Anonymous
Anonymous asked in Science & MathematicsMathematics · 6 years ago

A person removes two aces and a king from a deck of 52 playing cards, and draws, without replacement, two more cards from the deck.?

Find the probability that the person will draw two aces, two kings, or an ace and a king?????

1 Answer

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  • J. J..
    Lv 7
    6 years ago
    Favorite Answer

    Ok so 2 aces and a king are removed so that leaves 52 - 3 = 49 cards in the deck

    AA

    ----

    There are 49 cards in the deck and now there are only two aces. Therefore

    P(first card is an ace) = 2/49

    As we do not replave the ace, there are now 48 cards in the deck and there is only 1 ace

    Therefore the probability that the second card is also an ace, given that the first was an ace, = 1/48

    Therefore as we need both these events to occur, you MULTIPLY the individual probabilities

    2/49 * 1/48 = 2 / 2352

    = 1 / 1176

    KK

    ----

    There are 49 cards and 3 kings

    P(first is a king) = 3/49

    There now remain 48 cards and 2 kings

    P(second is a king), given the first is also a king = 2/48

    P(two kings) = 3/49 * 2/48 = 6/2352

    = 1 / 392

    A&K in either order

    --------------------------

    There are two ways that you can get an ace and a king

    Ace followed by king

    King followed by ace

    Ace then king = 2/49 * 3/48 = 6/2352 = 1/392

    King then ace = 3/49 * 2/48 = 6/2352 = 1/392

    Therefore P(ace and a king in either order) = 1/392 + 1/392 = 2/392 = 1/196

    FINALLY

    P(two aces or two kings or anace and a king) = P(two aces) + P(2 kings) + P(an ace and a king)

    = 1/1176 + 1/392 + 1/196

    LCM = 1176

    Therefore = (1 + 3 + 6) / 1176

    = 10/1176

    = 5/588

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