Math: Probability

Hi, could you please help me? I have been stick on this question for a week and I have tried getting help from my teacher and parents, but I still cant get it. TIA

A standard deck of 52 cards consists of 4 suits (hearts, clubs, diamonds, and spades). If a student deals 3 cards from a standard deck, determine the probability to the nearest hundredth, that at least one card is a heart. Use the method stated below:

A.) Direct Reasoning (2 marks)
B.) Indirect Reasoning (2 marks)

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  1. 52/4 = 13 hearts in the deck
    52-13 = 39 are not hearts

    probability that all 3 are H
    = 13/52 *12/51*11/50 = .01294

    probability that 2/3 are H
    =13/52 * 12/51* 39/50 = .04588
    {39 is number of non hearts}

    probability that 1/3 is H
    = 13/52 *39/51*38/50 = .14529
    add
    A) 0.2041

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  2. call h a heart
    call x another suit
    d = 52*51*50 = 132600
    combos we could get
    hhh =13*12*11/d = 1716/d
    hhx =13*12*39/d = 6084/d
    hxx =13*39*38/d = 19266/d
    hxh =13*39*12/d = 6084/d
    xxh =39*38*13/d = 19266/d
    xhx =39*13*38/d = 19266/d
    xhh =39*13*12/d = 6084/d
    add to get
    77766/d= 77766/132600 = .58647
    That is the hard way
    now
    to get ALL x, no hearts
    39/52*38/51*37/50 = 54834/d =.41353
    1 - .41353 = .58647 !!! whew, the easy way

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  3. When I did it the easy way, I got a different answer, so went back and plodded through the hard way.

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  4. P(some hearts) = 1-P(no hearts)
    = 1 - (39/52)(38/51)(37/50) = 0.58647

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