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StArAnGeL

  • 2 years ago

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?

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  1. ChrisV
    • 2 years ago
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    1/((52)(51)(50))

  2. ChrisV
    • 2 years ago
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    1/32600

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

  4. ChrisV
    • 2 years ago
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    really?

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

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

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

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

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

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

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

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

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

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

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

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

  17. ChrisV
    • 2 years ago
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    yes

  18. StArAnGeL
    • 2 years ago
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    Um Why ?

  19. ChrisV
    • 2 years ago
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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

  20. ChrisV
    • 2 years ago
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    occur

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

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

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

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

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

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

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

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

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

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

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

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

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

  34. ChrisV
    • 2 years ago
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    yes

  35. Directrix
    • 2 years ago
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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.

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