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Math Help - Pack of Cards

  1. #1
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    Pack of Cards

    3 cards are given from a pack of 52 playing cards. What is the probability that exactly one is a king?

    (4 kings in a pack)

    I did:

    4/52 * 48/51 * 47/50 * 4

    = 0.2...

    Is this correct?
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  2. #2
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    Quote Originally Posted by Aquafina View Post
    3 cards are given from a pack of 52 playing cards. What is the probability that exactly one is a king? I did:

    4/52 * 48/51 * 47/50 * 4
    = 0.2... Is this correct?
    I don't see what is being done above.

    Answer
    \frac{_4C_1\cdot _{48}C_2}{_{52}C_3}
    That gives the same answer and makes it clear how it works.
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  3. #3
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    Hello, Aquafina!

    3 cards are drawn from a pack of 52 playing cards.
    What is the probability that exactly one is a King?

    There are: 4 Kings and 48 Others.

    Three cards are drawn: X. Y, Z.

    The King can be in any of {\color{red}3} positions.
    . . (King-Other-Other), (Other-King-Other), (Other-Other-King)

    Suppose it is: (King-Other-Other)
    . . The probability is: . \frac{4}{52}\cdot\frac{48}{51}\cdot\frac{47}{50} \;=\;{\color{red}\frac{376}{5525}}

    Therefore: . P(\text{one King}) \;=\;3\cdot\frac{376}{5525} \;=\;\boxed{\frac{1128}{5525}}

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  4. #4
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    Quote Originally Posted by Soroban View Post
    Hello, Aquafina!


    There are: 4 Kings and 48 Others.

    Three cards are drawn: X. Y, Z.

    The King can be in any of {\color{red}3} positions.
    . . (King-Other-Other), (Other-King-Other), (Other-Other-King)

    Suppose it is: (King-Other-Other)
    . . The probability is: . \frac{4}{52}\cdot\frac{48}{51}\cdot\frac{47}{50} \;=\;{\color{red}\frac{376}{5525}}

    Therefore: . P(\text{one King}) \;=\;3\cdot\frac{376}{5525} \;=\;\boxed{\frac{1128}{5525}}

    Thank you

    Quote Originally Posted by Plato View Post
    I don't see what is being done above.

    Answer
    \frac{_4C_1\cdot _{48}C_2}{_{52}C_3}
    That gives the same answer and makes it clear how it works.
    Sorry I meant what Soroban did, but wrote 4 at the end by mistake instead of 3.

    Why is it that the binomial coefficient calculation works?

    Is it like:

    (Number of ways of choosing 1 king out of 4)*(Number of ways of choosing 2 others out of 58)/(Number of ways of choosing 3 cards out of 52)

    Is this conditional probability?
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