What is the probability we need k rolls of a fair sided die to get at least one five and one six? I HAVE checked the questions and none of them seem to specify any method to compensate for at least one 5 AND one 6. 
Here is how I tried it. 
$$ P(k\ Rolls \ and \ last \ is \ a \ six)= P(\ k -1 \ rolls \ have \ at \ least\ one \ 5 )*P(\ k \ is \ a \ six)$$
$$ = \sum_{j=1}^{k-1} {k-1\choose j}(\frac{1}{6})^j (\frac{4}{6})^j  $$
$$ = (\frac{5}{6})^{k-1}- {k-1\choose0}(\frac{1}{6})^0(\frac{4}{6})^{k-1}  $$
$$ =(\frac{5}{6})^4 - (\frac{4}{6})^4$$
 A: We can split up the "we need $k$ rolls event" into sub-cases: for each $j \in \{1, 2, \dots, k-1\}$, there's the sub-case where we rolled a 5 or 6 on roll $j$, then it took us until roll $k$ to roll the other one.
The probability of this subcase is $$ \left(\frac 46\right)^{j-1} \cdot \frac26 \cdot \left(\frac56\right)^{k-1-j} \cdot \frac16$$ because we have:


*

*$j-1$ initial rolls where we roll neither a 5 nor a 6,

*the $j^{\text{th}}$ roll, on which we roll one of the numbers we need,

*the next $k-1-j$ rolls, on which we fail to get the other needed number,

*the $k^{\text{th}}$ roll, on which we get the other number we needed.


We can simplify this to $$\frac1{10} \cdot \left(\frac56\right)^k \cdot \left(\frac45\right)^j.$$ Summing over all values of $j$, we get
$$\frac1{10} \cdot \left(\frac56\right)^k \cdot \sum_{j=1}^{k-1} \left(\frac45\right)^j = \frac1{10} \cdot \left(\frac56\right)^k \cdot \frac{4/5 - (4/5)^k}{1 - 4/5} = \frac25 \cdot \left(\frac56\right)^k \cdot \left(1 - \left(\frac45\right)^{k-1}\right).$$
A: It's not clear to me if the question is whether $k$ rolls suffice or whether exactly $k$ rolls are required.  I try to cover both possibilities below.
Let's say the first roll on which we have at least one 5 and at least one 6 is roll $K$.
Let's also say $A$ is the event of rolling no 5 in $k$ rolls and $B$ is the event of rolling no 6 in $k$ rolls.  Then the probability that we have at least one 5 and at least one 6 by roll $k$ is
$$\begin{align}
\Pr(K \le k) &=\Pr(A^c \cap B^c) \\
&= 1 - \Pr(A \cup B) \\
 &= 1 - \Pr(A) - \Pr(B) + \Pr(A \cap B) \\
 &= 1 - 2 \left( \frac{5}{6} \right)^k + \left( \frac{4}{6} \right)^k \\
\end{align}$$
So the probability that the first roll on which we have at least one 5 and at least one 6 is roll $k$ is
$$\Pr(K = k) = \Pr(K \le k) - \Pr(K \le k-1) = \frac{1}{3} \left( \frac{5}{6} \right)^{k-1} - \frac{1}{3} \left( \frac{4}{6} \right)^{k-1}$$
