Conditional probability of continuous throwing a dice

Question

A person will keep throwing a dice until he gets $6 .$ If he gets 6 then the experiment will be stopped.

The probability that experiment ends at 6 th trial, given that prime number does not appear is (A) $\left(\frac{2}{5}\right)^{5}\left(\frac{1}{6}\right)$ (B) $\left(\frac{2}{3}\right)^{5}\left(\frac{1}{3}\right)$ (C) $^{6} \mathrm{C}_{1}\left(\frac{2}{3}\right)^{5}\left(\frac{1}{3}\right)$ (D) $^{6} \mathrm{C}_{2}\left(\frac{2}{3}\right)^{5}\left(\frac{1}{3}\right)$

My approach:

$\begin{aligned} A=& \text { throwing a dice until } 6 \text { occurd and } \\ \text { prime } & \text { no. does not appear. } \end{aligned}$ $B=$ exp. ends at $6^{\text {th }}$ trial. $P(B / A)=\frac{P(B \cap A)}{P(A)}$ $=\frac{\left(\frac{2}{6}\right)^{5} \frac{1}{6}}{\frac{1}{6}+\frac{2}{6} \cdot \frac{1}{6}+\left(\frac{2}{6}\right)^{2} \frac{1}{6}+\cdots}$ $=\frac{\left(\frac{1}{3}\right)^{5} \frac{1}{6}}{\frac{116}{1-1 / 3}}=\left(\frac{1}{3}\right)^{5} \cdot \frac{1}{6} \cdot \frac{2 / 3}{16}$ No correct ans in option. Where did I go wrong.

Answer 1

Context is a bit confusingly worded to me... isn't it just simply $(B)$ because if no prime numbers appear then $2,3$ and $5$ won't appear, reducing the dice to only 3 possibilities of which only 1 of those 3 is a 6 so isn't it simply just $({2 \over 3})^5*(1/3)$? I could be wrong however...

Answer 2

In a single roll of the dice, the only way a prime number does not appear is, if the die lands $2,4,6$. Restricting our attention to these three outcomes, the probability that the game does not end, conditional on a prime not appearing is $2/3$, as you are interested only in $2,4$. All trials are independent. So, if we extend this to $5$ trials, we get $(2/3)^5$. Further, in the sixth trial, you would like a $6$. So, that $(1/3)$.

Hence, it is $(2/3)^5 (1/3)$.