We can use binomial distribution because we only have two cases for each dice:1) (1/6) of chance to get ace; 2) (5/6) of chances to get no ace.
[(1/6) + (5/6)]^6
a) C(6,1)(1/6)(5/6)^5 = 0.40188
b) 1 - C(6,6)(5/6)^6 = 0.66510
c) C(6,2)(1/6)^2(5/6)^4 = 0.20094
2006-12-28 18:52:02
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answer #1
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answered by sahsjing 7
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You have 6 dice with 6 sides each, so the total number of possible rolls is 6 to the 6th power. There are 6 possible ways to get only one ace, so the probability is 6/6^6, or 1/6^5.
2006-12-28 18:07:59
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answer #2
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answered by mamills47 2
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a)
P(exactly1 Ace) = 6(1/6)(5/6)^5 = 0.4
b)
P(at least 1 Ace) = (1 - (5/6)^6) = 0.665
c)
P(exactly 2 Aces) = 2*4(1/6)^2(5/6)^4 = 0.107
2006-12-28 18:19:36
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answer #3
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answered by Helmut 7
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"0" as u don't have aces in True Dice..if u define any one of the sequence of the sides as the ace then its
a) 1/6
b) 1/6
c) 1/3
2006-12-28 18:18:25
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answer #4
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answered by Alex 2
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a) 6 * (1/6)(5/6)^6 = (5/6)^6 <- Helmut and sahsjing agree
b) 1 - (5/6)^6 <- Helmut and sahsjing agree
c) (6 * 5 / 2) * (1/6)^2(5/6)^4 <- sahsjing agrees
2006-12-28 18:04:06
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answer #5
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answered by ? 6
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I think it sounds like the game YAHTZEE..... not sure about the "probability" part though... no... wait, yahtzee only has 5 die... sorry....
2006-12-28 18:02:55
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answer #6
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answered by ravenhm 2
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what is an ace in this context |?
2006-12-28 17:59:32
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answer #7
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answered by gjmb1960 7
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