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Math Help - [SOLVED] Depend event 2

  1. #1
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    Thumbs down [SOLVED] Depend event 2

    this is another problem for me.. i always stuck on playing card problems:
    hope some one can guide me on this

    Question:
    From a pack of 52 playing card, two card are selected without replacement. What is the probability that:
    (a) both card are diamond (ans: 1/17)
    (b) both card are same suit(ans: 4/17)
    (c) one card is a spade and the other a club (ans: 13/102)
    (d) both card are different suits? (ans: 13/17)

    thank for your time
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  2. #2
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    Hello,nikk!

    From a pack of 52 playing card, two card are selected without replacement.
    What is the probability that:

    (a) both cards are Diamonds? (ans: 1/17)

    The first card must be a Diamond: .probability = \tfrac{13}{52}

    . . There are 13 Diamonds among the 52 cards.

    The second card must be a Diamond: .probability = \tfrac{12}{51}

    . . There are 12 remaining Diamonds among the remaining 51 cards.

    Therefore: . P(\text{both Diamonds}) \:=:\frac{13}{52}\cdot\frac{12}{51} \:=\:\frac{1}{17}



    (b) both cards are same suit ? (ans: 4/17)

    The first card can be ANY card: . \tfrac{52}{52} \,=\,1

    The second card must match that suit: . \tfrac{12}{51}

    . . There are 12 remaining cards of that suit.

    Therefore: . P(\text{same suit}) \:=\:1\cdot\frac{12}{51} \:=\:\frac{4}{17}



    (c) one card is a Spade and the other a Club? (ans: 13/102)
    There are two cases to consider . . .

    (1) The first is a Spade: . \tfrac{13}{52}
    . . .The second is a Club: . \tfrac{13}{51}

    (2) The first is a Club: . \tfrac{13}{52}
    . . .The second is a Spade: . \tfrac{13}{51}


    Therefore: . P(\text{Spade \& Club}) \:=\:\frac{13}{52}\cdot\frac{13}{51} + \frac{13}{52}\cdot\frac{13}{51} \;=\;\frac{13}{102}



    (d) both cards are different suits? (ans: 13/17)

    The first card can be ANY card: . \tfrac{52}{52} \,=\,1

    The second card must not match the first: . \frac{39}{51}

    . . There are 39 cards of another suit among the 51 remaining cards.

    Therefore: . P(\text{different suits}) \:=\:1\cdot\frac{39}{51} \:=\:\frac{13}{17}

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