marigirl
  • marigirl
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??
Mathematics
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marigirl
  • marigirl
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??
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
interesting question.thanks for posting,,.needs an interpretation.
anonymous
  • anonymous
you start with the fact there are 4 dice
marigirl
  • marigirl
i was looking at all possible outcomes, 6*5*4*3*2*1=720 for each dice

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anonymous
  • anonymous
understand where you are coming from, but I dont think that will provide a solution
marigirl
  • marigirl
okay,
anonymous
  • anonymous
consider the first dice. it has 6 possible alternatives.
anonymous
  • anonymous
Now, once the first possiblity is chosen there are 5 possible alternatives for the next 3 dice
marigirl
  • marigirl
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anonymous
  • anonymous
Now you've got it!... So what is the overall probability?
marigirl
  • marigirl
0.277 thanks so much
anonymous
  • anonymous
it seems you multiplied the individual probabilites?
marigirl
  • marigirl
no wait... .
marigirl
  • marigirl
i multiplied it and got 1/360
marigirl
  • marigirl
but that doesnt sound right
anonymous
  • anonymous
|dw:1438900871076:dw|
marigirl
  • marigirl
i mean, we multiply and get 1/360.. then ?
anonymous
  • anonymous
correct
anonymous
  • anonymous
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.
marigirl
  • marigirl
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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