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# When a die that has one of six consecutive integers on each

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When a die that has one of six consecutive integers on each [#permalink]

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04 Oct 2004, 17:56
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When a die that has one of six consecutive integers on each of its sides is rolled twice, what is the probability of getting the number 1 on both rolls?
(1) the probability of NOT getting an eight is 1
(2) the probability of NOT getting a seven is 25/36
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04 Oct 2004, 18:53
1. Since prob of not getting a 8 is 1. Then there is no number 8 in the die. We dont know if number 1 is still there. So, Insuff.

2. The prob of not getting 7 is 25/36. Essentially means there is number 7. 25/36 is because in the first roll the prob of not getting 7 is (1-1/6) which is 5/6 and similarly in the second roll toalling to 25/36. Since 7 is there and the numbers are CONSECUTIVE, the numbers have to be (in a worst case scenario) 2-7 or can be 3-8 or anything else that includes 7. - but definitely does not include 1. So, Suff.

B.
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04 Oct 2004, 18:58
great explanation venksune. I totally agree.
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Paul

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04 Oct 2004, 22:21
Good point...I like this problem, very good one.
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05 Oct 2004, 03:55
Forgot to log in..

BTW: The question should state that there can only be 1 digit on each side, not just an Integer.
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11 Oct 2004, 14:32
gr8 detailed explanation
thank u
gr8 :idea:
-anish

[quote="venksune"]1. Since prob of not getting a 8 is 1. Then there is no number 8 in the die. We dont know if number 1 is still there. So, Insuff.

2. The prob of not getting 7 is 25/36. Essentially means there is number 7. 25/36 is because in the first roll the prob of not getting 7 is (1-1/6) which is 5/6 and similarly in the second roll toalling to 25/36. Since 7 is there and the numbers are CONSECUTIVE, the numbers have to be (in a worst case scenario) 2-7 or can be 3-8 or anything else that includes 7. - but definitely does not include 1. So, Suff.

B.[/quote]
[#permalink] 11 Oct 2004, 14:32
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# When a die that has one of six consecutive integers on each

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