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StArAnGeL

Three cards are drawn successively, without replacement from a pack of 52 well shuffled cards. What is the probability that first two cards are kings and the third card drawn is an ace?

  • 2 years ago
  • 2 years ago

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  1. ChrisV
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    1/((52)(51)(50))

    • 2 years ago
  2. ChrisV
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    1/32600

    • 2 years ago
  3. StArAnGeL
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    Sorry Thats not the answer

    • 2 years ago
  4. ChrisV
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    really?

    • 2 years ago
  5. StArAnGeL
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    Yeah ! The answer is 2/5525 But i need to know how these kind of problems are solved

    • 2 years ago
  6. ChrisV
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    of well the fact is your chance of the first card being a king is 1/52

    • 2 years ago
  7. ChrisV
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    the second card being a king would be 1/51

    • 2 years ago
  8. ChrisV
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    and the third card being an ace would be 1/50

    • 2 years ago
  9. ChrisV
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    how they came up with that answer i do not know

    • 2 years ago
  10. ChrisV
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    unless they are somehow figuring in that there are 4 of each

    • 2 years ago
  11. StArAnGeL
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    Oh Ok there are 4kings So chance of the first card being a king is 4/52

    • 2 years ago
  12. StArAnGeL
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    and then 3/51

    • 2 years ago
  13. StArAnGeL
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    and ace will be 4/50

    • 2 years ago
  14. StArAnGeL
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    So multiply and its 2/5525

    • 2 years ago
  15. gogind
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    ah you got it...nvm

    • 2 years ago
  16. StArAnGeL
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    But Btw do you know Why we are multiplying ?

    • 2 years ago
  17. ChrisV
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    yes

    • 2 years ago
  18. StArAnGeL
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    Um Why ?

    • 2 years ago
  19. ChrisV
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    well to find the probability of something happening you have to multiply the number of times it could possibly occure then divide it by the total number of possibilities

    • 2 years ago
  20. ChrisV
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    occur

    • 2 years ago
  21. ChrisV
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    like if you have 10 books and 2 are school books

    • 2 years ago
  22. ChrisV
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    whats the probability that the first book you pull out of your bag would be a school book

    • 2 years ago
  23. ChrisV
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    2/10=1/5

    • 2 years ago
  24. StArAnGeL
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    yes But we didnt multiply anything here ?

    • 2 years ago
  25. ChrisV
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    so say you have taken 1 school book out already

    • 2 years ago
  26. ChrisV
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    whats the chance of pulling both school book out of the bag in a row

    • 2 years ago
  27. ChrisV
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    well we already know the first one was 2/10

    • 2 years ago
  28. ChrisV
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    well that leaves 1/.9

    • 2 years ago
  29. ChrisV
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    so you multiply 2/10 x 1/9

    • 2 years ago
  30. ChrisV
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    the probability of pulling them both out in order would be 2/90 or 1/45

    • 2 years ago
  31. ChrisV
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    so out of every 90 times you try it, the probability of succedding is 2 times

    • 2 years ago
  32. StArAnGeL
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    OH Ohkay Thanks a lot =)

    • 2 years ago
  33. FoolForMath
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    we are multiplying because these are independent events. \( P(A\cap B)=P(A) \times P(B) \)

    • 2 years ago
  34. ChrisV
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    yes

    • 2 years ago
  35. Directrix
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    Disclaimer: I know this problem has been solved. I am just practicing here. --------------------------------------------------- 4 kings, want 2 4 aces, want 1 no replacement 52 cards in deck P(K) = 4/52. The 52-count drops to 51 with no replacement. King count drops to 3 P(2ndK) = 3/51. The 51 card count drops to 50. King count drops to 2. Ace count remains at 4. P(Ace) = 4 / 50 Using the Multiplication Principle, the probability of these 3 events happening is (4/52) (3/51) (4/ 50) = 2/5525 . Multiplication Principle states: If an event occurs in m ways and another event occurs independently in n ways, then the two events can occur in m × n ways.

    • 2 years ago
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