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agbyoung

  • 4 years ago

If you throw a dice 6 times, what's the chance that you'd get a six on: a: exactly one of the throws. b: one or more of the throws.

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  1. amistre64
    • 4 years ago
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    thats a pretty big sample space :)

  2. agbyoung
    • 4 years ago
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    ? I don't understand what your saying... lol

  3. amistre64
    • 4 years ago
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    1 6 15 20 15 6 1 neither do i yet :)

  4. dimbeni
    • 4 years ago
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    a. (1/6)*(5/6)^5*6 = (5/6)^5 b. (5/6)^6

  5. agbyoung
    • 4 years ago
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    LOL! I was a fool in my math class so I had extra homework... ( threw an apple at the teacher)

  6. dimbeni
    • 4 years ago
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    er, b. 1-(5/6)^6

  7. amistre64
    • 4 years ago
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    P(6 a / a / a / a / a / ) = P(6)*P(\)^5, where / means not a 6

  8. amistre64
    • 4 years ago
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    P(6) = 1/6; P(/) = 5/6

  9. agbyoung
    • 4 years ago
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    Thanks!

  10. amistre64
    • 4 years ago
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    satellite is good at these, he does them instead of the crosswords during breakfast :)

  11. anonymous
    • 4 years ago
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    exactly one: \[\dbinom{6}{1}\frac{1}{6}\times (\frac{5}{6})^5\]

  12. amistre64
    • 4 years ago
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    he has rooms filled with unthrown dice just waiting to be explored lol

  13. anonymous
    • 4 years ago
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    they are thrown.

  14. anonymous
    • 4 years ago
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    since 6 choose 1 is 6, this answer is really \[(\frac{5}{6})^5\]

  15. agbyoung
    • 4 years ago
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    Dang! I didn't think I would get so much attention :P

  16. anonymous
    • 4 years ago
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    one or more throws means not no sixes. the probability you get no sixes is \[(\frac{5}{6})^6\] so your answer is \[1-(\frac{5}{6})^6\]

  17. agbyoung
    • 4 years ago
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    Do my other question please: http://openstudy.com/groups/mathematics/updates/4e7a829b0b8b2cb239c38801#/groups/mathematics/updates/4e7a87380b8b2cb239c3ad98

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