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ALGEBRA II

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Using a standard deck of cards, find the probability that:

1) You choose two cards and one is a face card and the other is black

2) A black jack and then an ace are drawn

  • ALGEBRA II - ,

    I'll solve #2 first, where the two cards requested are mutually exclusive.
    Since there are two black jacks, and 4 aces, we have
    P(BJ,A)=2/52*(4/51)=8/2652=2/663

    ====================================

    For problem #1, we denote the events as:
    B=black card, B'=red card
    F=face card, F'=non-face card

    The possible successful outcomes can be split into two stages, first and second card, denoted by the couple ( , ).

    First a face card then a black card:
    P(B'F,B)=(6/52)(26/51) (red face,black)
    P(BF,BF')=(6/52)(20/51) (black face, black non-face)
    Note: (black face, black face) is included in the next case.
    Then a black card followed by a face card:
    (BF,F)=(6/52)(11/51) (black face,any face)
    (BF',F)=(20/52)(12/51) (black non-face, any face)

    All these add up to 97/442.

    We can also calculate the unsuccessful outcomes and subtract from 1 to get the successful outcomes.

    The unsuccessful outcomes are:
    P(F',F')=(40/52)(39/52) (two non-face)
    P(B',B')=(26/52)(25/52) (two reds)
    P(BF,B'F')=(6/52)(20/52) (black face,red non-face)
    P(B'F',BF)=(20/52)(6/52) (red non-face, black face)

    Out of this, we have double-counted, and therefore should subtract
    P(F'B',F'B')=(20/52)(19/51) (both cards red non face)

    Summing them all up we get 345/442 for the probability of unsuccessful outcomes.
    Subtract from 1 to get 97/442 as the probability of successful outcomes. This is the same probability as we calculated above.

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