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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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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 $P(A)=1-P(A^c)$. – Daniel Montealegre Mar 1 '12 at 9:20
I saw this on wiki where they do it for eight tosses which is basically the same: – Daniel Montealegre Mar 1 '12 at 9:21
*a die. Dice is always plural. – Tom Artiom Fiodorov Mar 1 '12 at 9:21
I corrected dice/die thingy, thanks for the input. @DanielMontealegre your comment made Alex Becker's answer even more clear. – Dorin Mar 1 '12 at 9:55
up vote 5 down vote accepted

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 $1-(5/6)^6=\frac{31031}{46656}$.

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$1-(5/6)^6$, 6 rolls – Tom Artiom Fiodorov Mar 1 '12 at 9:23
@ArtiomFiodorov Whoops. – Alex Becker Mar 1 '12 at 9:28
Great, thanks. Problem solved! – Dorin Mar 1 '12 at 9:53

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