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Old 12-19-2005, 10:04 PM
Victor Victor is offline
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Default easy probability question

you have 4 dice. you roll them all. what are the chances that exactly 1 die shows a 3 or a 4 (and no other 3s or 4s are showing)?
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  #2  
Old 12-19-2005, 10:38 PM
MagicFlea MagicFlea is offline
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Default Re: easy probability question

(1/3) * (2/3)^3 * 4 = 32/81

first, consider the first die coming up 3 or 4 and the rest coming up not... then multiply by four to account for other possibilities.
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Old 12-19-2005, 11:20 PM
Victor Victor is offline
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Default Re: easy probability question

thanks a lot.
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