of ways of getting 2 red cards from the 26 is: . {\color{blue}_{26}C_2}  \,=\, 325
    And similiarly for black cards: . {\color{blue}_{26}C_2} \,=\,325

    The total number of different ways of taking 4 cards from 52 is: . {\color{blue}_{52}C_4}  \:=\: 270,\!725

    So the probability of getting 2 red cards, and 2 black cards, is: . \frac{(_{26}C_2)(_{26}C_2)}{_{52}C_4} \;=\; \frac{325\cdot325}{270,\!725} \;=\;0.3901
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