Results 1 to 5 of 5

Math Help - 31.14 - Playing cards

  1. #1
    Junior Member
    Joined
    Apr 2008
    Posts
    56

    Smile 31.14 - Playing cards

    Two cards are sequentially drawn (without replacement) from a standard deck of 52 cards (well shuffled). Let A be the event that the two cards have the same value. Let B be the event that the first card drawn is an Ace. Are these events independent?

    Definition: Two events are said to be "independent" provided that P(A intersection B) = P(A)P(B).
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Bar0n janvdl's Avatar
    Joined
    Apr 2007
    From
    Meh
    Posts
    1,630
    Thanks
    6
    Quote Originally Posted by ccdelia7 View Post
    Two cards are sequentially drawn (without replacement) from a standard deck of 52 cards (well shuffled). Let A be the event that the two cards have the same value. Let B be the event that the first card drawn is an Ace. Are these events independent?

    Definition: Two events are said to be "independent" provided that P(A intersection B) = P(A)P(B).

    For number 1:

    Well the first card can be any card... The second card just needs to be the same value as the first one.

    So for example, if you drew one of the 4 Aces, then there are 3 Aces and 51 cards in total remaining. You thus have a \frac{3}{51} chance of drawing the same value card again.



    I'll think a bit more about number 2.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Number two..

    P(A intersection B)=probability that the first card is an ace AND the first two cards have the same value.
    This is equivalent to "probability that the two first cards are aces".

    First card is an ace -> 4/52
    Second card is an ace -> 3/51
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Apr 2008
    Posts
    56

    Smile

    Yes - I understand the probabilities of both situation (thanks for clarifying). Now, however, I'm concerned with the intersection of the two probabilities. I don't know how to think of the intersection of the two constituent probabilities. Thanks.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Apr 2008
    Posts
    56
    Gotcha! I see now how to look at it! thanks!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Playing cards
    Posted in the Statistics Forum
    Replies: 8
    Last Post: August 14th 2011, 07:31 AM
  2. 12 Cards are delath from 52 playing cards
    Posted in the Statistics Forum
    Replies: 7
    Last Post: November 13th 2009, 10:36 AM
  3. Playing cards-again :-)
    Posted in the Statistics Forum
    Replies: 1
    Last Post: October 22nd 2009, 01:04 PM
  4. Playing cards
    Posted in the Statistics Forum
    Replies: 5
    Last Post: October 22nd 2009, 12:20 PM
  5. three playing cards
    Posted in the Algebra Forum
    Replies: 1
    Last Post: May 11th 2007, 11:22 AM

Search Tags


/mathhelpforum @mathhelpforum