I got on the first part of the question fine but I'm really confused about the expression "the last of the dice to show a six does so on the $r$th roll" in the second part of the question. I interpret this line as in the following situations but none make sense.
If this is to ask the result of rolling the very last dice on the $r$th roll, then wouldn't it be independent from $n$ and $r$, and result simply be $p$ ?
If this is to ask the probability of at least one six shows up on the $r$th roll which we consider as the last roll, then wouldn't the probability simply be $1-q^n$, which is independent from $r$
If this is to ask the probability that result of six would never occur on the rolls after the $r$th roll, then wouldn't it simply be zero since the number of rolls would be infinity and no occurrence of a six would be impossible?
(i) A set of $n$ dice is rolled repeatedly. For each die the probability of showing a six is $p$. Show that the probability that the first of the dice to show a six does so on the $r$th roll is $$ q^{nr}\left ( q^{-n} - 1\right ) $$ where $q = 1 − p$. Determine, and simplify, an expression for the probability generating function for this distribution, in terms of $q$ and $n$. The first of the dice to show a six does so on the Rth roll. Find the expected value of $R$ and show that, in the case $n = 2$, $p = 1/6$, this value is 36/11.
(ii) Show that the probability that the last of the dice to show a six does so on the rth roll is $$ \left ( 1-q^{r} \right )^{n}-\left ( 1-q^{r-1} \right )^{n}. $$ Find, for the case $n = 2$, the probability generating function. The last of the dice to show a six does so on the $S$th roll. Find the expected value of $S$ and evaluate this when $p = 1/6$.
I'm afraid none of your interpretations are correct. We have $n$ dices. Say, after the $r-1$th roll, some dice(es) have not yet shown a six even once. But, after the $r$th roll, all the dice have shown a six at least once. We can translate this into:
Let $P(X)$ be the probability that all dice have shown at least one six by the $X$th roll. So, $P(r)$ is the probability that all the dice have shown a six by the $r$th roll. For the second condition, we need to subtract the probability that all the dice have shown a six by the $r-1$th roll(since it would imply no dice has shown a six for the first time on the $r$th roll). Thus, the required probability becomes: $$P(r)-P(r-1)$$ It now remains to find an expression for $P(X)$. We know that $q$ is the probability that a six will not appear. So, $q^X$ is the probability that a six will not appear for $X$ rolls. The complement (at least one six will appear in $X$ rolls) is $1-q^X$. The probability is the same for all $n$ dice. Thus, the probability that at least one six will appear in $X$ rolls is$$P(X) = (1-q^X)^n$$Now finding the required probability: $$P(r)-P(r-1)=\color{green}{(1-q^r)^n - (1-q^{r-1})^n}$$The expected value is easy from here.