2
$\begingroup$

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.

  • 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 $P(A)=1-P(A^c)$.2012-03-01
  • 0
    I saw this on wiki where they do it for eight tosses which is basically the same: http://en.wikipedia.org/wiki/Complementary_event2012-03-01
  • 0
    *a die. Dice is always plural.2012-03-01
  • 0
    I corrected dice/die thingy, thanks for the input. @DanielMontealegre your comment made Alex Becker's answer even more clear.2012-03-01

1 Answers 1

5

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

  • 0
    $1-(5/6)^6$, 6 rolls2012-03-01
  • 0
    @ArtiomFiodorov Whoops.2012-03-01
  • 0
    Great, thanks. Problem solved!2012-03-01