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Statistics
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A fair die is rolled 4 times. What is the probability of having no 1 and no 2 among the rolls? Round your answer to three decimal places.

May 31st, 2015

A fair die is rolled 6 times. What is the probability of having no 1 and no 2 among the rolls? 
The probability of getting a 3, 4, 5 or 6 the 1st time is 4 ways out of 6, which is 4%2F6 or 2%2F3
multiplied by
the probability of getting a 3, 4, 5 or 6 the 2nd time, or 2%2F3
multiplied by
the probability of getting a 3, 4, 5 or 6 the 3rd time, or 2%2F3

multiplied by
the probability of getting a 3, 4, 5 or 6 the 4th time, or 2%2F3
multiplied by
the probability of getting a 3, 4, 5 or 6 the 5th time, or 2%2F3
multiplied by
the probability of getting a 3, 4, 5 or 6 the 6th time, or 2%2F3 
So it's %282%2F3%29%5E8 or about .088 

thus

P(no1 and no2 in 6 rolls) = (2/3)^6 = 0.088



May 31st, 2015

Your answer is wrong.  I no longer need help. Thanks

May 31st, 2015

sorry, the answer is 

A fair die is rolled 4 times. What is the probability of having no 1 and no 2 among the rolls? 
The probability of getting a 3, 4, 5 or 6 the 1st time is 4 ways out of 6, which is 4%2F6 or 2%2F3
multiplied by
the probability of getting a 3, 4, 5 or 6 the 2nd time, or 2%2F3
multiplied by
the probability of getting a 3, 4, 5 or 6 the 3rd time, or 2%2F3

multiplied by
the probability of getting a 3, 4, 5 or 6 the 4th time, or 2%2F3
(2/3)^4

thus

P(no1 and no2 in 6 rolls) = (2/3)^4 = 0.198


May 31st, 2015

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