Probability distribution.. If $Z$ represents number of times $6$ appeared in two independent throws of a die, what would be the probability of $A =${maximum of two throws was $2$ and $Z = 0$}. $Z = 0$ means $6$ appeared $0$ time. -My try- Probability of $6$ appears $ 0$ time = $\frac{25}{36}$ maximum of two throws would be $(1,2)$ or $(2,1)$ or $(2,2)$. So there are $3$ ways to get maximum of two throws was $2$. Therefore $P(A) = \frac3{36} \times \frac{25}{36}$ ???

I am not sure that you are reporting the problem correctly. The maximum of the two throws is $2$ **and** there are no $6$ precisely if the maximum of the two throws is $2$. So the probability is $\dfrac{3}{36}$.

**Maybe** you were asked for the probability that the maximum of the throws is $2$ **given** that there was no $6$. Then by using the ordinary conditional probability procedure, we find that the probability the maximum is $2$ given there is no $6$ is $\dfrac{\frac{3}{36}}{\frac{25}{36}}$, which is $\dfrac{3}{25}$.