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Hesan

Two cards are drawn at random from a deck of well shuffled cards. Determine the probability that both the cards are a king and a queen.

  • one year ago
  • one year ago

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  1. Hesan
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    no

    • one year ago
  2. Hesan
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    two cards, remember

    • one year ago
  3. Nick2019
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    there are 4 suites of 13 cards each , so the probability of getting two cards in each suit is 2/13 there are four suits ,so the probability is 2\13*4 = 8\13 -(use calculator to make it to decimals !!)

    • one year ago
  4. Hesan
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    i m getting the wrong answer

    • one year ago
  5. terenzreignz
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    I'm not getting the question :( The two cards, are they supposed to be both kings or both queens or one king and one queen?

    • one year ago
  6. Benjammin
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    Well, there are 52 cards in a deck and 8 of them are Kings and Queens. So the probability of one random draw is 8/52. Assuming you draw one, there are 7/51 remaining, so you just need to multiply that out to get the answer. |dw:1341587716616:dw| ... by my calculations, about 2.11%

    • one year ago
  7. Hesan
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    @Benjammin: I did the same thing but getting the wrong answer

    • one year ago
  8. Zarkon
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    are you supposed to get one of each card..one king and one queen?

    • one year ago
  9. Hesan
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    @Benjammin: your calculations give the answer: 14/663

    • one year ago
  10. Hesan
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    @Zarkon: yes

    • one year ago
  11. Zarkon
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    \[\frac{{4\choose 1}{4\choose1}}{{52\choose 2}}\]

    • one year ago
  12. Zarkon
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    \[\frac{8}{663}\]

    • one year ago
  13. Hesan
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    yeah right.. How?

    • one year ago
  14. Zarkon
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    I gave the formula above

    • one year ago
  15. Zarkon
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    this is a classic hypergeometric distribution problem

    • one year ago
  16. Zarkon
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    you can look at it this way \(\frac{4}{52}\) for the first card and \(\frac{4}{51}\) for the second card but the order we pick them doesn't matter so the answer is \[2\cdot\frac{4}{52}\cdot\frac{4}{51}\]

    • one year ago
  17. Hesan
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    but there shoulb be 8 cards right? 4 for kings and 4 for queens

    • one year ago
  18. Zarkon
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    yes..hence the two 4's in the solution

    • one year ago
  19. Hesan
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    okey, means 8/52, now how did u get 4/51?

    • one year ago
  20. SmoothMath
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    The way I see it, it's simply: 8/52 * 4/51

    • one year ago
  21. Zarkon
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    if you choose the cards from the 8 (4K and 4Q) you might pick up 2 kings or 2 queens, but you only want one of each.

    • one year ago
  22. Zarkon
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    yes...same thing

    • one year ago
  23. Zarkon
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    pick from the 8...then don't pick a matching card...hence only 4 left to choose from

    • one year ago
  24. SmoothMath
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    Right, for the first card, a king OR a queen is a good outcome. However, assuming you pull one of those, for the second card, only one of those would be a good outcome.

    • one year ago
  25. Hesan
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    no i dont get it.. i was thinking it 7/51

    • one year ago
  26. Hesan
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    see we've drawn 1 card, now 7 are left, so it would be 7/51

    • one year ago
  27. Zarkon
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    but by doing that you could get a matching pair...2 kings or 2 queens

    • one year ago
  28. Hesan
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    fine, suppose that on drawing first card, we've got a queen, so we'll exclude queen and now we'll take 4 kings.. am i right?

    • one year ago
  29. Zarkon
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    that would work...that is what SmoothMath did

    • one year ago
  30. Hesan
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    right. one thing more..

    • one year ago
  31. Hesan
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    if we take 4 cards for 2nd turn, will it not be 49 in the denominator? 4/48

    • one year ago
  32. Benjammin
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    Oooh, I see. What I gave was to draw two cards, each of which would be a king or queen.

    • one year ago
  33. Zarkon
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    @Hesan no the experiment is drawing 2 cards from 52....so it is 52 then 51 as above

    • one year ago
  34. Hesan
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    i m getting it now...

    • one year ago
  35. Hesan
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    okey, now what is the probability that the cards are both jack?

    • one year ago
  36. Benjammin
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    (4*52)*(3*51)

    • one year ago
  37. Hesan
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    the second will always be /51 if we draw 2 cards?

    • one year ago
  38. Zarkon
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    yes...if you are drawing without replacement

    • one year ago
  39. Zarkon
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    because you pick one of the 52 cards...there are only 51 cards remaining

    • one year ago
  40. Hesan
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    alright, thanks.. i get it now!

    • one year ago
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