The problem goes like this:
A fair die is rolled till 6 shows up. The first 16 throws were unsuccessful(i.e. 6 did not show up in any of the first 16 throws), the probability that the first 6 will show up in the 19th throw is?
My attempt :
$P[Not\ getting\ a\ 6]=5/6$
For calculation of total possibilities considering the first 16 unsuccessful throws, I considered the last three throws as binomial random variable $X$ where the number of $6's$ can be $0,1,2$ and $3$.
Therefore, total probability is = $(5/6)^{16}*[\binom{3}{0}*(5/6)^{3}+\binom{3}{1}*(5/6)^{2}*(1/6)+\binom{3}{2}*(5/6)^{1}*(1/6)^{2}+\binom{3}{3}*(1/6)^{3}]$
$P[getting \ 6 \ in\ the\ 19th\ throw]$=
$\frac{(5/6)^{18}*(1/6)}{(5/6)^{16}*[\binom{3}{0}*(5/6)^{3}+\binom{3}{1}*(5/6)^{2}*(1/6)+\binom{3}{2}*(5/6)^{1}*(1/6)^{2}+\binom{3}{3}*(1/6)^{3}]}$= $(5/6)^{2}*(1/6)$
Is my approach to the problem correct? Please help.