I'm trying to figure out the probability of rolling a 7 with two dice, over 3 rolls.
Is it P=(18/36), or P=(6/36)x(6/36)x(6/36), or just P=(3/36), or something else entirely?
I'm so confused.
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$\begingroup$I will assume that you are rolling two dice, computing their sum, and then asking the question of if the sum is equal to 7, and then repeating this process.
The only time you add probabilities is if the events are mutually exclusive (no overlap).
Let $E_1$ be the event that the first pair of dice add to seven, $E_2$ be the event that the second pair of dice add to seven, and $E_3$ be the event that the third pair of dice add to seven.
If we are curious about "What is the probability that in three attempts, at least one of the times we get a sum of seven", then we are asking the question $Pr(E_1\cup E_2\cup E_3)$. In the case that they were mutually exclusive, then yes $Pr(E_1\cup E_2\cup E_3) = Pr(E_1) + Pr(E_2) + Pr(E_3)$, however that is not the case here (it is possible that you roll a total of seven in the first attempt and also in the second attempt), so we do not add.
The ideal way to proceed is via inclusion-exclusion (in its simplest form) which says $Pr(F) = 1 - Pr(F^c)$ (here I use $^c$ to denote complement, i.e. the opposite event). So, $Pr(E_1\cup E_2\cup E_3) = 1 - Pr(E_1^c\cap E_2^c\cap E_3^c)$ (where $E_i^c$ is the event of not rolling a sum of 7 on the $i^{th}$ attempt).
As we are dealing with rolling dice, we are tacitly assuming each roll to be independent of one another, so it is $=1 - Pr(E_1^c)\cdot Pr(E_2^2)\cdot Pr(E_3^c)$.
Worded in plainer English, the probability that you roll a sum of seven at least once is opposite the probability that you roll a sum of seven none of the times.
Complete the problem by solving for $Pr(E_i), Pr(E_i^c),$ and $1 - Pr(E_1^c)\cdot Pr(E_2^2)\cdot Pr(E_3^c)$
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