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Math Help - 2 probability questions...

  1. #1
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    Thumbs down 2 probability questions...

    1. You are dealt 5 cards from a standard 52 card deck, what is the condition probability that the hand contains no pairs and no spades?
    2. The probability of snow is 20%, and the probability of a traffic accident is 10%. Suppose further that the conditional probability of an accident, given that it snows, is 40%. What is the conditional probability that it snows, given that there is an accident?
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  2. #2
    Member Traveller's Avatar
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    1) \frac{(39)X(38-2)X(37-4)X(36-6)X(35-8)}{52X51X50X49X48}

    2) Is it 80% ?
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  3. #3
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    Hello, Sneaky!

    I agree with Traveller's answers.


    1. You are dealt 5 cards from a standard 52-card deck, .What is
    the probability that the hand contains no pairs and no Spades?

    I had to baby-talk my way through this one . . .


    We begin with a deck with 39 cards . . . only Hearts, Clubs, and Diamonds.

    The 1st card can be any of the 39 cards: . \boxed{39}\text{ choices}
    (Suppose it is the 3\heartsuit.)

    We draw a 2nd card from the remaining 38 cards in the deck,
    . . but we do not want the 3\clubsuit\text{ or }3\diamondsuit:\;\boxed{36}\text{ choices}
    (Suppose it is the 6\clubsuit.)

    We draw a 3rd card from the remaining 37 cards in the deck.
    . . but we do not want: 3\clubsuit,\:3\diamondsuit,\:6\heartsuit,\,6\diamo  ndsuit :\;\boxed{33}\text{ choices}
    (Suppose it is the 9\diamondsuit,)

    We draw a 4th card from the remaining 36 cards in the deck.
    . . We do not want: 3\clubsuit,\:3\diamondsuit,\:6\heartsuit,\:6\diamo  ndsuit,\:9\heartsuit,\:9\clubsuit:\;\boxed{30}\tex  t{ choices}
    (Suppose it is the J\heartsuit.)

    We draw a 5th card from the remaining 35 cards in the deck.
    . . We do not want: 3\clubsuit,\:3\diamondsuit,\:6\heartsuit,\:6\diamo  ndsuit,\:9\heartsuit,\:9\clubsuit,\:J\clubsuit,\:J  \diamondsuit: \;\boxed{27}\text{ choices}


    Hence, there are: . 39\cdot36\cdot33\cdot30\cdot27 \:=\:37,\!528,\!920 hands
    . . with no pairs and no Spades.

    And there are: . 52\cdot51\cdot50\cdot49\cdot48 \:=\:311,\!875,\!200 possible hands.


    \text{Therefore: }\;P(\text{no pairs and no Spades}) \;=\;\dfrac{37,\!528,\!920}{311,\!875,\!200} \;=\;\dfrac{8,\!019}{66,\!640}
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