Probability and dice rolls 
Two fair six sided die are rolled fairly and the scores are noted. What is the probability that it takes 3 or more rolls to get a score of 7?

Here is what I have got so far:
The probability that it takes $3$ or more rolls to get a score of $7$ is equal to $$1-P(\text{you get a score of $7$ on less than $3$ rolls})$$
So $$P(\text{get a $7$ on zero rolls})=0\\P(\text{get a $7$ on $1$ roll})=\frac16\\P(\text{get a $7$ on the $2$ roll})=\frac16$$
So $$1-\left(0+\dfrac16+\dfrac16\right)=1-\dfrac26=\dfrac46$$

The basic approach in the question is a good one,
except for an error in combining the 
probability that the first roll is a $7$ and the
probability that the second roll is a $7$.
In general, for two events $A$ and $B$,
$$P(A \cup B) = P(A) + P(B) - P(A \cap B).$$
If $A$ is the event of rolling $7$ on the first roll,
and $B$ is the event of rolling $7$ on the second roll,
then $P(A) = P(B) = \frac 16$ but $P(A \cup B) < \frac 13$
because $P(A \cap B) > 0$.
You need to find the probability that both the first and second rolls will be $7$,
and then you can apply the formula above.

A similar but slightly different approach is to let $A$ be the probability
that the first roll is a $7$
and let $B$ be the probability that the first roll is not $7$
but the second roll is a $7$.
You can confirm that any sequence of rolls that gets $7$ before the third roll
has either event $A$ or event $B$.
But now $A$ and $B$ are disjoint, that is, $P(A \cap B) = 0,$
so you can simply add the probabilities to get $P(A \cup B),$
and the solution is found by evaluating $1 - (P(A) + P(B))$.
This is similar to the attempted solution, except that $P(B) < \frac16.$
