1
$\begingroup$

A fair die is rolled until a $6$ appears. What is the probability that is must be cast more than 5 times?

So this is $1- P(\text{dice has to be cast less than or equal to}\ 5 \ \text{times})$. So this probability is equal to $$ P =\frac{1}{6}+ \frac{5}{6} \frac{1}{6}+\left(\frac{5}{6} \right)^{2} \frac{1}{6} + \cdots + \left(\frac{5}{6} \right)^{4} \frac{1}{6}$$

So just add $-1$ to this?

  • 1
    If you add $-1$ to $P$ you will get a negative number.2011-12-25
  • 0
    @AndréNicolas: I meant I do the following: $-P+1$?2011-12-25
  • 0
    That's fine. Your calculation is then longer than the simple $(5/6)^n$, but perfectly correct.2011-12-25

3 Answers 3