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What is the probability of getting at least one six in a

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VP
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New post 13 Mar 2005, 19:10
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

What is the probability of getting at least one six in a single throw of three unbiased dice?

(a) 1 / 6
(b) 125 / 216
(c) 1 / 36
(d) 91 / 216
(e) 11 / 38
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New post 13 Mar 2005, 20:34
P(getting a 6) = 1/6
P(not getting a 6 on one throw) = 5/6
P (not getting a 6 on three throws) = 5/6*5/6*5/6 = 125/216

P(getting at least a 6) = 1-125/216= 91/216 <--- (D)
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New post 13 Mar 2005, 20:44
One 6 = 3*25/6^3
Two 6 = 3*5/6^3
Three 6 = 1/6^3

p(e) = 91/6^3
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New post 14 Mar 2005, 15:01
atleast.. again the way i would go is "think minus 1"

P(not getting any 6 three times would be 125/216

1-125/216= 91/216
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New post 14 Mar 2005, 17:29
:-D OA is 91/216
  [#permalink] 14 Mar 2005, 17:29
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