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anonymous

  • 5 years ago

kay tears off labels of all 10 soup cans her mother knows there are 2 tomato and 8 vegetable she selects 4 at random. what is the probability a. exactly one of the 4 cans is tomato b. none of the 4 cans is tomato c. at least one of the 4 is tomato?

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  1. anonymous
    • 5 years ago
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    a) 2C1/10C4

  2. anonymous
    • 5 years ago
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    b) 2C0/10C4

  3. anonymous
    • 5 years ago
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    c) (2C1 + 2C2)/10C4

  4. anonymous
    • 5 years ago
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    i am very confused...and i know the answers to these problems...i just dont know how to do them?

  5. anonymous
    • 5 years ago
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    on top of the fraction,the numerator is the no of ways o selecting 1 tomato can out of a possible 2 , and the denominator is the total no. of selections ie 10C4, selecting any 4 out of 10 cans

  6. anonymous
    • 5 years ago
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    get it or nt?

  7. anonymous
    • 5 years ago
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    no not really........the answer to the first is .8 so that is why i am confused

  8. anonymous
    • 5 years ago
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    cant be

  9. anonymous
    • 5 years ago
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    how?

  10. amistre64
    • 5 years ago
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    10! -------- sounds plausible? 2!(10-2)!

  11. anonymous
    • 5 years ago
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    check my solution...shes sayin its 0.8 the answer

  12. amistre64
    • 5 years ago
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    then it aint 45 lol

  13. amistre64
    • 5 years ago
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    2 2 2 2 ------- perchance? 10 9 8 7

  14. amistre64
    • 5 years ago
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    combo and perms are still new to me :)

  15. anonymous
    • 5 years ago
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    ha same

  16. amistre64
    • 5 years ago
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    id have to draw up a tree or something and count them out lol

  17. anonymous
    • 5 years ago
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    yeah...i am not sure if it is a combo though? def probablility

  18. anonymous
    • 5 years ago
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    its using combos

  19. anonymous
    • 5 years ago
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    m 95% sure my methods right nelly

  20. amistre64
    • 5 years ago
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    there are 45 ways to select 4 cans right? or did i mess that up too

  21. anonymous
    • 5 years ago
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    10C4

  22. amistre64
    • 5 years ago
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    10C4....what is that in longhand tho?

  23. amistre64
    • 5 years ago
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    10! --- ?? 4!6!

  24. anonymous
    • 5 years ago
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    10!/(6!4!)

  25. amistre64
    • 5 years ago
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    210?

  26. anonymous
    • 5 years ago
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    i got 120960?

  27. amistre64
    • 5 years ago
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    10 9 8 7 6 5 4 3 2 ---------------- 4 3 2 6 5 4 3 2 10 9 8 7 -------- 4 3 2 5 3 2 7 = 15(14) = 210 right?

  28. anonymous
    • 5 years ago
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    yeah

  29. amistre64
    • 5 years ago
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    so we have 210 ways to do this lol

  30. anonymous
    • 5 years ago
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    great

  31. amistre64
    • 5 years ago
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    how many ways can we have 2 tcans?

  32. anonymous
    • 5 years ago
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    672

  33. amistre64
    • 5 years ago
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    or is it easier to say how many ways to pick no tcans? 8 can ways then...

  34. anonymous
    • 5 years ago
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    no no

  35. anonymous
    • 5 years ago
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    pick one out of 2 cans is 2C1

  36. amistre64
    • 5 years ago
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    but that doesnt gives us a total ways for tcans does it?

  37. anonymous
    • 5 years ago
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    336

  38. amistre64
    • 5 years ago
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    if there are 8 ways to choose 4 cans 8! --- ?? 4!4!

  39. amistre64
    • 5 years ago
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    70?

  40. amistre64
    • 5 years ago
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    8765432 -------- 432.432 8765 ----- 432 725 = 14*5 = 70

  41. amistre64
    • 5 years ago
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    70/210 = 7/21 = 1/3; 33.33% ??

  42. amistre64
    • 5 years ago
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    so the chance that 1 is at least a tcan could be 66' 2/3% does that seem reasonable?

  43. amistre64
    • 5 years ago
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    33' 1/3% chance we got no tcans 66' 2/3% chance we got at least 1 tcan seems good to me

  44. anonymous
    • 5 years ago
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    nope the anwser is .8 blah

  45. amistre64
    • 5 years ago
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    .8 is 80% which answer is .8? a,b or c?

  46. amistre64
    • 5 years ago
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    10C4 is the total number of ways to choose 4 cans.... 8C4 is the total number of ways to choose no tcans.... casue we removed them from the equation.... % = part/total % = 8c4/10c4 for no tcans 100- no tcan% = at least 1 tcan.... is my reasoning

  47. anonymous
    • 5 years ago
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    a

  48. amistre64
    • 5 years ago
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    whats the answers to b and c :)

  49. anonymous
    • 5 years ago
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    b .333333 c. .66666

  50. amistre64
    • 5 years ago
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    yay!! i was right :)

  51. amistre64
    • 5 years ago
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    exactly one of the 4 cans is tomato lets see if we take away one tcan; just for the sake of argument; we can do this then

  52. amistre64
    • 5 years ago
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    9! ---- = 3.3.2.7 = 9(14) = 126 ways to possible ways to 4!5! to choose at least 1 tcan

  53. anonymous
    • 5 years ago
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    i am still very confused how to do the first as i get 122/210 = .53333

  54. amistre64
    • 5 years ago
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    126 ways for 1 possible tcan and we still have 70 ways for no tcans..; i wonder

  55. amistre64
    • 5 years ago
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    126-70 would give us the number of ways for 1 tcan right? 126 70 ---- 56 ways that at least 1 tcan is in it... i hope lol

  56. amistre64
    • 5 years ago
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    56/210 . --------- 105/28.0000 ......not gonna be .8 lol

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