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10 November, 21:40

What is the probability that all three cards are face cards when you do not replace each card before selecting the next card round your answer to the nearest tenth

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  1. 10 November, 22:04
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    Define three events A, B, C such that

    A = first card is a face card

    B = second card is a face card (assuming event A happens)

    C = third card is a face card (assuming events A & B happen)

    No replacements are made

    There are 3 face cards (J, K, Q) per suit. There are 4 suits. In total, there are 3*4 = 12 face cards out of 52 total

    P (A) = probability that event A happens

    P (A) = 12/52

    P (A) = 3/13

    After event A happens, we have 12-1 = 11 face cards out of 52-1 = 51 total

    P (B) = 11/51

    After B happens, we have 11-1 = 10 face cards out of 51-1 = 50 left over

    P (C) = 10/50

    P (C) = 1/5

    Based on how the events A, B, C are set up, we can form this equation

    P (A and B and C) = P (A) * P (B) * P (C)

    P (A and B and C) = (3/13) * (11/51) * (1/5)

    P (A and B and C) = 11/1105

    P (A and B and C) = 0.00995475113122

    Rounded to the nearest tenth, the answer is simply 0.0 which is a lousy answer in my opinion because it implies that the answer is truly 0 instead of something really small. Though to be fair, the result is very close to 0. A much better approach is to round to the nearest hundredth to get 0.01; though I would ask your teacher for clarification and guidance.

    note: a probability of 0 means that the event is impossible to happen, yet it is not impossible to pull out three face cards in a row. Is it unlikely? Yes. Because the result is roughly 0.00995 which is a really small decimal close to 0. But it's not 0 itself.
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