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  1. You want to count the ways to get the desired hand with 3 aces and 2 kings. From the 4 aces, pick 3 of them: ...

    1 Answers · Science & Mathematics · 06/12/2016

  2. Event A: First card is Ace and fourth card is King. Event B: Third...

    2 Answers · Science & Mathematics · 05/12/2016

  3. Can you please clarify whether you must draw *exactly* one ace or *at least* one ace? Also, I'm not clear...s first consider the case that you *don't* get any aces. P(no aces) = 0.9 x 0.9 x 0.9 x 0.9...

    2 Answers · Science & Mathematics · 02/12/2016

  4. I take it that the three of a kind are not aces or kings, then. So there are 4 possible aces, 4 possible...

    1 Answers · Science & Mathematics · 30/11/2016

  5. Yes, the sample space could be A X B or B X A. Those are just sets of ordered pairs. There's nothing that requires A or B to "go first" in the ordered pair.

    1 Answers · Science & Mathematics · 30/11/2016

  6. Here's the logic. If you draw an ace (probability = 4/52 = 1/13), you get 9 points. If you draw...you should be getting 12 points for drawing an ace for the game to be fair. Got it?

    3 Answers · Science & Mathematics · 10/11/2016

  7. You answered your own question. Practice as much as you can.

    1 Answers · Science & Mathematics · 05/11/2016

  8. number of possible hands = C(52, 5) = 52! / (47! 5!) number of possible successful hands = C(50, 3) = 50! / (47! 3!) probability of success = C(50, 3) / C(52, 5) = [50! / (47! 3!)] / [52! / (47! 5!)] = (50! 5!) / (52! 3!) = (5)(4) / [(52)(51)] = 5 / [(13)(51)] = 5 / 663 ≈ 0.00754

    3 Answers · Science & Mathematics · 04/11/2016

  9. 1) P(first card drawn is an ace) = P(A) = 4/52 P(second card drawn is not...

    1 Answers · Science & Mathematics · 17/12/2016

  10. Search for practice problems online and do them!

    1 Answers · Science & Mathematics · 22/10/2016

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