Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

Hesan

  • 3 years ago

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.

  • This Question is Closed
  1. Hesan
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    no

  2. Hesan
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    two cards, remember

  3. Nick2019
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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 !!)

  4. Hesan
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i m getting the wrong answer

  5. terenzreignz
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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?

  6. Benjammin
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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%

  7. Hesan
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @Benjammin: I did the same thing but getting the wrong answer

  8. Zarkon
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    are you supposed to get one of each card..one king and one queen?

  9. Hesan
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @Benjammin: your calculations give the answer: 14/663

  10. Hesan
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @Zarkon: yes

  11. Zarkon
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\frac{{4\choose 1}{4\choose1}}{{52\choose 2}}\]

  12. Zarkon
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\frac{8}{663}\]

  13. Hesan
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yeah right.. How?

  14. Zarkon
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I gave the formula above

  15. Zarkon
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    this is a classic hypergeometric distribution problem

  16. Zarkon
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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 \[2\cdot\frac{4}{52}\cdot\frac{4}{51}\]

  17. Hesan
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    but there shoulb be 8 cards right? 4 for kings and 4 for queens

  18. Zarkon
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    yes..hence the two 4's in the solution

  19. Hesan
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    okey, means 8/52, now how did u get 4/51?

  20. SmoothMath
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    The way I see it, it's simply: 8/52 * 4/51

  21. Zarkon
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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.

  22. Zarkon
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    yes...same thing

  23. Zarkon
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    pick from the 8...then don't pick a matching card...hence only 4 left to choose from

  24. SmoothMath
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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.

  25. Hesan
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    no i dont get it.. i was thinking it 7/51

  26. Hesan
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    see we've drawn 1 card, now 7 are left, so it would be 7/51

  27. Zarkon
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    but by doing that you could get a matching pair...2 kings or 2 queens

  28. Hesan
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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?

  29. Zarkon
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    that would work...that is what SmoothMath did

  30. Hesan
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    right. one thing more..

  31. Hesan
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    if we take 4 cards for 2nd turn, will it not be 49 in the denominator? 4/48

  32. Benjammin
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Oooh, I see. What I gave was to draw two cards, each of which would be a king or queen.

  33. Zarkon
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    @Hesan no the experiment is drawing 2 cards from 52....so it is 52 then 51 as above

  34. Hesan
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i m getting it now...

  35. Hesan
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    okey, now what is the probability that the cards are both jack?

  36. Benjammin
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    (4*52)*(3*51)

  37. Hesan
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    the second will always be /51 if we draw 2 cards?

  38. Zarkon
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    yes...if you are drawing without replacement

  39. Zarkon
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    because you pick one of the 52 cards...there are only 51 cards remaining

  40. Hesan
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    alright, thanks.. i get it now!

  41. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy