Results 1 to 2 of 2

Math Help - When to use which?

  1. #1
    wby
    wby is offline
    Newbie
    Joined
    Aug 2007
    Posts
    4

    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!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,804
    Thanks
    695
    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: . \frac{\left(_{26}C_3\right)\left(_6C_2\right)}{_{5  2}C_5}

    but I got: . \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 _{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 \frac{1}{2} each time,
    . . and the probability of getting a black face card is not \frac{6}{52} each time.


    Their answer is correct.

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

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

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

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

    Follow Math Help Forum on Facebook and Google+


/mathhelpforum @mathhelpforum