Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
2  
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: en.wikipedia.org/wiki/Complementary_event –  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
add comment

1 Answer

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

share|improve this answer
    
$1-(5/6)^6$, 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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.