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03 Jul 2006, 19:15
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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 two rolls?
(1) the probability of NOT getting an eight is 1
(2) the probability of NOT getting a seven is 25/36
[/i]
(1) the probability of NOT getting an eight is 1
(2) the probability of NOT getting a seven is 25/36
[/i]
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This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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03 Jul 2006, 19:57
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sharadGmat
First we do not know the first consecutive number of the 6 or the last of the 6.
st. 1 not getting an 8 is 100%, which means that the last consecutive number may be 7 or 6 (assuming no negative values) INSUFF. (B, C, E)
st. 2 we know that the P(not 7) = 25/36, therefore we know that 7 is part of the consecutive number. But this alone does not tell us where is at in the number sequence. Is 7 last or first? Either way, it does not matter because if 7 is the last, 2 will be the smallest number on the die.
P( getting 1 both times) = 0%
B is my answer...
03 Jul 2006, 22:09
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acfuture
Agreed also got B with same logic
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This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
