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marigirl

  • one year ago

when four dice are thrown (each numbered from 1 to 6), calculate the probability that all the numbers facing up are different . where do i start??

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  1. anonymous
    • one year ago
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    interesting question.thanks for posting,,.needs an interpretation.

  2. anonymous
    • one year ago
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    you start with the fact there are 4 dice

  3. marigirl
    • one year ago
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    i was looking at all possible outcomes, 6*5*4*3*2*1=720 for each dice

  4. anonymous
    • one year ago
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    understand where you are coming from, but I dont think that will provide a solution

  5. marigirl
    • one year ago
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    okay,

  6. anonymous
    • one year ago
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    consider the first dice. it has 6 possible alternatives.

  7. anonymous
    • one year ago
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    Now, once the first possiblity is chosen there are 5 possible alternatives for the next 3 dice

  8. marigirl
    • one year ago
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    |dw:1438900477187:dw|

  9. anonymous
    • one year ago
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    Now you've got it!... So what is the overall probability?

  10. marigirl
    • one year ago
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    0.277 thanks so much

  11. anonymous
    • one year ago
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    it seems you multiplied the individual probabilites?

  12. marigirl
    • one year ago
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    no wait... .

  13. marigirl
    • one year ago
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    i multiplied it and got 1/360

  14. marigirl
    • one year ago
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    but that doesnt sound right

  15. anonymous
    • one year ago
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    |dw:1438900871076:dw|

  16. marigirl
    • one year ago
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    i mean, we multiply and get 1/360.. then ?

  17. anonymous
    • one year ago
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    correct

  18. anonymous
    • one year ago
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    what this means is that there is a one in 360 chance that if you roll 4 die that they will all be the same.

  19. marigirl
    • one year ago
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    okay then from here: the total possible outcomes for the 4 dices with 6 sides 6*6*6*6=1296 then the prob that all the numbers facing up are different is: 360/1296

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