<s>[INST] Conditional probability of continuous throwing a dice 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.
 [/INST] 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...
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