How can I calculate the following probability:
Throwing a die 6 times, what is the probability of having face no. 1 showing at least one time.
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2$\begingroup$ Let $A$ be the event that at least one time face $1$ appeared in the first six tosses. Then the complement of $A$ is the event that $1$ did not appear a single time in the first six tosses. Now use $P(A)=1-P(A^c)$. $\endgroup$– Daniel MontealegreCommented Mar 1, 2012 at 9:20
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$\begingroup$ I saw this on wiki where they do it for eight tosses which is basically the same: en.wikipedia.org/wiki/Complementary_event $\endgroup$– Daniel MontealegreCommented Mar 1, 2012 at 9:21
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$\begingroup$ *a die. Dice is always plural. $\endgroup$– Tom Artiom FiodorovCommented Mar 1, 2012 at 9:21
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$\begingroup$ I corrected dice/die thingy, thanks for the input. @DanielMontealegre your comment made Alex Becker's answer even more clear. $\endgroup$– DorinCommented Mar 1, 2012 at 9:55
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1 Answer
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Consider the probability of having no 1 appear in each roll, which is $5/6$. Since the outcomes of each roll are independent, the probability of having no 1 appear in six rolls is $(5/6)^6$. Thus the probability of having a 1 appear at least once is $1-(5/6)^6=\frac{31031}{46656}$.
answered Mar 1, 2012 at 9:17