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kittenbalm

  • one year ago

a jar contains 4 green marbles, 5 red marbles, and 7 blue marbles. If two balls are selected randomly what is the probability of getting : 1. 1 green and 1 blue 2. 2 red 3. 1 blue and 1 red 4. 2 blue

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  1. kittenbalm
    • one year ago
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    hey help please?

  2. yrelhan4
    • one year ago
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    well i just got confused.. give me a minute.

  3. kittenbalm
    • one year ago
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    okay the title of lesson is combinations i got no idea how to do it lol

  4. kittenbalm
    • one year ago
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    @ganeshie8

  5. Zarkon
    • one year ago
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    are you sampling with or without replacement?

  6. kittenbalm
    • one year ago
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    without i think

  7. Zarkon
    • one year ago
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    then this is a hypergeometric distribution

  8. kittenbalm
    • one year ago
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    a what????? lol explain

  9. Zarkon
    • one year ago
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    suppose you have a box with a red and b green balls. you grab k balls the prob you get r red and g green balls (r+g=k) is given by \[\frac{ _aC_r _bC_g}{_{(a+b)}C_k}\]

  10. kittenbalm
    • one year ago
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    yeah i dont get it

  11. Zarkon
    • one year ago
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    \[\frac{_aC_r\cdot_b C_g}{_{a+b}C_k}\]

  12. Zarkon
    • one year ago
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    do you know what \[_nC_r\] is ?

  13. kittenbalm
    • one year ago
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    not at all

  14. Zarkon
    • one year ago
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    \[{n\choose r}\]?

  15. kittenbalm
    • one year ago
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    nope i dont know any of this stuff

  16. Zarkon
    • one year ago
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    permutations/combinations?

  17. yrelhan4
    • one year ago
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    i think it would be easier if we dont use combinations, zarkon.. it would be easier if if we do it like this.. for 1 green ball. 4/16.. 1 blue ball after 1 green ball hass been selected.. 7/15. and then multiply them.. whats your view?

  18. kittenbalm
    • one year ago
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    i knwo thats what this unit is called combinations

  19. yrelhan4
    • one year ago
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    eh right. so it would be like 4/16*7/15 + 7/16*4/15.. right?

  20. kittenbalm
    • one year ago
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    id think s if that was the case

  21. kittenbalm
    • one year ago
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    wait but youd get the same thing right 7/60 + 7/60

  22. yrelhan4
    • one year ago
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    yup.

  23. yrelhan4
    • one year ago
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    then for the second part.. 5/16*4/15.. you get why?

  24. kittenbalm
    • one year ago
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    which would equal 14/60 right because you dont change the denominator right or would it be 14/120

  25. yrelhan4
    • one year ago
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    yeah.

  26. kittenbalm
    • one year ago
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    so it is 14/120

  27. yrelhan4
    • one year ago
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    am i right with the second part @RolyPoly ?

  28. yrelhan4
    • one year ago
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    it should be 14/120..

  29. Zarkon
    • one year ago
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    no

  30. yrelhan4
    • one year ago
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    14/60.. lol

  31. Zarkon
    • one year ago
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    the answer is 7/30

  32. kittenbalm
    • one year ago
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    well idk how to do this thn i am so lost im gonna fail lol

  33. Zarkon
    • one year ago
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    or 14/60 if you like

  34. kittenbalm
    • one year ago
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    how

  35. Zarkon
    • one year ago
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    the solution is written above

  36. kittenbalm
    • one year ago
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    okay so will you help by showing work on two three and four maybe if i see it on three ill be able to do it

  37. RolyPoly
    • one year ago
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    P(blue first, then green) = P(blue) x P(green after drawing a blue) = 7/16 x 4/(16-1) = a P(green first, then blue) = P(green) x P(blue after drawing a green) = 4/16 x 7/(16-1) = b P(1 blue and 1 green) =P(blue first, then green) + P(green first, then blue) = a + b

  38. kittenbalm
    • one year ago
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    now im deffinately confused i just ned a simple explanation

  39. RolyPoly
    • one year ago
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    2 red, similar, despite the order of picking a red doesn't matter this time, since you don't the red ones are identical

  40. kittenbalm
    • one year ago
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    what omg i need a tutor

  41. RolyPoly
    • one year ago
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    Which part are you confused at?

  42. kbrece1
    • one year ago
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    2 BLUE

  43. kittenbalm
    • one year ago
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    2 red actually all of it really

  44. kittenbalm
    • one year ago
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    u gonna help k?

  45. kittenbalm
    • one year ago
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    no thats not the question @phi help please

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