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Thread: When to use which?

  1. #1
    wby
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    When to use which?

    if 5 card are dealt from 52 cards, determine the probabilty of getting 3 red cards and 2 black face cards.

    The answer is 26C3(6C2)/52C5, but I got 52C5(1/2)^5(6/52)^2, how do you know when to use which?

    Thank you!
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  2. #2
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    Hello, wby!

    If 5 cards are dealt from 52 cards,
    determine the probabilty of getting 3 red cards and 2 black face cards.

    The answer is: .$\displaystyle \frac{\left(_{26}C_3\right)\left(_6C_2\right)}{_{5 2}C_5}$

    but I got: .$\displaystyle \left(_{52}C_5\right)\left(\frac{1}{2}\right)^5\le ft(\frac{6}{52}\right)^2$ . ??

    How do you know when to use which?
    Sorry, but everything you wrote is incorrect . . .


    That $\displaystyle _{52}C_5$ is the number of possible 5-card hands.

    . . It belongs in the denominator.


    When they state "Five cards are dealt",

    . . we generally assume it is without replacement.

    So the probability of getting a certain color is not $\displaystyle \frac{1}{2}$ each time,
    . . and the probability of getting a black face card is not $\displaystyle \frac{6}{52}$ each time.


    Their answer is correct.

    You want to draw 3 red cards from the available 26 red cards.
    . . There are: .$\displaystyle _{26}C_3$ ways.

    You want to draw 2 black face cards from the available 6 black face cards.
    . . There are: .$\displaystyle _6C_2$ ways.

    . . Hence, there are: .$\displaystyle \left(_{26}C_3\right)\left(_6C_2\right)$ ways to draw 3 red cards and 2 black face cards.

    And that is over $\displaystyle _{52}C_5$, the number of possible 5-card hands.

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